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GCSE Maths · EdexcelChapter 13 · Quadratic and other graphs
Chapter 13 · Higher tier

GCSE Maths Quadratic and Other Graphs (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

  • Complete a table of values and plot a quadratic graph as a smooth curve
  • Find the roots, y-intercept, line of symmetry and turning point of a quadratic from its graph
  • Solve equations by reading a graph, including by drawing a line on the same grid
  • Recognise and draw cubic and reciprocal graphs, and match a graph to its equation
  • Read and use real-life curved graphs
  • Draw, read and interpret distance-time and speed-time graphs
  • Higher: sketch a quadratic and find its turning point by completing the square

You should already know

  • Substituting numbers into a rule, including negative numbers (Chapter 8)
  • Plotting a straight-line graph from a table of values (Chapter 12)
  • Squares, cubes and square roots (Chapter 2)
  • Speed = distance ÷ time (more in Chapter 16)

Where this leads

  • Solving quadratic equations by factorising and the formula (Chapter 14)
  • Simultaneous equations solved by graphs (Chapter 14)
  • Compound measures and proportion graphs (Chapter 16)
  • Gradients of curves, areas under graphs and transformations of graphs (Chapter 25)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: A11, A12, A14.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

Exercise P  Do you remember?

Non-calculator
1
FWork out (−3)².
2
FWork out the value of y when x = 4 for y = 2x − 5.
3
FA straight line passes through (0, 1) and (2, 7). Work out its gradient.
4
FA cyclist travels 45 km in 3 hours. Work out her average speed.
5
FWrite down the coordinates of the point on the line y = 3x + 1 where x = −2.
13.1

Tables of values and plotting quadratic graphs

Edexcel 1MA1: A11, A12
Grade 3-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

In Chapter 12 you drew straight-line graphs. Some rules do not give a straight line. A rule that contains x² (“x squared”) gives a curve. The highest power of x is 2, and a rule like this is called a quadratic. Here are some quadratic rules: y = x², y = x² + 3, y = x² − 2x − 3 and y = 5 − x². The curve has its own name: a parabola.

To draw a quadratic graph we build up the picture from separate points. First we make a table of values. This lists some x-values and works out the matching y-value for each one by substituting: replacing x by the number. Each pair (x, y) is a coordinate. Then we plot the coordinates and join them with a smooth curve.

Be very careful with negative x-values. When you square a negative number the answer is positive, because a negative times a negative is positive: (−3)² = (−3) × (−3) = 9. On a calculator always put the negative number in brackets.

There are two possible shapes. If the number in front of x² is positive the curve is U-shaped and has a lowest point. If it is negative the curve is n-shaped (an upside-down U) and has a highest point. The lowest or highest point is called the turning point. The curve is perfectly symmetrical: it is a mirror image of itself, so the y-values in a table go down and then come back up in exactly the same order (for example 5, 0, −3, −4, −3, 0, 5).

xy-3-2-1123-1123456789(0, 0)y = x² (U-shape)xy-3-2-1123-5-4-3-2-112345(0, 4)y = 4 − x² (n-shape)
A positive x² term gives a U-shape with a lowest point. A negative x² term gives an n-shape with a highest point.

Key ideas

  • A quadratic has an x² term and no higher power. Its graph is a smooth curve called a parabola.
  • Make a table of values: substitute each x into the rule to get y. Put negative numbers in brackets.
  • Plot the points (x, y) and join them with a smooth freehand curve (no ruler, no pointed corners).
  • Positive x² gives a U-shape; negative x² gives an n-shape.
  • The curve is symmetrical about a vertical line through its turning point.
Worked example 13.1.1
Complete the table of values for y = x2 + 2.
x−3−2−10123
y                     
▶ Watch this solution as a video

Key words. To substitute means to replace the letter by a number. A table of values lists x-values and the y-values that go with them. Each pair (x, y) is a coordinate that can be plotted.

Method. Put each x-value into the rule, one at a time, and work out y.

  1. Write the rule and the x-values. A table of values pairs every x-value with its y-value. Each y-value comes from putting that x-value into the rule.
    y = x2 + 2
    x-values: −3, −2, −1, 0, 1, 2, 3
  2. Substitute the x-values one at a time. Do the same calculation every time, only the number changes. Use brackets round negative numbers.
    x = −3:  y = (−3)2 + 2 = 9 + 2 = 11
    x = −2:  y = (−2)2 + 2 = 4 + 2 = 6
    x = −1:  y = (−1)2 + 2 = 1 + 2 = 3
    x = 0:  y = 02 + 2 = 0 + 2 = 2
    x = 1:  y = 12 + 2 = 1 + 2 = 3
    x = 2:  y = 22 + 2 = 4 + 2 = 6
    x = 3:  y = 32 + 2 = 9 + 2 = 11
  3. Write the y-values in the table. Each pair of numbers is a coordinate, ready to be plotted on a graph.
    y-values: 11, 6, 3, 2, 3, 6, 11
    Coordinates: (−3, 11), (−2, 6), (−1, 3), (0, 2), (1, 3), (2, 6), (3, 11)

Check. The gaps between neighbouring y-values are −5, −3, −1, 1, 3, 5. The gaps change by 2 every time (second differences are constant), which is what must happen for a quadratic. ✓

Watch out. Work out the x² part first. A negative number squared is positive.

Worked example 13.1.2
(a) Complete the table of values for y = x2 − 2x − 3. (b) Draw the graph of y = x2 − 2x − 3 for values of x from −2 to 4.
x−2−101234
y                     
▶ Watch this solution as a video

Key words. To substitute means to replace the letter by a number. A table of values lists x-values and the y-values that go with them. Each pair (x, y) is a coordinate that can be plotted.

Method. Make a table of values, plot the points, then join them with a smooth curve.

  1. Make a table of values. Each y-value comes from putting an x-value into the rule. Use brackets round negative numbers.
    x = −2:  y = (−2)2 − 2 × (−2) − 3 = 4 + 4 − 3 = 5
    x = −1:  y = (−1)2 − 2 × (−1) − 3 = 1 + 2 − 3 = 0
    x = 0:  y = 02 − 2 × 0 − 3 = 0 + 0 − 3 = −3
    x = 1:  y = 12 − 2 × 1 − 3 = 1 − 2 − 3 = −4
    x = 2:  y = 22 − 2 × 2 − 3 = 4 − 4 − 3 = −3
    x = 3:  y = 32 − 2 × 3 − 3 = 9 − 6 − 3 = 0
    x = 4:  y = 42 − 2 × 4 − 3 = 16 − 8 − 3 = 5
  2. List the coordinates. Each x with its y makes a point (x, y) to plot.
    (−2, 5), (−1, 0), (0, −3), (1, −4), (2, −3), (3, 0), (4, 5)
  3. Choose the scales for the axes. The x-values go from −2 to 4 and the y-values go from −4 to 5. The axes must cover all of these, and each square must always stand for the same amount.
    x-axis: −2 to 4.   y-axis: −5 to 6.
  4. Plot each point carefully. Go along the x-axis first, then up or down to the y-value. Mark each point with a small cross.
    Plot (−2, 5), (−1, 0), (0, −3), (1, −4), (2, −3), (3, 0), (4, 5).
  5. Join the points with a smooth curve. The graph of this rule is a smooth curve, not a set of straight lines. Draw it freehand in one flowing line, rounded at the turning point (no sharp corner).
    Label the curve y = x2 − 2x − 3.

Check. Every plotted point should lie on the curve, and the curve should pass through (−2, 5) and (4, 5) at the ends. It should be symmetrical about the line x = 1. ✓

Watch out. Do not use a ruler to join the points, and do not draw pointed corners. A quadratic is smooth and rounded at the turning point.

xy-2-11234-5-4-3-2-1123456
The finished graph of y = x² − 2x − 3 (Worked example 13.1.2). It is U-shaped, symmetrical, with its lowest point at (1, −4).
Worked example 13.1.3
Draw the graph of y = −x2 + 5 for values of x from −3 to 3. Is the curve U-shaped or n-shaped?
▶ Watch this solution as a video

Key words. To substitute means to replace the letter by a number. A table of values lists x-values and the y-values that go with them. Each pair (x, y) is a coordinate that can be plotted.

Method. Make a table of values, plot the points, then join them with a smooth curve.

  1. Make a table of values. Each y-value comes from putting an x-value into the rule. Use brackets round negative numbers.
    x = −3:  y = −(−3)2 + 5 = −9 + 5 = −4
    x = −2:  y = −(−2)2 + 5 = −4 + 5 = 1
    x = −1:  y = −(−1)2 + 5 = −1 + 5 = 4
    x = 0:  y = −(0)2 + 5 = 0 + 5 = 5
    x = 1:  y = −(1)2 + 5 = −1 + 5 = 4
    x = 2:  y = −(2)2 + 5 = −4 + 5 = 1
    x = 3:  y = −(3)2 + 5 = −9 + 5 = −4
  2. List the coordinates. Each x with its y makes a point (x, y) to plot.
    (−3, −4), (−2, 1), (−1, 4), (0, 5), (1, 4), (2, 1), (3, −4)
  3. Choose the scales for the axes. The x-values go from −3 to 3 and the y-values go from −4 to 5. The axes must cover all of these, and each square must always stand for the same amount.
    x-axis: −3 to 3.   y-axis: −5 to 6.
  4. Plot each point carefully. Go along the x-axis first, then up or down to the y-value. Mark each point with a small cross.
    Plot (−3, −4), (−2, 1), (−1, 4), (0, 5), (1, 4), (2, 1), (3, −4).
  5. Join the points with a smooth curve. The graph of this rule is a smooth curve, not a set of straight lines. Draw it freehand in one flowing line, rounded at the turning point (no sharp corner).
    Label the curve y = −x2 + 5.

Check. Every plotted point should lie on the curve, and the curve should pass through (−3, −4) and (3, −4) at the ends. It should be symmetrical about the line x = 0. ✓

Watch out. Do not use a ruler to join the points, and do not draw pointed corners. A quadratic is smooth and rounded at the turning point.

xy-3-2-1123-5-4-3-2-1123456
The graph of y = 5 − x² (Worked example 13.1.3). The x² term is negative, so it is n-shaped.
Worked example 13.1.4
Does the point (−1, 6) lie on the graph of y = x2 − x + 4? Show your working.
▶ Watch this solution as a video

Key words. To substitute means to replace the letter by a number. A table of values lists x-values and the y-values that go with them. Each pair (x, y) is a coordinate that can be plotted.

Method. A point lies on a graph if its coordinates make the rule true. Substitute the x-value and see whether we get the y-value.

  1. Choose the x-value from the point. The point is (−1, 6), so x = −1 and the claim is that y = 6.
    y = x2 − x + 4
  2. Substitute x = −1 into the rule. Use brackets round the negative number. (−1)² = 1 and −(−1) = +1.
    y = (−1)2 − (−1) + 4 = 1 + 1 + 4 = 6
  3. Compare with the y-value in the point. We found y = 6, which is the same as the y-value in the point.
    y = 6, so the point (−1, 6) does lie on the graph.

Check. Try a different point: x = 0 gives y = 4, so (0, 4) is on the graph. The point (0, 6) would not be. ✓

Watch out. A point is on the graph only if the y-value matches exactly.

Watch out

Do not join the points with a ruler. A quadratic is a smooth curve, rounded at the turning point. A pointed bottom or straight pieces will lose marks. Also check every negative x: (−2)² is +4, not −4.

Exam tip

In a table the first and last y-values are usually the biggest, and the table is symmetrical. If your values are not, check the negative x calculations again.

Exercise 13A  Tables of values and plotting quadratics

Non-calculator
1
FWork out the value of y for each rule when x takes the value shown. Show the calculation.
(a)y = x2 when x = −2
(b)y = x2 + 3 when x = −1
(c)y = x2 − 5 when x = 3
(d)y = 2x2 when x = −2
(e)y = x2 + x when x = −3
(f)y = x2 − 3x + 1 when x = −2
2
FComplete the table of values for y = x2 + 2. Some values are already filled in.
x−3−2−10123
y   6   2   6   
3
FComplete the table of values for y = x2 − 4.
x−3−2−10123
y                     
4
FComplete the table of values for y = x2 + x.
x−3−2−1012
y                  
5
F(a) Complete the table of values for y = x2 + 1. (b) Draw the graph of y = x2 + 1 for values of x from −3 to 3.
x−3−2−10123
y                     
6
F(a) Complete the table of values for y = x2 − 4x + 3. (b) Draw the graph of y = x2 − 4x + 3 for values of x from −1 to 5.
x−1012345
y                     
7
F(a) Complete the table of values for y = −x2 + 2x + 3. (b) Draw the graph of y = −x2 + 2x + 3 for values of x from −2 to 4.
x−2−101234
y                     
8
RAisha is working out y when x = −3 for y = x² + 2. She writes: “(−3)² = −9, so y = −9 + 2 = −7.” Explain her mistake and work out the correct value.
9
FFor each point, say whether it lies on the graph of y = x2 − 3.
(a)(2, 1)
(b)(−1, −4)
(c)(0, −3)
(d)(−3, 6)
10
FWithout drawing the graph, say whether each quadratic graph is U-shaped or n-shaped.
(a)y = x2 + 5
(b)y = −x2 + 5
(c)y = −x2 + 3
(d)y = x2 − 5x
11
PSThe point (k, 11) lies on the graph of y = x² + 2. Work out the two possible values of k.
12
F(a) Complete the table of values for y = 2x2 − 3x − 5. (b) Draw the graph of y = 2x2 − 3x − 5 for values of x from −2 to 3.
x−2−10123
y                  
13
F(a) Complete the table of values for y = x² − 3x. (b) Draw the graph for values of x from 0 to 3. (c) Write down the coordinates of the turning point.
x00.511.522.53
y                     
14
RExplain why the graph of y = x² never goes below the x-axis.
15
RBen draws the graph of y = x² − 2x − 3 by joining his plotted points with a ruler. Explain why this is wrong.
16
R(a) Complete the table of values for y = x² − 6x + 8. (b) The point (1, 3) is on the graph. Use the symmetry of the table to write down the x-value of another point with y = 3. (c) Write down the equation of the line of symmetry. (d) Explain how you know.
x0123456
y                     
(a)Complete the table.
(b)Another point with y = 3.
(c)Line of symmetry.
(d)Explain.
13.2

Roots, intercepts, symmetry and turning points

Edexcel 1MA1: A11
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

A quadratic graph is full of useful information. There are four things you must be able to read from the graph.

The roots are the x-values where the curve meets the x-axis. On the x-axis every point has y = 0, so the roots are the solutions of the equation y = 0. A quadratic can have two roots, one root (the curve just touches the axis) or no roots (the curve never reaches the axis).

The y-intercept is where the curve crosses the y-axis. On the y-axis every point has x = 0, so the y-intercept is the number on its own in the rule (for y = x² − 2x − 3 it is −3).

The turning point is the lowest point of a U-shaped curve (a minimum) or the highest point of an n-shaped curve (a maximum). Give it as a coordinate pair (x, y).

The line of symmetry is the vertical mirror line through the turning point. Its equation is x = (the x-value of the turning point). It is exactly halfway between the two roots, which is a handy way to find it.

xy-2-11234-5-4-3-2-112345rootrooty-interceptturning point
The graph of y = x² − 2x − 3. Roots: x = −1 and x = 3. y-intercept: −3. Turning point (minimum): (1, −4). Line of symmetry (dashed): x = 1.

Finding roots by algebra. You can also find roots without a graph. Put y = 0. If the rule is written as two brackets multiplied together, such as y = (x − 3)(x + 2), then (x − 3)(x + 2) = 0. Two numbers multiplied together give 0 only when one of them is 0, so x − 3 = 0 or x + 2 = 0. That gives x = 3 or x = −2. Notice that the sign flips. If the rule is y = x² − 16, put y = 0, so x² = 16, and x = 4 or x = −4 (both square to give 16).

Key ideas

  • Roots: where the curve crosses or touches the x-axis (y = 0). Give them as x = ... values.
  • y-intercept: where the curve crosses the y-axis (x = 0).
  • Turning point: the lowest point (U-shape, minimum) or highest point (n-shape, maximum). Give it as (x, y).
  • Line of symmetry: the vertical line x = (x-value of the turning point), halfway between the roots.
  • A quadratic has 0, 1 or 2 roots.
  • To find roots by algebra put y = 0: for (x + p)(x + q) = 0 each bracket in turn equals 0.
Worked example 13.2.1
The graph of a quadratic is drawn. Write down its roots.
xy-2-112345-2-112345
▶ Watch this solution as a video

Key words. A quadratic is a rule where the highest power of x is x² (x squared). Its graph is a smooth curve called a parabola. The turning point is where the curve changes direction (the lowest or highest point). A root is an x-value where the curve meets the x-axis (where y = 0). The y-intercept is where the curve crosses the y-axis (where x = 0).

Method. The roots are the x-values where the curve crosses the x-axis, so read them off the graph.

  1. Remember what a root is. A root is an x-value where the curve meets the x-axis. On the x-axis every point has y = 0, so roots are the solutions of y = 0.
  2. Find where the curve crosses the x-axis. Look along the horizontal axis for the places where the curve meets it. Read the x-coordinate of each place from the numbers along the bottom.
    The curve crosses the x-axis at (1, 0) and (3, 0).
  3. Write the roots. We only need the x-values (the roots are x-values, not coordinates).
    x = 1 or 3

Check. Substitute each root into the rule; y should be 0.
y = 12 − 4 × 1 + 3 = 1 − 4 + 3 = 0
y = 32 − 4 × 3 + 3 = 9 − 12 + 3 = 0 ✓

Watch out. Give x-values only: say x = 3, not (3, 0).

Worked example 13.2.2
The graph of y = −x² + 2x + 8 is drawn. Write down the coordinates of the turning point. Is it a maximum or a minimum?
xy-3-2-112345-4-3-2-112345678910
▶ Watch this solution as a video

Key words. A quadratic is a rule where the highest power of x is x² (x squared). Its graph is a smooth curve called a parabola. The turning point is where the curve changes direction (the lowest or highest point). A root is an x-value where the curve meets the x-axis (where y = 0). The y-intercept is where the curve crosses the y-axis (where x = 0).

Method. The turning point is the lowest point of a U-shaped curve, or the highest point of an n-shaped curve. Read both coordinates.

  1. Remember what a turning point is. The turning point is where the curve stops going down and starts going up (or the other way round). An n-shaped curve has a highest point.
  2. Find the highest point of the curve. Read the x-coordinate along the bottom and the y-coordinate up the side.
    The turning point is (1, 9).
  3. Write the answer as a coordinate pair. It is a maximum (highest point).
    (1, 9)

Check. Substitute x = 1: y = −(1)2 + 2 × 1 + 8 = −1 + 2 + 8 = 9. ✓

Watch out. Write both coordinates, x first: (x, y).

Worked example 13.2.3
Find the roots of y = (x − 3)(x + 2) without drawing a graph.
▶ Watch this solution as a video

Key words. A quadratic is a rule where the highest power of x is x² (x squared). Its graph is a smooth curve called a parabola. The turning point is where the curve changes direction (the lowest or highest point). A root is an x-value where the curve meets the x-axis (where y = 0). The y-intercept is where the curve crosses the y-axis (where x = 0).

Method. Put y = 0. When two brackets multiply to give 0, one of the brackets must be 0.

  1. Roots are where y = 0. The roots are the x-values where the curve meets the x-axis. On the x-axis y = 0, so put y = 0 in the equation.
    0 = (x − 3)(x + 2)
  2. Use the fact that a product is zero only if one factor is zero. If two things multiply to give 0, at least one of them must be 0. So at least one bracket must equal 0.
    (x − 3) = 0  or  (x + 2) = 0
  3. Solve each little equation. Get x on its own by doing the inverse operation to both sides.
    x − 3 = 0 → x = 3
    x + 2 = 0 → x = −2
  4. Write the roots. Both values make y = 0.
    x = 3 or −2

Check. Substitute: y = (−2)2 − (−2) − 6 = 4 + 2 − 6 = 0; y = 32 − 3 − 6 = 9 − 3 − 6 = 0 ✓

Watch out. The sign flips: the bracket (x + 3) gives the root x = −3, not +3.

Worked example 13.2.4
The graph of y = x² − 4x − 5 is drawn. Write down the equation of its line of symmetry.
xy-2-1123456-10-9-8-7-6-5-4-3-2-11234
▶ Watch this solution as a video

Key words. A quadratic is a rule where the highest power of x is x² (x squared). Its graph is a smooth curve called a parabola. The turning point is where the curve changes direction (the lowest or highest point). A root is an x-value where the curve meets the x-axis (where y = 0). The y-intercept is where the curve crosses the y-axis (where x = 0).

Method. The line of symmetry is the vertical line through the turning point: x = (the x-coordinate of the turning point).

  1. Remember what the line of symmetry is. A parabola is a mirror image of itself. The line of symmetry is the vertical mirror line that goes through the turning point.
  2. Find the turning point (the lowest point). The mirror line goes straight through it, so it has the same x-value as the turning point.
    Turning point = (2, −9)
  3. Write the equation of the line. A vertical line has an equation of the form x = number. Use the x-coordinate of the turning point.
    x = 2

Check. The line of symmetry is halfway between the roots: (−1 + 5) ÷ 2 = 2. ✓

Watch out. A vertical line is written x = number (not y = number). y = number is a horizontal line.

Watch out

The roots are x-values (written x = 3). The y-intercept is a y-value. A turning point is a coordinate pair. Do not swap the order: the turning point of the curve in the diagram is (1, −4), not (−4, 1). And a vertical line is x = number, not y = number.

Exam tip

The roots and the line of symmetry check each other: the line of symmetry is always halfway between the roots. Spot a mistake instantly by checking this.

Exercise 13B  Reading quadratic graphs

Non-calculator
1
FHere is the graph of y = x² − 2x − 3. Use it to write down (a) the roots, (b) the y-intercept, (c) the coordinates of the turning point, (d) the equation of the line of symmetry.
xy-2-11234-5-4-3-2-112345
(a)the roots
(b)the y-intercept
(c)the turning point
(d)the line of symmetry
2
FHere is the graph of y = x² − 4x + 3. (a) Write down the roots. (b) Write down the coordinates of the turning point. (c) Is the turning point a maximum or a minimum? Give a reason.
xy-2-112345-2-112345
(a)the roots
(b)the turning point
(c)maximum or minimum?
3
FHere is the graph of y = 4 − x². Write down (a) the y-intercept, (b) the roots, (c) the maximum value of y.
xy-3-2-1123-5-4-3-2-112345
(a)the y-intercept
(b)the roots
(c)the maximum value of y
4
FHere is the graph of y = x² + 2x − 8. (a) Write down the roots. (b) Write down the coordinates of the turning point. (c) Write down the minimum value of y. (d) Write down the equation of the line of symmetry.
xy-5-4-3-2-1123-10-9-8-7-6-5-4-3-2-11234
(a)the roots
(b)the turning point
(c)the minimum value of y
(d)the line of symmetry
5
FThe graph of y = x² − 6x + 9 touches the x-axis at one point. Write down the root of the equation x² − 6x + 9 = 0.
xy-2-112345-112345678910
6
FHere is the graph of y = x² + 2x + 3. How many roots does it have? Give a reason for your answer.
xy-4-3-2-1123-112345678
7
FUse the roots to write down the equation of the line of symmetry of each curve.
(a)Roots at 1 and 5
(b)Roots at −4 and 2
(c)Roots at −7 and −1
8
FWork out the roots of each quadratic without drawing a graph. (Put y = 0.)
(a)y = (x − 3)(x + 2)
(b)y = (x + 1)(x − 5)
(c)y = x(x − 4)
(d)y = x2 − 25
(e)y = x2 − 49
(f)y = (x − 6)(x − 6)
9
FHere is the graph of y = −x² + 6x − 5. Write down (a) the roots, (b) the coordinates of the turning point, (c) the maximum value of y, (d) the equation of the line of symmetry.
xy-1123456-6-5-4-3-2-112345
(a)the roots
(b)the turning point
(c)the maximum value of y
(d)the line of symmetry
10
PSThe graph of y = x² − 6x + 5 crosses the x-axis at (1, 0) and (5, 0). Work out the coordinates of the turning point.
11
PSA parabola crosses the x-axis at (−3, 0) and (7, 0). It is n-shaped and its highest point has y-value 25. Write down (a) the equation of its line of symmetry, (b) the coordinates of its turning point.
(a)the line of symmetry
(b)the turning point
12
RSam says: “A parabola with roots at 2 and 6 has its line of symmetry at x = 8.” Is Sam right? Explain.
13
RExplain why a quadratic graph can have no roots, one root or two roots, but it can never have three roots.
14
RPriya says the turning point of y = x² − 2x − 3 is (−4, 1). Explain why she is wrong and write down the correct coordinates.
15
FHere is the graph of y = x² − 6x + 4. Use the graph to find (a) the roots, to 1 decimal place, (b) the coordinates of the turning point.
xy-11234567-6-5-4-3-2-1123456
(a)the roots (1 d.p.)
(b)the turning point
16
FFactorise to find the roots of each quadratic.
(a)y = x2 − 5x + 6
(b)y = x2 + x − 12
(c)y = x2 − 7x + 10
(d)y = x2 + 6x + 9
17
RThe graph of y = x² − 2x − 3 has its turning point at (1, −4). (a) Write down the coordinates of the point where the curve crosses the y-axis. (b) Use symmetry to write down the coordinates of another point on the curve with the same y-value. (c) Explain how you used symmetry.
(a)the y-intercept
(b)another point with y = −3
(c)Explain.
13.3

Solving equations using graphs

Edexcel 1MA1: A11, A12
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

A graph can be used to solve an equation. This is useful when an equation is hard to solve in any other way, and it shows you what the solutions mean. The idea is always the same: a solution is the x-value of a point where two graphs meet.

Solving x² − 2x − 3 = 0. The curve y = x² − 2x − 3 has the rule on one side of the equation. The equation says that this rule equals 0. On the graph, y = 0 is the x-axis. So the solutions are the x-values where the curve crosses the x-axis. They are the roots.

Solving x² − 2x − 3 = 5. Now the rule must equal 5. Draw the horizontal line y = 5 across the grid. Where it meets the curve, the curve has y = 5. Read the x-values of those points. There can be two, one or no solutions.

xy-2-11234-5-4-3-2-112345
Solving x² − 2x − 3 = 5. The horizontal line y = 5 meets the curve at the two red points. Reading across the bottom: x = −2 and x = 4.

Solving x² = 2x + 3. The two sides are two different rules. Draw both as graphs on the same grid: the curve y = x² and the straight line y = 2x + 3. At a crossing point both graphs have the same x and the same y, so x² = 2x + 3 is true there. The solutions are the x-values of the crossing points.

xy-2-112342468101214
The curve y = x² (blue) and the line y = 2x + 3 (orange) cross at (−1, 1) and (3, 9). The solutions of x² = 2x + 3 are x = −1 and x = 3.

Answers read from a graph are only approximate: you can only read as accurately as the grid allows. If the crossing is between grid lines, estimate to 1 decimal place. Exam questions that need this will say “use the graph” or “give your answers to 1 decimal place”. Always show the line you drew.

Higher tier. Sometimes the equation you must solve is not exactly the rule that was graphed. For example, you are given the graph of y = x² − 2x − 3 and asked to solve x² − 3x − 4 = 0. Rearrange the equation so one side is the graphed rule: x² − 3x − 4 = 0 becomes x² − 2x − 3 = x + 1. The other side tells you which line to draw: y = x + 1.

Key ideas

  • A solution of an equation is the x-value of a point where two graphs meet.
  • To solve (rule) = 0, read where the curve crosses the x-axis.
  • To solve (rule) = k, draw the horizontal line y = k and read the x-values where it meets the curve.
  • To solve (rule 1) = (rule 2), graph both on the same grid and read the x-values where they cross.
  • Graph answers are approximate. Read the x-values, not the y-values.
Worked example 13.3.1
Here is the graph of y = x² − 2x − 3. Use it to solve x² − 2x − 3 = 0.
xy-2-11234-5-4-3-2-112345
▶ Watch this solution as a video

Key words. To solve an equation graphically means to read its answers from a graph. Where two graphs meet they have the same x and the same y, so the x-value at a crossing point makes the two rules equal. Graph readings are approximate because we can only read to the accuracy of the drawing.

Method. Draw the horizontal line y = (the number on the right). Where it crosses the curve, read the x-values.

  1. Spot the graph being used. The graph shows y = x2 − 2x − 3. The equation we want to solve is the same expression set equal to 0.
    x2 − 2x − 3 = 0 means y = 0 on this graph
  2. Draw the horizontal line y = 0. Every point on this line has a y-value of 0. Where the line meets the curve, the curve has y = 0, which is exactly when the equation is true.
    Start at 0 on the y-axis and draw straight across (a ruler helps).
  3. Find where the line crosses the curve. Mark each crossing point.
    The line meets the curve at (−1, 0) and (3, 0).
  4. Read the x-values. The solutions of the equation are the x-values of the crossing points (the numbers along the bottom axis).
    x = −1 or 3

Check. Substitute back: y = (−1)2 − 2 × (−1) − 3 = 1 + 2 − 3 = 0; y = 32 − 2 × 3 − 3 = 9 − 6 − 3 = 0 ✓

Watch out. Read the x-values of the crossing points, not the y-values. The y-value is just the number you started with.

Worked example 13.3.2
Use the same graph of y = x² − 2x − 3 to solve x² − 2x − 3 = 5.
▶ Watch this solution as a video

Key words. To solve an equation graphically means to read its answers from a graph. Where two graphs meet they have the same x and the same y, so the x-value at a crossing point makes the two rules equal. Graph readings are approximate because we can only read to the accuracy of the drawing.

Method. Draw the horizontal line y = (the number on the right). Where it crosses the curve, read the x-values.

  1. Spot the graph being used. The graph shows y = x2 − 2x − 3. The equation we want to solve is the same expression set equal to 5.
    x2 − 2x − 3 = 5 means y = 5 on this graph
  2. Draw the horizontal line y = 5. Every point on this line has a y-value of 5. Where the line meets the curve, the curve has y = 5, which is exactly when the equation is true.
    Start at 5 on the y-axis and draw straight across (a ruler helps).
  3. Find where the line crosses the curve. Mark each crossing point.
    The line meets the curve at (−2, 5) and (4, 5).
  4. Read the x-values. The solutions of the equation are the x-values of the crossing points (the numbers along the bottom axis).
    x = −2 or 4

Check. Substitute back: y = (−2)2 − 2 × (−2) − 3 = 4 + 4 − 3 = 5; y = 42 − 2 × 4 − 3 = 16 − 8 − 3 = 5 ✓

Watch out. Read the x-values of the crossing points, not the y-values. The y-value is just the number you started with.

Worked example 13.3.3
The graphs of y = x² and y = 2x + 3 are drawn on the same grid. Use them to solve x² = 2x + 3.
xy-2-112342468101214
▶ Watch this solution as a video

Key words. To solve an equation graphically means to read its answers from a graph. Where two graphs meet they have the same x and the same y, so the x-value at a crossing point makes the two rules equal. Graph readings are approximate because we can only read to the accuracy of the drawing.

Method. The solutions are the x-values of the points where the two graphs cross.

  1. Say which two graphs are being used. The equation has the rule of the curve on one side and the rule of the line on the other. Each side is drawn as its own graph.
    Curve: y = x2
    Line: y = 2x + 3
  2. Explain why crossing points give the answers. At a crossing point both graphs have the same x and the same y, so the two rules give the same value. That is what the equation says.
    Solve x² = 2x + 3: the curve y = x² and the line y = 2x + 3 are both drawn.
  3. Find the crossing points. Look for every place where the two graphs meet.
    They cross at (−1, 1) and (3, 9).
  4. Read the x-values. The solutions are the x-coordinates of the crossing points.
    x = −1 or 3

Check. Substitute each x-value into both sides.
x = −1: left = 1, right = 1. Equal. ✓
x = 3: left = 9, right = 9. Equal. ✓

Watch out. Write only the x-values. (The y-values at the crossings are not the answer.)

Worked example 13.3.4
Here is the graph of y = x². Use it to find estimates of the solutions of x² = 10. Give your answers to 1 decimal place.
xy-4-3-2-11234246810121416
▶ Watch this solution as a video

Key words. To solve an equation graphically means to read its answers from a graph. Where two graphs meet they have the same x and the same y, so the x-value at a crossing point makes the two rules equal. Graph readings are approximate because we can only read to the accuracy of the drawing.

Method. Draw the horizontal line y = (the number on the right). Where it crosses the curve, read the x-values.

  1. Spot the graph being used. The graph shows y = x2. The equation we want to solve is the same expression set equal to 10.
    x2 = 10 means y = 10 on this graph
  2. Draw the horizontal line y = 10. Every point on this line has a y-value of 10. Where the line meets the curve, the curve has y = 10, which is exactly when the equation is true.
    Start at 10 on the y-axis and draw straight across (a ruler helps).
  3. Find where the line crosses the curve. Mark each crossing point.
    The line meets the curve at (−3.2, 10) and (3.2, 10).
  4. Read the x-values. The solutions of the equation are the x-values of the crossing points (the numbers along the bottom axis).
    x ≈ −3.2 or 3.2 (accept answers within 0.1)

Check. Substitute back: at x = −3.2 the rule gives 10.00; at x = 3.2 the rule gives 10.00 ✓

Watch out. Read the x-values of the crossing points, not the y-values. The y-value is just the number you started with.

Worked example 13.3.5
The graph of y = x² − 2x − 3 is drawn. By drawing a suitable straight line, solve x² − 3x − 4 = 0.
▶ Watch this solution as a video

Key words. To solve an equation graphically means to read its answers from a graph. Where two graphs meet they have the same x and the same y, so the x-value at a crossing point makes the two rules equal. Graph readings are approximate because we can only read to the accuracy of the drawing.

Method. Rearrange the equation so that one side is the rule of the graph you are given. The other side is the line to draw.

  1. Compare the equation with the graph that is drawn. The graph is y = x2 − 2x − 3. We try to make the left side of the equation look like x2 − 2x − 3.
    x2 − 3x − 4 = 0
  2. Split the left side into the graph's expression plus what is left over. Use the x² and x terms so that the graph's rule appears exactly. Whatever is left is the extra part.
    x2 − 3x − 4 = (x2 − 2x − 3) + (−x − 1)
  3. Move the left-over part to the other side. Do the same to both sides: take away the left-over part (every term changes sign when it crosses to the other side).
    x2 − 2x − 3 = x + 1
  4. Decide which line to draw on the grid. The curve already shows y = x2 − 2x − 3. The right-hand side gives the second graph.
    Draw the line y = x + 1.
  5. Read the crossing points. The line and curve cross where the two sides are equal, which is where the original equation is true.
    They cross at (−1, 0) and (4, 5).
  6. Write the x-values. The solutions are the x-values of the crossing points.
    x = −1 or 4

Check. Substitute into the original equation: x = −1 gives x2 − 3x − 4 = 0; x = 4 gives x2 − 3x − 4 = 0 ✓

Watch out. Watch the signs when you move terms to the other side: each one changes sign.

Watch out

Read the x-values where the graphs meet. The y-value of a crossing point is not the answer. For x² = 2x + 3 the crossing points are (−1, 1) and (3, 9), but the solutions are x = −1 and x = 3. Also remember there can be two solutions: look for both crossings.

Exam tip

Draw the line you use in a different colour with a ruler, and put a small cross at each crossing point before you read down to the x-axis. Write “x = ... or x = ...”.

Exercise 13C  Solving equations using graphs

Non-calculator
1
FHere is the graph of y = x² − 2x − 3. Use it to solve (a) x² − 2x − 3 = 0, (b) x² − 2x − 3 = 5, (c) x² − 2x − 3 = −3, (d) x² − 2x − 3 = −4.
xy-2-11234-5-4-3-2-112345
(a)x² − 2x − 3 = 0
(b)x² − 2x − 3 = 5
(c)x² − 2x − 3 = −3
(d)x² − 2x − 3 = −4
2
FHere is the graph of y = x². Use it to solve each equation. Give any approximate answers to 1 decimal place.
xy-4-3-2-11234246810121416
(a)x² = 4
(b)x² = 9
(c)x² = 6.25
(d)x² = 10
3
FHere is the graph of y = x² − 4x + 3. Use it to solve (a) x² − 4x + 3 = 0, (b) x² − 4x + 3 = 3, (c) x² − 4x + 3 = 8, (d) x² − 4x + 3 = −1.
xy-2-112345-2-112345
(a)x² − 4x + 3 = 0
(b)x² − 4x + 3 = 3
(c)x² − 4x + 3 = 8
(d)x² − 4x + 3 = −1
4
FHere is the graph of y = 4 − x². Use it to solve (a) 4 − x² = 0, (b) 4 − x² = 3, (c) 4 − x² = −5, (d) 4 − x² = 4.
xy-3-2-1123-5-4-3-2-112345
(a)4 − x² = 0
(b)4 − x² = 3
(c)4 − x² = −5
(d)4 − x² = 4
5
FThe graphs of y = x² and y = 2x + 3 are drawn on the grid. Use the graphs to solve x² = 2x + 3.
xy-2-112342468101214
6
FThe graphs of y = x² and y = x + 2 are drawn on the grid. Use the graphs to solve x² = x + 2.
xy-2-1123-112345678910
7
FThe graphs of y = x² − 4 and y = x − 2 are drawn on the grid. Use them to solve x² − 4 = x − 2.
xy-2-1123-5-4-3-2-1123456
8
PSUse the graph of y = x² − 2x − 3 (drawn above) to find estimates of the solutions of x² − 2x − 3 = 7. Give your answers to 1 decimal place.
xy-2-11234-5-4-3-2-112345
9
PSHere is the graph of y = x² − 3x. Use it to find estimates of the solutions of x² − 3x = 1. Give your answers to 1 decimal place.
xy-2-112345-3-2-11234567
10
RHana says: “The solutions of x² − 2x − 3 = 0 are the roots of the graph of y = x² − 2x − 3.” Explain why she is right.
11
REmma solves x² = 2x + 3 by drawing both graphs. The graphs cross at (−1, 1) and (3, 9). Emma writes: “The solutions are 1 and 9.” Explain her mistake and give the correct solutions.
12
RUse the graph of y = x² to explain why the equation x² = −5 has no solutions.
13
FThe graph of y = x² − 2x − 3 is drawn. By drawing a suitable straight line on the graph, solve each equation.
xy-2-11234-5-4-3-2-112345
(a)x² − 3x − 4 = 0
(b)x² − 2x − 6 = 0
(c)x² − 4x = 0
14
PSThe graph of y = x² is drawn. Show how to use it to solve x² − x − 6 = 0. State the equation of the straight line you need to draw.
xy-4-3-2-11234246810121416
15
PSThe graph of y = x² − 4x + 1 is drawn. Use it to find estimates of the solutions of x² − 4x + 1 = 4. Give your answers to 1 decimal place.
xy-2-1123456-4-3-2-1123456789
16
RThe graph of y = x² − 2x − 3 has its lowest point at (1, −4). Write down the number of solutions of x² − 2x − 3 = k when (a) k = 2, (b) k = −4, (c) k = −6. Explain how you know.
(a)k = 2
(b)k = −4
(c)k = −6
13.4

Cubic and reciprocal graphs

Edexcel 1MA1: A12
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Cubic graphs. A cubic rule has x³ (“x cubed”) as its highest power, for example y = x³ or y = x³ − 4x. You draw it in the same way as a quadratic: make a table of values, plot the points, join them with a smooth curve. Be careful with negatives: (−2)³ = (−2) × (−2) × (−2) = −8. An odd number of negative signs gives a negative answer, so a cube of a negative number is negative (unlike a square).

The graph of y = x³ is a smooth S-shaped curve. It goes up from the bottom left to the top right, passing through (0, 0). It is steeper the further you go from the origin. With a negative sign, y = −x³ is the same shape flipped so that it goes down from the top left to the bottom right. Many cubics also have a small “wiggle”, with one hump and one dip, as in y = x³ − 4x. A cubic can cross the x-axis once, twice or three times.

xy-3-2-1123-8-6-4-22468y = x³xy-3-2-1123-8-6-4-22468y = x³ − 4x
Cubic graphs: the simple S-shape of y = x³ (left) and a cubic with a hump and a dip (right).

Reciprocal graphs. The reciprocal of a number is 1 divided by it. The graph of y = 1 ÷ x (written 1x) has a very different shape. You cannot divide by 0, so there is no point at x = 0. The graph has two separate branches: for positive x the values of y are positive and get smaller as x grows (1, 0.5, 0.25, …); for negative x the values of y are negative. Near x = 0 the curve shoots off steeply. The curve gets closer and closer to the axes but never touches them: the axes are called asymptotes.

The graph of y = k ÷ x (for example y = 6 ÷ x) has the same shape, further from the axes when k is bigger. If k is negative the two branches swap: they are in the top-left and bottom-right.

xy-4-3-2-11234-4-3-2-11234y = 1/xxy-4-3-2-11234-4-3-2-11234y = −1/x
The two branches of y = 1/x (left) are in the top-right and bottom-left. For y = −1/x (right) they are in the top-left and bottom-right.

Key ideas

  • A cubic has x³ as its highest power. A negative number cubed is negative: (−2)³ = −8.
  • The graph of y = x³ is S-shaped, up from bottom-left to top-right, through (0, 0).
  • y = −x³ goes down from top-left to bottom-right. Cubics can also have a hump and a dip.
  • The graph of y = k ÷ x has two branches and no point at x = 0.
  • Positive k: branches in top-right and bottom-left. Negative k: top-left and bottom-right.
  • The curve gets close to the axes (asymptotes) but never touches them.
Worked example 13.4.1
Complete the table of values for y = x3.
x−3−2−10123
y                     
▶ Watch this solution as a video

Key words. To substitute means to replace the letter by a number. A table of values lists x-values and the y-values that go with them. Each pair (x, y) is a coordinate that can be plotted.

Method. Put each x-value into the rule, one at a time, and work out y.

  1. Write the rule and the x-values. A table of values pairs every x-value with its y-value. Each y-value comes from putting that x-value into the rule.
    y = x3
    x-values: −3, −2, −1, 0, 1, 2, 3
  2. Substitute the x-values one at a time. Do the same calculation every time, only the number changes. Use brackets round negative numbers.
    x = −3:  y = (−3)3 = −27
    x = −2:  y = (−2)3 = −8
    x = −1:  y = (−1)3 = −1
    x = 0:  y = 03 = 0
    x = 1:  y = 13 = 1
    x = 2:  y = 23 = 8
    x = 3:  y = 33 = 27
  3. Write the y-values in the table. Each pair of numbers is a coordinate, ready to be plotted on a graph.
    y-values: −27, −8, −1, 0, 1, 8, 27
    Coordinates: (−3, −27), (−2, −8), (−1, −1), (0, 0), (1, 1), (2, 8), (3, 27)

Check. First gaps: 19, 7, 1, 1, 7, 19. Gaps of the gaps: −12, −6, 0, 6, 12. Gaps of those: 6, 6, 6, 6, always the same, which is the pattern for a cubic. ✓

Watch out. A negative number cubed is negative: (−2)³ = −8.

Worked example 13.4.2
Draw the graph of y = x3 − 4x for values of x from −3 to 3.
x−3−2.5−2−10122.53
y                           
▶ Watch this solution as a video

Key words. To substitute means to replace the letter by a number. A table of values lists x-values and the y-values that go with them. Each pair (x, y) is a coordinate that can be plotted.

Method. Make a table of values, plot the points, then join them with a smooth curve.

  1. Make a table of values. Each y-value comes from putting an x-value into the rule. Use brackets round negative numbers.
    x = −3:  y = (−3)3 − 4 × (−3) = −27 + 12 = −15
    x = −2.5:  y = (−2.5)3 − 4 × (−2.5) = −15.625 + 10 = −5.625
    x = −2:  y = (−2)3 − 4 × (−2) = −8 + 8 = 0
    x = −1:  y = (−1)3 − 4 × (−1) = −1 + 4 = 3
    x = 0:  y = 03 − 4 × 0 = 0 + 0 = 0
    x = 1:  y = 13 − 4 × 1 = 1 − 4 = −3
    x = 2:  y = 23 − 4 × 2 = 8 − 8 = 0
    x = 2.5:  y = 2.53 − 4 × 2.5 = 15.625 − 10 = 5.625
    x = 3:  y = 33 − 4 × 3 = 27 − 12 = 15
  2. List the coordinates. Each x with its y makes a point (x, y) to plot.
    (−3, −15), (−2.5, −5.625), (−2, 0), (−1, 3), (0, 0), (1, −3), (2, 0), (2.5, 5.625), (3, 15)
  3. Choose the scales for the axes. The x-values go from −3 to 3 and the y-values go from −15 to 15. The axes must cover all of these, and each square must always stand for the same amount.
    x-axis: −3 to 3.   y-axis: −20 to 20.
  4. Plot each point carefully. Go along the x-axis first, then up or down to the y-value. Mark each point with a small cross.
    Plot (−3, −15), (−2.5, −5.625), (−2, 0), (−1, 3), (0, 0), (1, −3), (2, 0), (2.5, 5.625), (3, 15).
  5. Join the points with a smooth curve. The graph of this rule is a smooth curve, not a set of straight lines. Draw it freehand in one flowing line, passing through every point.
    Label the curve y = x3 − 4x.

Check. Every plotted point should lie on the curve, and the curve should pass through (−3, −15) and (3, 15) at the ends. Check one point by substituting. ✓

Watch out. Join the points with a smooth curve, not with straight segments.

xy-3-2-1123-20-15-10-55101520
The graph of y = x³ − 4x (Worked example 13.4.2). It crosses the x-axis three times, at x = −2, 0 and 2.
Worked example 13.4.3
Draw the graph of y = 1x for values of x from −4 to −0.5 and from 0.5 to 4.
x−4−2−1−0.50.5124
y                        
▶ Watch this solution as a video

Key words. To substitute means to replace the letter by a number. A table of values lists x-values and the y-values that go with them. Each pair (x, y) is a coordinate that can be plotted.

Method. Make a table of values, plot the points, then join them with a smooth curve.

  1. Make a table of values. Each y-value comes from putting an x-value into the rule. Use brackets round negative numbers.
    x = −4:  y = 1 ÷ (−4) = −0.25
    x = −2:  y = 1 ÷ (−2) = −0.5
    x = −1:  y = 1 ÷ (−1) = −1
    x = −0.5:  y = 1 ÷ (−0.5) = −2
    x = 0.5:  y = 1 ÷ 0.5 = 2
    x = 1:  y = 1 ÷ 1 = 1
    x = 2:  y = 1 ÷ 2 = 0.5
    x = 4:  y = 1 ÷ 4 = 0.25
  2. List the coordinates. Each x with its y makes a point (x, y) to plot.
    (−4, −0.25), (−2, −0.5), (−1, −1), (−0.5, −2), (0.5, 2), (1, 1), (2, 0.5), (4, 0.25)
  3. Choose the scales for the axes. The x-values go from −4 to 4 and the y-values go from −2 to 2. The axes must cover all of these, and each square must always stand for the same amount.
    x-axis: −4 to 4.   y-axis: −3 to 3.
  4. Plot each point carefully. Go along the x-axis first, then up or down to the y-value. Mark each point with a small cross.
    Plot (−4, −0.25), (−2, −0.5), (−1, −1), (−0.5, −2), (0.5, 2), (1, 1), (2, 0.5), (4, 0.25).
  5. Join the points with a smooth curve. The graph of this rule is a smooth curve, not a set of straight lines. Draw it freehand in one flowing line, in two separate branches, one on each side of x = 0.
    Label the curve y = 1x.

Check. Every plotted point should lie on the curve, and the curve should pass through (−4, −0.25) and (4, 0.25) at the ends. Check one point by substituting. ✓

Watch out. The two branches never join up: there is no point at x = 0.

xy-4-3-2-11234-3-2-1123
The graph of y = 1/x (Worked example 13.4.3): two separate branches.
Worked example 13.4.4
Here is the graph of y = 6x. Use it to find x when 6x = 2.5.
xy-7-6-5-4-3-2-11234567-7-6-5-4-3-2-11234567
▶ Watch this solution as a video

Key words. To solve an equation graphically means to read its answers from a graph. Where two graphs meet they have the same x and the same y, so the x-value at a crossing point makes the two rules equal. Graph readings are approximate because we can only read to the accuracy of the drawing.

Method. Draw the horizontal line y = (the number on the right). Where it crosses the curve, read the x-values.

  1. Spot the graph being used. The graph shows y = 6x. The equation we want to solve is the same expression set equal to 2.5.
    6x = 2.5 means y = 2.5 on this graph
  2. Draw the horizontal line y = 2.5. Every point on this line has a y-value of 2.5. Where the line meets the curve, the curve has y = 2.5, which is exactly when the equation is true.
    Start at 2.5 on the y-axis and draw straight across (a ruler helps).
  3. Find where the line crosses the curve. Mark each crossing point.
    The line meets the curve at (2.4, 2.5).
  4. Read the x-values. The solutions of the equation are the x-values of the crossing points (the numbers along the bottom axis).
    x = 2.4

Check. Substitute back: y = 6 ÷ 2.4 = 2.5 ✓

Watch out. Read the x-values of the crossing points, not the y-values. The y-value is just the number you started with.

Watch out

A cube of a negative number is negative, but a square of a negative number is positive. (−2)³ = −8 but (−2)² = 4. And on a reciprocal graph never join the two branches, and never draw a point at x = 0.

Exam tip

To check a reciprocal table, multiply x by y. For y = 6 ÷ x every pair must multiply to 6: 2 × 3 = 6, (−3) × (−2) = 6.

Exercise 13D  Cubic and reciprocal graphs

Non-calculator
1
FComplete the table of values for y = x3 + 2.
x−3−2−10123
y      1   3      
2
FComplete the table of values for y = 2x3.
x−2−1012
y               
3
FComplete the table of values for y = −x3.
x−2−1012
y               
4
F(a) Complete the table of values for y = x3. (b) Draw the graph of y = x3 for values of x from −2 to 2.
x−2−1.5−1−0.500.511.52
y                           
5
F(a) Complete the table of values for y = x3 − 4x. (b) Draw the graph of y = x3 − 4x for values of x from −3 to 3.
x−3−2−10123
y                     
6
FComplete the table of values for y = 1 ÷ x. Give y as a decimal.
x−4−2−1−0.50.5124
y                        
7
F(a) Complete the table of values for y = 4x. (b) Draw the graph of y = 4x for values of x from −8 to 8 (but not 0).
x−8−4−2−11248
y                        
8
F(a) Complete the table of values for y = −2x. (b) Draw the graph of y = −2x for values of x from −4 to 4 (but not 0).
x−4−2−1124
y                  
9
RGemma tries to complete a table for y = 1 ÷ x and cannot work out y when x = 0. Explain why, and say what this means for the graph.
10
FHere is the graph of y = 6 ÷ x. Use the graph to find (a) y when x = 2, (b) y when x = −3, (c) x when y = 1.5, (d) x when y = −2.
xy-7-6-5-4-3-2-11234567-7-6-5-4-3-2-11234567
(a)y when x = 2
(b)y when x = −3
(c)x when y = 1.5
(d)x when y = −2
11
PSTwo numbers x and y multiply to give 12, so y = 12 ÷ x. (a) Complete the table of values. (b) Draw the graph. (c) Use your graph to find x when y = 5.
x−12−6−4−3−2−11234612
y                                    
12
RWrite true or false for each statement and explain your answer.
(a)The graph of y = x³ passes through the origin.
(b)The graph of y = 1 ÷ x crosses the y-axis.
(c)The graph of y = x³ is U-shaped.
(d)When x is negative, y = x³ is negative.
13
PSUse the graph of y = x³ − 4x to (a) write down the roots of x³ − 4x = 0, (b) find estimates of the solutions of x³ − 4x = 1, to 1 decimal place.
xy-3-2-1123-8-6-4-22468
(a)the roots
(b)x³ − 4x = 1
14
RThe curves y = 1 ÷ x and the line y = x are drawn on the grid. (a) Write down the coordinates of the points where they cross. (b) Show that these are the solutions of 1 ÷ x = x.
xy-4-3-2-11234-4-3-2-11234
(a)the crossing points
(b)show they solve 1/x = x
13.5

Recognising graphs from their equations

Edexcel 1MA1: A12
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

There are four families of graph in this chapter and in Chapter 12. You should be able to recognise each one from its equation and from its shape, and to sketch it. A sketch is a quick drawing that shows the correct shape and any important points, such as where it crosses the axes. It does not need a table of values or an accurate scale.

xyy = x + 1xyy = 2 − xxyy = x²xyy = −x²xyy = x³xyy = −x³xyy = 1/xxyy = −1/x
The four families: straight lines, quadratics (U or n), cubics (S-shape) and reciprocals (two branches), each with a positive and a negative version.

To name a graph from its equation, look at the highest power of x, or whether x is on the bottom of a fraction:

Once you know the family, use any numbers added or subtracted to find the y-intercept: put x = 0 into the rule. For y = x² − 3 the intercept is −3, so the U-shaped curve crosses the y-axis at (0, −3).

Key ideas

  • Highest power x: straight line. Highest power x²: parabola (U if positive, n if negative).
  • Highest power x³: cubic S-shape (up if positive, down if negative).
  • x on the bottom of a fraction: reciprocal, two branches, never touching the axes.
  • The y-intercept comes from putting x = 0 into the rule.
  • A sketch shows the shape and the key points (intercepts, turning point) without an accurate scale.
Worked example 13.5.1
Name the type of graph given by y = x³ − 2 and describe its shape. Where does it cross the y-axis?
▶ Watch this solution as a video

Key words. A cubic is a rule where the highest power of x is x³ (x cubed). A reciprocal is 1 divided by a number, so the reciprocal of 4 is 1 ÷ 4 = 0.25. An asymptote is a line that a curve gets closer and closer to but never touches.

Method. Look at the highest power of x to name the graph, then use the sign of that term and the constant term.

  1. Find the highest power of x. In x³ − 2 the highest power is 3.
    Highest power: x³
  2. Name and describe the graph. A graph with x³ is a cubic. The number in front of x³ is +1 (positive), so it goes from bottom-left to top-right.
    Cubic, rising from bottom-left to top-right
  3. Find the y-intercept. On the y-axis x = 0. Put x = 0 into the rule.
    y = 0³ − 2 = −2, so it crosses at (0, −2)

Check. The point (0, −2) fits the rule, and a cubic does cross the y-axis once. ✓

Watch out. Do not call a cubic a quadratic.

Worked example 13.5.2
Match each equation to its graph (A to D): (1) y = 2x + 1, (2) y = x² + 1, (3) y = x³, (4) y = 4 ÷ x.
xyAxyBxyCxyD
▶ Watch this solution as a video

Key words. A cubic is a rule where the highest power of x is x³ (x cubed). A reciprocal is 1 divided by a number, so the reciprocal of 4 is 1 ÷ 4 = 0.25. An asymptote is a line that a curve gets closer and closer to but never touches.

Method. Name the graph from each equation, then find the matching picture.

  1. Name each graph. Look at the highest power of x, or x on the bottom.
    1: 2x + 1 is linear.
    2: x² + 1 is quadratic.
    3: x³ is cubic.
    4: 4 ÷ x is reciprocal.
  2. Match the shapes. Linear is a straight line (B). Quadratic is a U (D). Cubic is an S (A). Reciprocal has two branches (C).
    A: S-shape, B: straight line, C: two branches, D: U-shape
  3. Write the matches.
    1 → B, 2 → D, 3 → A, 4 → C

Check. Check one: y = x² + 1 is always at least 1, so it never goes below the x-axis: the U in D. ✓

Watch out. Name the graph first.

Worked example 13.5.3
This graph crosses the y-axis at 3. Which equation could it be? A: y = x² + 3, B: y = 3 − x², C: y = x³ + 3, D: y = 3 ÷ x.
xy-3-2-1123-5-4-3-2-112345
▶ Watch this solution as a video

Key words. A cubic is a rule where the highest power of x is x³ (x cubed). A reciprocal is 1 divided by a number, so the reciprocal of 4 is 1 ÷ 4 = 0.25. An asymptote is a line that a curve gets closer and closer to but never touches.

Method. Use the shape to decide the family and the direction, then the y-intercept to decide between the rest.

  1. Describe the shape. The curve is an n-shape (opening downwards) with its highest point on the y-axis.
    n-shaped parabola, so a quadratic with a negative x² term
  2. Remove the wrong families. C is a cubic, D is a reciprocal and A has a positive x² term (a U-shape).
    Only B has a negative x² term: 3 − x² = −x² + 3
  3. Check the y-intercept. Put x = 0 into B.
    y = 3 − 0 = 3, which matches the graph
  4. Write the answer.
    B: y = 3 − x²

Check. Test the root: x² = 3 gives x = ±1.7 (about), where the curve meets the x-axis in the picture. ✓

Watch out. Check the sign of the x² term, not just the number 3.

Worked example 13.5.4
Sketch the graph of y = −x³.
▶ Watch this solution as a video

Key words. A cubic is a rule where the highest power of x is x³ (x cubed). A reciprocal is 1 divided by a number, so the reciprocal of 4 is 1 ÷ 4 = 0.25. An asymptote is a line that a curve gets closer and closer to but never touches.

Method. Name the family, then decide its direction.

  1. Name the graph. The highest power is x³, so it is a cubic.
    Cubic
  2. Find the direction. The number in front of x³ is negative (−1), so the graph goes from top-left to bottom-right.
    Falls from top-left to bottom-right
  3. Find a key point. Put x = 0 in the rule.
    y = 0, so it passes through the origin (0, 0)
  4. Draw. Draw smooth axes, then a smooth S-shaped curve that falls through the origin.
    See the sketch below.

Check. At x = 1, y = −1 (below the axis) and at x = −1, y = 1 (above): consistent with falling. ✓

Watch out. Label the origin and keep the curve smooth.

xy
Sketch of y = −x³: falls from top-left to bottom-right through the origin.

Watch out

A graph is quadratic only if the highest power is x². Do not mix up the shapes: x³ is S-shaped, not U-shaped. And a reciprocal graph never touches either axis. Always check the sign: a minus in front of the power turns the picture upside down (or flips it).

Exam tip

Decide the family first (highest power), then the sign, then use the y-intercept (put x = 0) to choose between similar graphs.

Exercise 13E  Recognising and sketching graphs

Non-calculator
1
FName the type of graph given by each equation (linear, quadratic, cubic or reciprocal) and describe its shape.
(a)y = 3x − 2
(b)y = x2 + 4
(c)y = x3 − 1
(d)y = 5x
(e)y = −2x2
(f)y = −3x
2
FMatch each equation to its graph.
xyAxyBxyCxyD
(a)y = 2x + 1
(b)y = x2 + 1
(c)y = x3
(d)y = 4x
3
FMatch each equation to its graph (A to D).
xyAxyBxyCxyD
(a)y = −x + 3
(b)y = −x2
(c)y = −x3
(d)y = −2x
4
FSketch each graph. Your sketch should show the correct shape.
(a)Sketch y = x2
(b)Sketch y = −x2
(c)Sketch y = x3
(d)Sketch y = −x3
(e)Sketch y = 2x
(f)Sketch y = −2x
5
FWhich graph could be y = x² − 3? A: crosses the y-axis at (0, 3), U-shaped. B: crosses the y-axis at (0, −3), U-shaped. C: crosses the y-axis at (0, −3), n-shaped.
6
RWrite true or false for each statement. Give a reason.
(a)A quadratic graph is always U-shaped.
(b)The graph of y = 2x + 5 is a curve.
(c)The graph of y = x³ goes through (−1, −1).
(d)The graph of y = 4 ÷ x has a point with x = 0.
7
RWhich of these graphs has a turning point: y = 2x + 1, y = x² − 3, y = x³, y = 1 ÷ x?
8
PSWhich of these points lie on the graph of y = x³?
(a)(2, 8)
(b)(−2, −8)
(c)(1, 2)
(d)(−1, 1)
9
PSA graph is n-shaped, crosses the y-axis at 4 and crosses the x-axis at −2 and 2. Which equation could it be? y = x² − 4, y = 4 − x², y = x² + 4, y = 4 ÷ x.
10
RRavi says the graph of y = x² + 1 passes through the origin. Show that he is wrong.
11
RUse the values of x = 2 and x = −2 to explain the difference between the shapes of y = x² and y = x³.
12
PSA parabola crosses the x-axis at (−3, 0) and (1, 0) and crosses the y-axis at (0, 3). Which equation could it be? y = (x − 1)(x + 3), y = −(x − 1)(x + 3), y = (x + 1)(x − 3), y = −(x + 1)(x − 3).
13
PSSketch the graph of (a) y = x³ − 4x and (b) y = 4x − x³. For each, show where it crosses the x-axis (you may use x(x² − 4) = 0).
(a)y = x³ − 4x
(b)y = 4x − x³
14
RZara says: “Every graph of the form y = ax² + c is symmetrical about the y-axis.” Show that she is right by comparing the y-values for x and −x.
13.6

Real-life curved graphs

Edexcel 1MA1: A14
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Many real situations are described by curved graphs: the path of a ball, the area of a field, the profit of a business, the depth of water draining from a tank. You meet them in exams as a graph that is already drawn (or a rule from which you draw it), and you answer questions by reading values and by interpreting what the graph tells you.

Always check which axis is which, and the units. To find the value on one axis from a value on the other, draw a straight line from the given value to the curve, then straight across or down to the other axis. Use a ruler and use the grid carefully.

1234510152025Height (m)Time (seconds)
A ball is thrown upwards. Its height h metres after t seconds is h = 20t − 5t². The ball lands after 4 seconds.

Some special points on a curve have a meaning in the real situation:

If you are given the rule you can also find the exact value by substitution. For example when t = 1: h = 20 × 1 − 5 × 1² = 15. A reciprocal graph also appears in real life. If a coach costing £240 is shared equally between n people, each person pays C = 240 ÷ n pounds. The more people, the less each pays, but the cost never reaches 0.

Key ideas

  • Read the axis labels and units first.
  • To read a value: go from the known value to the curve, then to the other axis, using straight lines.
  • The highest or lowest point of the curve gives the maximum or minimum.
  • A horizontal line can meet a curve twice: look for both answers.
  • Where the curve meets the horizontal axis the quantity is 0.
  • Readings are approximate; check with the rule if you are given one.
Worked example 13.6.1
A ball is thrown upwards. Its height h metres after t seconds is shown on the graph. Use the graph to find the height of the ball after 1 second.
1234510152025Height (m)Time (seconds)
▶ Watch this solution as a video

Key words. To read from a graph means going from one axis to the curve and across to the other axis. Readings are approximate, so give your answer with its unit. The maximum is the highest value and the minimum is the lowest value.

Method. Go up from the time to the curve, then across to the height.

  1. Find 1 s on the horizontal axis. The horizontal axis shows the time.
    Locate t = 1.
  2. Go straight up to the curve, then straight across to the vertical axis. The vertical axis shows the height. Use a ruler to keep the lines straight.
    The curve is at height 15.
  3. Read the height. Read the number on the vertical axis and write the unit.
    15 m

Check. Use the rule: t = 1:  h = −5 × (1)2 + 20 × 1 = −5 + 20 = 15. ✓

Watch out. Read from the correct axis. The question tells you which quantity is given and which you need.

Worked example 13.6.2
Use the same graph to find the times when the ball is 15 m above the ground.
▶ Watch this solution as a video

Key words. To read from a graph means going from one axis to the curve and across to the other axis. Readings are approximate, so give your answer with its unit. The maximum is the highest value and the minimum is the lowest value.

Method. Go across from the height to the curve, then down to the time. Check for more than one crossing.

  1. Find 15 m on the vertical axis. The vertical axis shows the height.
    Locate h = 15.
  2. Draw a horizontal line across to the curve. Use a ruler. Every point on this line has a height of 15 m. Mark every place where it meets the curve (there may be more than one).
    The line meets the curve at (1, 15) and (3, 15).
  3. Go straight down from each point to the horizontal axis and read the time. The horizontal axis shows the time.
    1 and 3 s

Check. Use the rule: t = 1:  h = −5 × (1)2 + 20 × 1 = −5 + 20 = 15; t = 3:  h = −5 × (3)2 + 20 × 3 = −45 + 60 = 15. ✓

Watch out. A horizontal line can cross a curve twice. Look for both crossing points.

Worked example 13.6.3
Use the graph to find the greatest height reached by the ball, and the time when it happens.
▶ Watch this solution as a video

Key words. To read from a graph means going from one axis to the curve and across to the other axis. Readings are approximate, so give your answer with its unit. The maximum is the highest value and the minimum is the lowest value.

Method. The maximum is at the turning point of the curve.

  1. Find the highest point of the curve. This is the turning point. A U-shaped curve has a lowest point; an n-shaped curve has a highest point.
    The turning point is (2, 20).
  2. Read both coordinates. The horizontal coordinate is the time at which it happens. The vertical coordinate is the maximum height.
    Time = 2 s, height = 20 m
  3. Write the answer in words. Always say what the numbers mean and give units.
    The maximum height is 20 m, at 2 s.

Check. Use the rule: t = 2:  h = −5 × (2)2 + 20 × 2 = −20 + 40 = 20. ✓

Watch out. The question may ask for the maximum value (a height) or for when it happens (a time). Read the right coordinate.

Worked example 13.6.4
A coach costing £240 is shared equally by n people. The cost per person is C = 240 ÷ n. Use the graph to find how many people are needed so that each person pays £15.
4812162024283210203040506070Cost per person (£)Number of people
▶ Watch this solution as a video

Key words. To read from a graph means going from one axis to the curve and across to the other axis. Readings are approximate, so give your answer with its unit. The maximum is the highest value and the minimum is the lowest value.

Method. Go across from the cost per person to the curve, then down to the number of people. Check for more than one crossing.

  1. Find £15 on the vertical axis. The vertical axis shows the cost per person.
    Locate C = 15.
  2. Draw a horizontal line across to the curve. Use a ruler. Every point on this line has a cost per person of £15. Mark every place where it meets the curve (there may be more than one).
    The line meets the curve at (16, 15).
  3. Go straight down from each point to the horizontal axis and read the number of people. The horizontal axis shows the number of people.
    16 people

Check. Use the rule: n = 16:  C = 240 ÷ 16 = 15. ✓

Watch out. A horizontal line can cross a curve twice. Look for both crossing points.

Watch out

Check what each axis shows. On the ball graph the horizontal axis is time and the vertical axis is height. The highest point is (2, 20): the ball is highest after 2 seconds, and the greatest height is 20 m. Do not give only one answer when the question asks when and a horizontal line crosses the curve twice. Always include the unit.

Exam tip

Draw your reading lines on the diagram with a ruler (the examiner can then give credit even if the final number is slightly out), and write the units in your answer.

Exercise 13F  Real-life graphs

Non-calculator
1
FThe graph shows the height of a ball thrown upwards, h = 20t − 5t². Use it to find
1234510152025Height (m)Time (seconds)
(a)the height of the ball after 1 second
(b)the height of the ball after 3 seconds
(c)the greatest height reached by the ball
(d)how long the ball is in the air
2
FUse the ball graph to find the time(s) when the height is (a) 15 m, (b) 20 m, (c) 10 m. Give (c) to 1 decimal place.
1234510152025Height (m)Time (seconds)
(a)height 15 m
(b)height 20 m
(c)height 10 m
3
FA farmer has 24 m of fencing for a rectangular pen against a wall. The width is x m and the area is A m² where A = 12x − x². Use the graph to find (a) the area when x = 3, (b) the largest area and the width that gives it, (c) the widths that give an area of 20 m².
123456789101112510152025303540Area (m²)Width (m)
(a)the area when x = 3
(b)the largest area
(c)widths giving area 20 m²
4
FWater drains from a tank. The depth d cm after t minutes is shown on the graph. Use it to find (a) the depth after 4 minutes, (b) the depth after 2 minutes, (c) when the depth is 25 cm, (d) when the tank is empty.
12345678910102030405060708090100Depth (cm)Time (minutes)
(a)the depth after 4 minutes
(b)the depth after 2 minutes
(c)when the depth is 25 cm
(d)when the tank is empty
5
FA coach costing £240 is shared equally between n people, so each pays C = 240 ÷ n pounds. Use the graph to find (a) the cost per person when 8 people go, (b) the cost per person when 15 people go, (c) the number of people if each pays £12, (d) the number if each pays £10.
4812162024283210203040506070Cost per person (£)Number of people
(a)8 people
(b)15 people
(c)each pays £12
(d)each pays £10
6
PSA school disco sells tickets at x pounds. The profit is P pounds where P = −5x² + 50x − 80. Use the graph to find (a) the profit when the ticket price is £3, (b) the ticket price that gives the greatest profit and the profit it gives, (c) the two ticket prices at which the school makes no profit (breaks even).
12345678910-80-60-40-20204060Profit (£)Ticket price (£)
(a)the profit at £3
(b)the best ticket price
(c)break-even prices
7
PSUse the profit graph (above) to find the ticket prices for which the profit is at least £40. (Find where the profit is £40 first.)
12345678910-80-60-40-20204060Profit (£)Ticket price (£)
8
PSDan says that the ball is more than 18 m above the ground for more than 1 second. Use the ball graph to decide whether he is right. Give a reason.
1234510152025Height (m)Time (seconds)
9
PSEach person on the coach trip can pay at most £20. Use the graph of C = 240 ÷ n to find the smallest number of people needed.
4812162024283210203040506070Cost per person (£)Number of people
10
PSRead from the tank graph the depth after 0 minutes and after 5 minutes. Work out the average rate at which the depth falls in the first 5 minutes, in cm per minute.
12345678910102030405060708090100Depth (cm)Time (minutes)
11
RSasha says: “The ball takes 2 seconds to reach its highest point, so the whole flight takes 4 seconds.” Use the symmetry of the graph to explain why she is right.
12
RA student looks at the tank graph and says the water drains at a constant rate. Use values from the graph to show that this is wrong.
13
PSThe bumpy road profile has height h cm at distance x m, with h = x³ − 6x² + 9x. Use the graph to find (a) the height at x = 2, (b) the height at x = 4, (c) the distances at which the height is 2 cm, to 1 decimal place.
1234123456Height (cm)Distance (m)
(a)x = 2
(b)x = 4
(c)height 2 cm
14
RUse the ball graph to find the height at 0.5 seconds and at 3.5 seconds. Explain why the two heights are the same.
1234510152025Height (m)Time (seconds)
(a)height at t = 0.5
(b)height at t = 3.5
(c)Explain
15
PSUse the area graph to find the widths for which the area of the pen is 30 m². Give your answers to 1 decimal place. Explain why there are two answers.
123456789101112510152025303540Area (m²)Width (m)
(a)widths for 30 m²
(b)Explain
16
RAnother ball is thrown with height h = 30t − 5t². Without drawing a graph, find (a) when it lands, (b) the time when it is at its highest, (c) the greatest height.
(a)when it lands
(b)time of greatest height
(c)greatest height
13.7

Distance-time graphs

Edexcel 1MA1: A14
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

A distance-time graph shows a journey. Time goes along the horizontal axis and the distance from the start goes up the vertical axis. You can read how far someone has travelled at any time and when they reached any distance.

The important idea is that the gradient (steepness) of the line is the speed. Remember from Chapter 12 that gradient = change up ÷ change across. On this graph that is change in distance ÷ change in time, which is exactly the formula for speed.

1235101520253035Distance (km)Time (hours)
A cyclist rides for 1 hour, stops for half an hour, then rides on. Part 1 and part 3 slope up (moving). Part 2 is flat (stopped). Part 3 is steeper than part 1, so it is faster.

Finding a speed. Pick a straight part. Work out the change in distance and the change in time, then divide. For the first hour: 15 km in 1 hour is 15 ÷ 1 = 15 km/h. For the last part: 20 km in 1 hour is 20 km/h.

Average speed. The average speed of a whole journey is total distance travelled ÷ total time taken. The time includes stops. For the cyclist: 35 km in 2.5 hours = 14 km/h.

Units. Distance in kilometres and time in hours give a speed in km/h. If the graph uses minutes, change to hours (divide minutes by 60) before you give a speed in km/h. 30 minutes = 0.5 hours.

Key ideas

  • Gradient of a distance-time graph = speed = change in distance ÷ change in time.
  • Steeper line = faster. Horizontal line = stopped. Line sloping down = returning.
  • Average speed = total distance travelled ÷ total time taken (including stops).
  • Use units that match: km and hours give km/h; m and seconds give m/s.
  • Two journeys on the same graph meet where the lines cross.
Worked example 13.7.1
The graph shows the cyclist’s journey. How far had she cycled after 1 hour, and how far after 2 hours?
1235101520253035Distance (km)Time (hours)
▶ Watch this solution as a video

Key words. A distance-time graph shows how far something has travelled from its start (up the side) as time passes (along the bottom). The gradient (steepness) of the line is the speed, because gradient = change up ÷ change across = distance ÷ time. A flat line means stationary (not moving).

Method. Go up from the time to the graph, then across to the distance axis.

  1. Find the time on the horizontal axis. The horizontal axis shows time (Time (hours)).
    Find 2 on the bottom axis.
  2. Go straight up to the line, then straight across. Draw a vertical line up to the graph, then a horizontal line across to the vertical (distance) axis.
    The line meets the graph at height 25.
  3. Read the distance. Read the number on the distance axis, and include the unit.
    25 km

Check. This time is between 1.5 and 2.5, where the distance goes from 15 to 35; 25 lies between them. ✓

Watch out. Read from the right axis: time is along the bottom, distance is up the side.

Worked example 13.7.2
Work out the speed of the cyclist during the last part of her journey, from 1.5 hours to 2.5 hours.

Key words. A distance-time graph shows how far something has travelled from its start (up the side) as time passes (along the bottom). The gradient (steepness) of the line is the speed, because gradient = change up ÷ change across = distance ÷ time. A flat line means stationary (not moving).

Method. Speed is the gradient of a distance-time graph: speed = change in distance ÷ change in time.

  1. Pick the part of the graph. Speed is the gradient of the line. We need the change in distance and the change in time for this part.
    From (1.5, 15) to (2.5, 35)
  2. Find the change in distance and the change in time. Subtract the start values from the end values.
    Change in distance = 35 − 15 = 20 km
    Change in time = 2.5 − 1.5 = 1 hours
  3. Divide distance by time. Speed = distance ÷ time.
    20 ÷ 1 = 20 km/h

Check. Distance = speed × time = 20 × 1 = 20 km, the change we read from the graph. ✓

Watch out. Use the change in distance and the change in time for the same part of the journey, not the totals from the start.

Worked example 13.7.3
A bus travels as shown. Find its speed during the first 20 minutes, in km/h.
51015202530354045505560510152025303540Distance (km)Time (minutes)

Key words. A distance-time graph shows how far something has travelled from its start (up the side) as time passes (along the bottom). The gradient (steepness) of the line is the speed, because gradient = change up ÷ change across = distance ÷ time. A flat line means stationary (not moving).

Method. Speed is the gradient of a distance-time graph: speed = change in distance ÷ change in time.

  1. Pick the part of the graph. Speed is the gradient of the line. We need the change in distance and the change in time for this part.
    From (0, 0) to (20, 10)
  2. Find the change in distance and the change in time. Subtract the start values from the end values.
    Change in distance = 10 − 0 = 10 km
    Change in time = 20 − 0 = 20 minutes
  3. Divide distance by time. Speed = distance ÷ time.
    10 ÷ 20 = 0.5 km/min
  4. Change to the units asked for. The question wants a different unit, so convert.
    There are 60 minutes in an hour, so multiply by 60: 0.5 × 60 = 30 km/h
    Speed = 30 km/h

Check. Distance = speed × time: 30 km/h for 20 minutes (after converting the units) gives 10 km. ✓

Watch out. Use the change in distance and the change in time for the same part of the journey, not the totals from the start.

Worked example 13.7.4
Work out the average speed of the bus for the whole 60 minutes, in km/h.
▶ Watch this solution as a video

Key words. A distance-time graph shows how far something has travelled from its start (up the side) as time passes (along the bottom). The gradient (steepness) of the line is the speed, because gradient = change up ÷ change across = distance ÷ time. A flat line means stationary (not moving).

Method. Average speed = total distance ÷ total time. Change the time to hours so the answer is in km/h.

  1. Find the total distance travelled. Add the distances of the parts where the vehicle moved.
    10 + 30 = 40 km
  2. Find the total time taken, in hours. Speed in km/h needs the time in hours. There are 60 minutes in an hour, so divide the minutes by 60.
    60 minutes = 60 ÷ 60 = 1 hours
  3. Divide distance by time. Average speed = total distance ÷ total time.
    40 ÷ 1 = 40 km/h

Check. In 1 hours at 40 km/h the distance is 40 × 1 = 40 km. ✓

Watch out. Do not divide kilometres by minutes and call it km/h: convert the minutes to hours first.

Worked example 13.7.5
Priya cycles from town A to town B, 80 km away, at a steady speed. Quinn leaves B one hour later and travels towards A. Use the graph to find when they meet and how far they are from A.
12341020304050607080Distance (km)Time (hours)
▶ Watch this solution as a video

Key words. A distance-time graph shows how far something has travelled from its start (up the side) as time passes (along the bottom). The gradient (steepness) of the line is the speed, because gradient = change up ÷ change across = distance ÷ time. A flat line means stationary (not moving).

Method. They meet where the two lines cross, because at that point they are at the same distance at the same time.

  1. Find the crossing point. Look for where the two lines meet.
    The lines cross at (2, 40).
  2. Read the time. The time is the horizontal coordinate.
    Time = 2 hours after Priya set off
  3. Read the distance. The distance from A is the vertical coordinate.
    Distance = 40 km
  4. Write the answer.
    They meet after 2 hours, 40 km from A.

Check. Priya: 20 km/h × 2 h = 40 km. Quinn has travelled 1 hour at 40 km/h = 40 km from B, which is 80 − 40 = 40 km from A. ✓

Watch out. The crossing point gives both the time and the distance.

Watch out

A horizontal line on a distance-time graph means the person is stopped, not moving at constant speed. A line sloping down means returning to the start, and the speed is still the size of the gradient. Also use total distance travelled ÷ total time for the average speed, and include the time spent stopped.

Exam tip

When the graph is in minutes and the answer is in km/h, convert the time to hours first, or multiply a speed in km per minute by 60.

Exercise 13G  Distance-time graphs

Non-calculator
1
FThe graph shows a cycle ride. Use it to find (a) the distance travelled after 1 hour, (b) the distance after 2 hours, (c) the time when the distance was 35 km, (d) how long the cyclist stopped.
1235101520253035Distance (km)Time (hours)
(a)after 1 hour
(b)after 2 hours
(c)the time when the distance was 35 km
(d)how long she stopped
2
FUse the cycle ride graph to work out (a) the speed during the first hour, (b) the speed during the last hour, (c) the average speed for the whole ride.
1235101520253035Distance (km)Time (hours)
(a)the speed in the first hour
(b)the speed in the last hour
(c)the average speed for the whole ride
3
FA car makes a journey to a shop and back. Use the graph to find (a) the greatest distance from the start, (b) how long the car stopped at the shop, (c) the speed on the way there, (d) the speed on the way back, (e) the average speed for the whole journey.
123451015202530354045505560Distance (km)Time (hours)
(a)the greatest distance
(b)how long the car stopped
(c)the speed on the way there
(d)the speed on the way back
(e)the average speed
4
FThe graph shows a bus journey. Distance is in kilometres and time is in minutes. Find (a) the speed during the first 20 minutes in km/h, (b) how long the bus stopped, (c) the speed in the last 30 minutes in km/h.
51015202530354045505560510152025303540Distance (km)Time (minutes)
(a)speed in the first 20 minutes
(b)how long the bus stopped
(c)speed in the last 30 minutes
5
FWork out the average speed of the bus in the previous question for the whole 60 minutes. Give your answer in km/h.
51015202530354045505560510152025303540Distance (km)Time (minutes)
6
PSJo walks to school. The graph shows distance in metres and time in minutes. Find Jo’s speed (a) in metres per minute during the first 10 minutes, (b) in km/h during the first 10 minutes.
510152025303520040060080010001200140016001800Distance (m)Time (minutes)
(a)in metres per minute
(b)in km/h
7
FMia leaves home and cycles 12 km in 1 hour. She rests for half an hour, then cycles home in 1.5 hours. Draw a distance-time graph of her journey. Use a horizontal axis from 0 to 3 hours and a vertical axis from 0 to 12 km.
8
PSSam drives 90 km in 1.5 hours, stops for 30 minutes, then drives a further 60 km in 1 hour. Draw a distance-time graph and work out his average speed for the whole journey.
9
PSPriya cycles from A to B (80 km) at a steady speed. Quinn leaves B one hour later and travels towards A. Use the graph to find (a) when they meet, (b) how far from A they meet, (c) who travels faster, and how you can tell.
12341020304050607080Distance (km)Time (hours)
(a)when they meet
(b)how far from A
(c)who is faster
10
RTom says: “A horizontal line on a distance-time graph means the person is travelling at a constant speed.” Explain his mistake.
11
ROn a journey graph the line slopes down from (2, 60) to (3, 0). Ali says the car is slowing down. Explain why he is wrong.
12
PSPriya’s line is d = 20t. Quinn’s line is d = 120 − 40t (for 1 ≤ t ≤ 3). Work out when they meet by solving 20t = 120 − 40t.
13
RIn a 200 m race Amy runs at a steady 8 m/s from the start. Ben starts 4 seconds later and runs at a steady 10 m/s. Draw both on a distance-time graph and decide who wins and by how long.
13.8

Speed-time graphs

Edexcel 1MA1: A14
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

A speed-time graph (also called a velocity-time graph) shows how the speed of something changes. Time is along the horizontal axis and speed is up the vertical axis. This is different from a distance-time graph, so be careful to read the labels on the axes.

The gradient of a speed-time graph is the acceleration: change in speed ÷ change in time. With speed in m/s and time in s the unit is m/s² (metres per second per second). A gradient of 2 means the speed rises by 2 m/s every second. A negative gradient means the speed is falling. This is a deceleration.

The area under the graph (between the line and the time axis) is the distance travelled. This works because distance = speed × time: speed is the height and time is the width. Split the area into rectangles, triangles and trapezia. (Chapter 25 returns to areas under curved graphs.)

246810121416182022246810121416Speed (m/s)Time (seconds)
A car speeds up for 8 s, travels at 16 m/s for 10 s, then slows to a stop in 4 s. The three coloured areas (triangle, rectangle, triangle) are the distances travelled in each stage.

Key ideas

  • Speed-time graph: gradient = acceleration = change in speed ÷ change in time (unit m/s²).
  • Sloping up = accelerating. Flat = constant speed. Sloping down = decelerating (negative acceleration).
  • Area under the graph = distance travelled. Use a rectangle (flat), a triangle (starts or ends at 0) or a trapezium (area = ½(a + b) × width).
  • Average speed = total distance ÷ total time.
Worked example 13.8.1
A car's journey is shown on the speed-time graph. Find its speed after 4 seconds.
246810121416182022246810121416Speed (m/s)Time (seconds)
▶ Watch this solution as a video

Key words. A speed-time graph shows the speed (up the side) as time passes (along the bottom). The gradient is the acceleration (how fast the speed changes), and the area under the graph is the distance travelled. Negative acceleration (the speed is dropping) is called deceleration.

Method. Go up from the time to the line, then across to the speed axis.

  1. Find the time on the horizontal axis. The horizontal axis shows time in seconds.
    Find 4 s along the bottom.
  2. Go straight up to the line, then straight across. The vertical axis shows speed.
    The line is at height 8.
  3. Read the speed. Write the unit.
    8 m/s

Check. The speed at this time is between the speeds at the two corners either side of it. ✓

Watch out. Read the vertical axis for speed.

Worked example 13.8.2
Work out the acceleration of the car during the first 8 seconds.
▶ Watch this solution as a video

Key words. A speed-time graph shows the speed (up the side) as time passes (along the bottom). The gradient is the acceleration (how fast the speed changes), and the area under the graph is the distance travelled. Negative acceleration (the speed is dropping) is called deceleration.

Method. Acceleration is the gradient of a speed-time graph: change in speed ÷ change in time.

  1. Pick the part of the graph. Acceleration is the gradient of the line, so we use the change in speed and the change in time.
    From (0, 0) to (8, 16)
  2. Find the change in speed and the change in time. Subtract the start values from the end values.
    Change in speed = 16 − 0 = 16 m/s
    Change in time = 8 − 0 = 8 s
  3. Divide. Acceleration = change in speed ÷ change in time.
    16 ÷ 8 = 2
  4. Write the answer with its unit. The unit is metres per second per second, written m/s².
    Acceleration = 2 m/s²

Check. The speed changes by 2 m/s every second, for 8 s: 2 × 8 = 16 m/s, the change we found. ✓

Watch out. Acceleration is the gradient. The height of the line is the speed, not the acceleration.

Worked example 13.8.3
Work out the deceleration of the car in the last 4 seconds.
▶ Watch this solution as a video

Key words. A speed-time graph shows the speed (up the side) as time passes (along the bottom). The gradient is the acceleration (how fast the speed changes), and the area under the graph is the distance travelled. Negative acceleration (the speed is dropping) is called deceleration.

Method. Acceleration is the gradient of a speed-time graph: change in speed ÷ change in time.

  1. Pick the part of the graph. Acceleration is the gradient of the line, so we use the change in speed and the change in time.
    From (18, 16) to (22, 0)
  2. Find the change in speed and the change in time. Subtract the start values from the end values.
    Change in speed = 0 − 16 = −16 m/s
    Change in time = 22 − 18 = 4 s
  3. Divide. Acceleration = change in speed ÷ change in time.
    −16 ÷ 4 = −4
  4. Interpret the sign. The line slopes downwards, so the speed is falling. This is a deceleration. As a gradient it is negative.
    Acceleration = −4 m/s² (a deceleration of 4 m/s²)

Check. The speed changes by −4 m/s every second, for 4 s: −4 × 4 = −16 m/s, the change we found. ✓

Watch out. Acceleration is the gradient. The height of the line is the speed, not the acceleration.

Worked example 13.8.4
Work out the total distance travelled by the car in the 22 seconds.
▶ Watch this solution as a video

Key words. A speed-time graph shows the speed (up the side) as time passes (along the bottom). The gradient is the acceleration (how fast the speed changes), and the area under the graph is the distance travelled. Negative acceleration (the speed is dropping) is called deceleration.

Method. Distance travelled = area under the speed-time graph. Split it into rectangles, triangles and trapezia.

  1. Remember: distance is the area under the graph. Distance = speed × time. On the graph, speed is the height and time is the width, so distance is the area between the line and the time axis.
  2. Split the area into simple shapes. Use a rectangle where the line is flat, a triangle where it starts or ends at 0, and a trapezium otherwise.
    From 0 s to 8 s, a triangle: ½ × 8 × 16 = 64
    From 8 s to 18 s, a rectangle: 16 × 10 = 160
    From 18 s to 22 s, a triangle: ½ × 4 × 16 = 32
  3. Add the areas. The total area is the total distance travelled.
    64 + 160 + 32 = 256 m

Check. Sense check: the vehicle never goes faster than 16 m/s and travels for 22 s, so the distance must be less than 352 m. Our 256 m is smaller. ✓

Watch out. The unit of the area is the unit of distance (m), not m/s. Area = height × width means speed × time.

Worked example 13.8.5
A train's speed is shown. Work out its average speed over the whole 40 seconds.
5101520253035402468101214161820Speed (m/s)Time (seconds)
▶ Watch this solution as a video

Key words. A speed-time graph shows the speed (up the side) as time passes (along the bottom). The gradient is the acceleration (how fast the speed changes), and the area under the graph is the distance travelled. Negative acceleration (the speed is dropping) is called deceleration.

Method. Average speed = total distance ÷ total time. Find the distance from the area under the graph.

  1. Find the area of each part. Triangle, rectangle, triangle.
    Triangle: ½ × 10 × 20 = 100
    Rectangle: 20 × 20 = 400
    Triangle: ½ × 10 × 20 = 100
  2. Add to get the total distance.
    100 + 400 + 100 = 600 m
  3. Divide by the total time. The time is 40 seconds.
    600 ÷ 40 = 15 m/s

Check. 15 m/s for 40 s = 600 m. ✓

Watch out. Average speed is not the top speed (20 m/s).

Watch out

On a speed-time graph a flat line means constant speed (not stopped). A line sloping down means slowing down, not going backwards. And a gradient is acceleration here, not speed. The area under the graph has units of distance (m), not m/s.

Exam tip

Label each area you use (A, B, C) with its shape and its calculation, and add them at the end. Check the units of every axis before you start.

Exercise 13H  Speed-time graphs

Non-calculator
1
FThe graph shows the speed of a car. Use it to find (a) the speed after 4 seconds, (b) how long the car travelled at a constant speed, (c) the acceleration in the first 8 seconds, (d) the deceleration in the last 4 seconds.
246810121416182022246810121416Speed (m/s)Time (seconds)
(a)speed after 4 seconds
(b)time at constant speed
(c)acceleration in the first 8 seconds
(d)deceleration in the last 4 seconds
2
FA train’s journey is shown. Find (a) its acceleration in the first 10 seconds, (b) its speed while moving steadily, (c) how long it takes to stop at the end, (d) its deceleration at the end.
5101520253035402468101214161820Speed (m/s)Time (seconds)
(a)acceleration
(b)steady speed
(c)time to stop
(d)deceleration
3
FUse the car graph to work out the distance travelled (a) in the first 8 seconds, (b) in the 10 seconds at constant speed, (c) in the whole 22 seconds.
246810121416182022246810121416Speed (m/s)Time (seconds)
(a)first 8 seconds
(b)the 10 seconds at constant speed
(c)the whole 22 seconds
4
FWork out the total distance travelled by the train in the graph (40 seconds).
5101520253035402468101214161820Speed (m/s)Time (seconds)
5
FA cyclist’s speed is shown. Find (a) the acceleration in the first 6 seconds, (b) the distance travelled in the 10 seconds.
123456789101234567891011121314Speed (m/s)Time (seconds)
(a)acceleration
(b)distance in 10 seconds
6
FA lorry travels at a steady speed, then brakes to a stop. Use the graph to find (a) the speed of the lorry before braking, (b) the deceleration, (c) the total distance travelled in the 10 seconds.
12345678910123456789101112Speed (m/s)Time (seconds)
(a)speed before braking
(b)deceleration
(c)total distance
7
FA car starts from rest and speeds up to 12 m/s in 6 seconds. It then travels at 12 m/s for 10 seconds, then slows down to a stop in 4 seconds. Draw a speed-time graph for the car, then work out its total distance.
8
PSUse the train graph to find the train’s average speed over the 40 seconds.
5101520253035402468101214161820Speed (m/s)Time (seconds)
9
PSThe train’s speed increases steadily from 0 to 20 m/s in 10 seconds. How far has it travelled when its speed first reaches 15 m/s?
5101520253035402468101214161820Speed (m/s)Time (seconds)
10
RLayla says: “The line on a speed-time graph slopes down, so the car is going backwards.” Explain her mistake.
11
RComplete the table to compare what a flat line and a sloping-up line mean on each type of graph.
(a)A flat line on a distance-time graph
(b)A flat line on a speed-time graph
(c)A sloping-up line on a speed-time graph
12
PSA train’s speed is in km/h and time is in minutes: it accelerates steadily from 0 to 60 km/h in 10 minutes, then travels at 60 km/h for 10 minutes. Work out the distance travelled in km.
246810121416182051015202530354045505560Speed (km/h)Time (minutes)
13
PSUse the car graph to find the times when the car’s speed is 10 m/s.
246810121416182022246810121416Speed (m/s)Time (seconds)
14
RCar A accelerates at 2 m/s² from rest for 10 seconds and then travels at constant speed. Car B accelerates steadily from rest at 1 m/s² for 20 seconds. Both start together. Draw both on a speed-time graph. Which car has travelled further after 20 seconds, and by how much?
13.9

Sketching quadratics and completing the square

Edexcel 1MA1: A11, A12
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

A sketch shows the shape of a graph and its key points without plotting a table of values. For a quadratic you need: (1) the shape (U or n, from the sign of x²), (2) the y-intercept (put x = 0), (3) the roots (put y = 0 and solve, usually by factorising), and (4) the turning point. Label each key point with its coordinates and draw a smooth, symmetrical curve through them.

xy(1, −4)(−1, 0)(3, 0)(0, −3)
A sketch of y = x² − 2x − 3: U-shaped, y-intercept (0, −3), roots (−1, 0) and (3, 0), turning point (1, −4).

Finding the turning point by completing the square. The turning point comes from rewriting x² + bx + c in the form (x + p)² + q. To see why, think about (x + p)². A number squared is never negative, so (x + p)² is smallest (it is 0) when x + p = 0, that is when x = −p. At that point y = 0 + q = q. So the minimum point (the turning point) of y = (x + p)² + q is (−p, q).

How to complete the square (when the number in front of x² is 1): halve the number in front of x to get p. Write (x + p)². Expanding it gives x² + 2px + p², so it contains an extra p²: subtract p² and add the original constant. For x² − 6x + 4: p = −3, (x − 3)² = x² − 6x + 9, so x² − 6x + 4 = (x − 3)² − 9 + 4 = (x − 3)² − 5. The turning point is (3, −5).

If the number in front of x² is not 1, take it out of the x² and x terms first. If it is negative the curve is n-shaped, and the turning point is a maximum.

Exact roots. If a quadratic does not factorise, complete the square, set y = 0 and square root both sides. For (x − 3)² − 5 = 0 we get (x − 3)² = 5, so x = 3 ± √5 (about 0.76 and 5.24).

xy(3, −5)(0.8, 0)(5.2, 0)(0, 4)
A sketch of y = x² − 6x + 4 = (x − 3)² − 5. The roots 3 − √5 and 3 + √5 are about 0.8 and 5.2.

Key ideas

  • To sketch a quadratic: shape (sign of x²), y-intercept (x = 0), roots (y = 0), turning point. Label the coordinates.
  • Completing the square: x² + bx + c = (x + b/2)² + c − (b/2)².
  • The turning point of y = (x + p)² + q is (−p, q): minimum if the curve is U-shaped, maximum if n-shaped.
  • Roots by completing the square: (x + p)² = number, so x = −p ± the square root.
  • If the number in front of x² is not 1, take it out of the x² and x terms first.
Worked example 13.9.1
Sketch the graph of y = x² − 2x − 3. Show the points where the curve crosses the axes and its turning point.
▶ Watch this solution as a video

Key words. To sketch a graph means to draw its shape with the key points labelled, without plotting every point. The intercepts are where the graph crosses the axes.

Method. Find the shape, the y-intercept, the roots and the turning point, then draw a smooth curve through them.

  1. Decide the shape. The sign of the x² term tells us: positive gives a U-shape, negative gives an n-shape.
    The number in front of x2 is 1, so the curve is U-shaped.
  2. Find the y-intercept. The curve crosses the y-axis where x = 0, which is the constant term.
    x = 0:  y = 02 − 2 × 0 − 3 = 0 + 0 − 3 = −3
    The y-intercept is (0, −3).
  3. Find the roots. The roots are where the curve meets the x-axis, so put y = 0 and solve. Factorise, then use "if two brackets multiply to 0, one of them is 0".
    Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -3.
    We need two numbers that multiply to -3 (the last number) and add to -2 (the middle number). List the pairs that multiply to -3:
    1 × -3 = -3,   add: 1 + -3 = -2 ✓ this pair works
    Put the pair into two brackets: (x + 1)(x − 3).
    Check by expanding: x² − 3x + 1x − 3 = x2 − 2x − 3. ✓
    Set (x − 3)(x + 1) = 0, so x = −1 or x = 3
    The curve crosses the x-axis at (−1, 0) and (3, 0).
  4. Find the turning point. The turning point is on the line of symmetry. Complete the square, or use the midpoint of the roots.
    y = (x − 1)2 − 4
    Turning point = (1, −4)
  5. Draw the axes and the curve. A sketch does not need an accurate scale. It must show the right shape and label the key points: the y-intercept, any roots and the turning point.
    Label (0, −3), the roots, and the turning point (1, −4).

Check. The curve is symmetrical: the turning point lies halfway between the two roots, and the y-intercept is on the curve (substitute x = 0 to see). ✓

Watch out. A sketch needs labelled coordinates, not a perfect drawing. Make sure the curve is smooth and symmetrical.

Worked example 13.9.2
Write x² − 6x + 4 in the form (x + p)² + q. Hence write down the coordinates of the turning point of y = x² − 6x + 4.
▶ Watch this solution as a video

Key words. Completing the square means writing x² + bx + c as (x + p)² + q. The turning point is the lowest point (minimum) of a U-shape or the highest point (maximum) of an n-shape.

Method. Complete the square: write the quadratic as a(x + p)² + q. Then the turning point is (−p, q).

  1. Write the quadratic and look at the numbers. We want it in the form a(x + p)² + q. That form shows the turning point at a glance.
    y = x2 − 6x + 4
  2. Halve the number in front of x. In (x + p)² = x² + 2px + p², the middle term is 2p. So p is half of the number in front of x.
    −6 ÷ 2 = −3, so the bracket is (x − 3)
  3. Square the bracket and see what it gives. Expanding the bracket gives the x² and x terms we want, but it also adds an extra number p².
    (x − 3)2 = x2 − 6x + 9
  4. Take away the extra number and add the original constant. We have an extra 9, so subtract it to keep the value the same. Then put back the constant term.
    x2 − 6x + 4 = (x − 3)2 − 9 + 4
  5. Tidy up the numbers. Combine the two plain numbers.
    y = (x − 3)2 − 5
  6. Find the turning point. A squared bracket is never negative, and it equals 0 when the bracket is 0. So y is at its minimum when (x − 3) = 0, that is when x = 3. Then y = −5.
    Turning point = (3, −5)   (minimum value of y is −5)

Check. Expand the answer: (x − 3)2 − 5 = x2 − 6x + 4, which is the original. ✓

Watch out. The turning point has the opposite sign to the number in the bracket: (x − 3)² + 1 has its turning point at x = +3.

Worked example 13.9.3
Solve x² − 6x + 4 = 0 by completing the square. Give your answers as surds.
▶ Watch this solution as a video

Key words. A surd is a root that does not simplify to a whole number or fraction, such as √5. Exact means leaving it as √5 instead of a rounded decimal.

Method. Complete the square, then undo it: square root both sides (positive and negative).

  1. Put y = 0. Roots are where the curve meets the x-axis.
    0 = x2 − 6x + 4
  2. Complete the square. Halve the number in front of x (−6 ÷ 2 = −3), square the bracket, and correct the constant.
    0 = (x − 3)2 − 5
  3. Get the bracket on its own. Move the 5 to the other side (do the inverse operation).
    (x − 3)2 = 5
  4. Take the square root of both sides. There are two square roots: positive and negative. Leave the answer as a surd (an exact root), simplified if possible.
    x − 3 = ±√5
  5. Get x on its own. Do the inverse of subtracting 3.
    x = 3 ± √5

Check. Decimals: 3 + 2.24 = 5.24 and 3 − 2.24 = 0.76. Substituting 5.24 into the rule gives about 0. ✓

Watch out. Do not forget the ±. Leave surds exact unless the question asks for decimals.

Worked example 13.9.4
Find the coordinates of the turning point of y = −x² + 4x + 1 by completing the square. Is it a maximum or a minimum?
▶ Watch this solution as a video

Key words. Completing the square means writing x² + bx + c as (x + p)² + q. The turning point is the lowest point (minimum) of a U-shape or the highest point (maximum) of an n-shape.

Method. Complete the square: write the quadratic as a(x + p)² + q. Then the turning point is (−p, q).

  1. Write the quadratic and look at the numbers. We want it in the form a(x + p)² + q. That form shows the turning point at a glance.
    y = −x2 + 4x + 1
  2. Take the number −1 out of the x² and x terms. This leaves x² with a 1 in front, so we can complete the square inside the bracket. The constant stays outside.
    y = −[x2 − 4x] + 1
  3. Complete the square inside the bracket. Halve the number in front of x: −4 ÷ 2 = −2. Then x2 − 4x = (x − 2)2 − 4.
    y = −[(x − 2)2 − 4] + 1
  4. Multiply the outside number into what is being subtracted. The −1 multiplies both parts of the square bracket.
    y = −(x − 2)2 + 4 + 1
  5. Tidy up the numbers. Combine the plain numbers.
    y = −(x − 2)2 + 5
  6. Find the turning point. A squared bracket is never negative, and it equals 0 when the bracket is 0. So y is at its maximum when (x − 2) = 0, that is when x = 2. Then y = 5.
    Turning point = (2, 5)   (maximum value of y is 5)

Check. Expand the answer: −(x − 2)2 + 5 = −x2 + 4x + 1, which is the original. ✓

Watch out. The turning point has the opposite sign to the number in the bracket: (x − 3)² + 1 has its turning point at x = +3.

Worked example 13.9.5
Write 2x² − 12x + 7 in the form a(x + p)² + q and hence find the minimum value of 2x² − 12x + 7.
▶ Watch this solution as a video

Key words. Completing the square means writing x² + bx + c as (x + p)² + q. The turning point is the lowest point (minimum) of a U-shape or the highest point (maximum) of an n-shape.

Method. Complete the square: write the quadratic as a(x + p)² + q. Then the turning point is (−p, q).

  1. Write the quadratic and look at the numbers. We want it in the form a(x + p)² + q. That form shows the turning point at a glance.
    y = 2x2 − 12x + 7
  2. Take the number 2 out of the x² and x terms. This leaves x² with a 1 in front, so we can complete the square inside the bracket. The constant stays outside.
    y = 2[x2 − 6x] + 7
  3. Complete the square inside the bracket. Halve the number in front of x: −6 ÷ 2 = −3. Then x2 − 6x = (x − 3)2 − 9.
    y = 2[(x − 3)2 − 9] + 7
  4. Multiply the outside number into what is being subtracted. The 2 multiplies both parts of the square bracket.
    y = 2(x − 3)2 − 18 + 7
  5. Tidy up the numbers. Combine the plain numbers.
    y = 2(x − 3)2 − 11
  6. Find the turning point. A squared bracket is never negative, and it equals 0 when the bracket is 0. So y is at its minimum when (x − 3) = 0, that is when x = 3. Then y = −11.
    Turning point = (3, −11)   (minimum value of y is −11)

Check. Expand the answer: 2(x − 3)2 − 11 = 2x2 − 12x + 7, which is the original. ✓

Watch out. The turning point has the opposite sign to the number in the bracket: (x − 3)² + 1 has its turning point at x = +3.

Watch out

The turning point has the opposite sign to the number inside the bracket: (x − 3)² + 1 has its turning point at x = +3. And do not forget to subtract p² when you complete the square, otherwise the expression has changed. A minus in front of the bracket makes the turning point a maximum.

Exam tip

Always check by expanding your answer. If (x − 3)² − 5 expands to x² − 6x + 4, you are right.

Exercise 13I  Sketching quadratics and completing the square

Non-calculator
1
FWrite each expression in the form (x + p)² + q.
(a)x2 + 4x + 1
(b)x2 − 6x + 2
(c)x2 + 10x
(d)x2 − 2x − 5
(e)x2 + 3x
(f)x2 − 8x + 20
2
FComplete the square to find the coordinates of the turning point of each graph.
(a)y = x2 − 4x + 7
(b)y = x2 + 6x + 5
(c)y = x2 − 10x + 30
(d)y = x2 + x + 1
3
FSketch each graph. Mark the coordinates of the y-intercept, any roots and the turning point.
(a)Sketch y = x2 − 4x + 3
(b)Sketch y = x2 + 2x − 8
(c)Sketch y = −x2 + 4
(d)Sketch y = x2 − 6x + 9
(e)Sketch y = x2 + 4x + 7
4
FSolve each equation by completing the square. Give your answers as surds.
(a)x2 − 4x + 1 = 0
(b)x2 + 6x + 2 = 0
(c)x2 − 2x − 5 = 0
5
PSSketch the graph of y = x² − 6x + 4, showing the exact coordinates of the points where it crosses the x-axis.
6
FWrite down the coordinates of the turning point of each graph, and say whether it is a maximum or a minimum. (No working out is needed.)
(a)y = (x − 5)2 + 3
(b)y = (x + 2)2 − 7
(c)y = −(x − 1)2 + 4
7
RShow that x² − 8x + 19 is always greater than 0.
8
PSThe curve y = x² + bx + c has its turning point at (3, −2). Find the values of b and c.
9
FWrite 2x² − 12x + 7 in the form a(x + p)² + q. Write down the minimum value of 2x² − 12x + 7 and the value of x when it happens.
10
FBy completing the square, find the coordinates of the maximum point of y = −x² + 6x − 5.
11
PSThe path of a ball has equation y = 20x − 5x². Complete the square to find the greatest height and where it happens.
12
RThe graph of y = x² + kx + 9 touches the x-axis at exactly one point, and k is positive. Find k and the coordinates of the point.
13
RPriya says that the turning point of y = (x + 4)² − 5 is (4, −5). Explain her mistake and give the correct coordinates.
14
RProve that the turning point of y = x² + bx + c is at x = −b/2, for any numbers b and c.

Chapter summary

SkillMethod
Table of valuesSubstitute each x into the rule (negatives in brackets), write the y-values, plot (x, y), join with a smooth curve
Quadratic shapex² positive: U-shape with a minimum. x² negative: n-shape with a maximum. Symmetrical about a vertical line
Roots, y-interceptRoots: where the curve meets the x-axis (y = 0). y-intercept: where it meets the y-axis (x = 0)
Turning point and symmetryTurning point (x, y) is the lowest or highest point. Line of symmetry x = x-value of the turning point, halfway between the roots
Roots by algebra(x + p)(x + q) = 0: each bracket equal to 0 in turn (signs flip). x² = k gives x = ±√k
Solving by graph(rule) = k: draw y = k and read the x-values. (rule 1) = (rule 2): read the x-values where the two graphs cross
Cubic graphsHighest power x³. y = x³ is S-shaped through the origin; negative x³ flips it. A cubic can cross the axis up to 3 times
Reciprocal graphsy = k ÷ x: two branches, no point at x = 0. Positive k: top-right and bottom-left. Negative k: top-left and bottom-right
Recognising graphsHighest power x: straight line. x²: parabola. x³: cubic. x on the bottom: reciprocal. Use the sign and the y-intercept
Real-life graphsCheck the axes and units. Read values with straight lines to the curve. Maximum = turning point. A horizontal line can meet the curve twice
Distance-time graphGradient = speed = change in distance ÷ change in time. Flat = stopped. Average speed = total distance ÷ total time
Speed-time graphGradient = acceleration (m/s²). Flat = constant speed. Area under the graph = distance travelled
Completing the square (H)x² + bx + c = (x + b/2)² + c − (b/2)². Turning point (−b/2, c − b²/4)
Sketching a quadratic (H)Shape, y-intercept, roots, turning point, with coordinates labelled and a smooth symmetrical curve

Can you…?

  • I can complete a table of values for a quadratic and plot it as a smooth curve.
  • I can say whether a quadratic is U-shaped or n-shaped from its equation.
  • I can read the roots, y-intercept, line of symmetry and turning point from a quadratic graph.
  • I can find roots by algebra when the rule is in brackets or is x² minus a number.
  • I can solve an equation by reading a graph, including by drawing a horizontal line or a second graph.
  • I can draw and recognise cubic and reciprocal graphs.
  • I can match a graph to its equation from its shape.
  • I can read values, maximum values and roots from real-life curved graphs.
  • I can read distances, times and speeds from a distance-time graph and find an average speed.
  • I can find acceleration and distance travelled from a speed-time graph.
  • Higher: I can sketch a quadratic and find its turning point by completing the square.

Key terms

Quadratic
A rule whose highest power of x is x². Its graph is a parabola.
Parabola
The U-shaped or n-shaped curve that is the graph of a quadratic.
Table of values
A list of x-values with the y-values worked out from the rule.
Root
An x-value where a graph meets the x-axis (where y = 0).
y-intercept
The point where a graph crosses the y-axis (where x = 0).
Turning point
The lowest point (minimum) or highest point (maximum) of a curve.
Line of symmetry
A mirror line that splits a graph into two matching halves. For a quadratic it is vertical through the turning point.
Cubic
A rule whose highest power of x is x³.
Reciprocal
One divided by a number. The graph of y = k/x has two branches.
Asymptote
A line that a curve gets closer to but never touches.
Sketch
A quick drawing of a graph that shows its shape and key points without an accurate scale.
Gradient
How steep a line is: change in y ÷ change in x.
Speed
Distance travelled per unit time. The gradient of a distance-time graph.
Acceleration
The rate at which speed changes. The gradient of a speed-time graph, in m/s².
Deceleration
Slowing down: a negative acceleration.
Completing the square
Writing x² + bx + c as (x + p)² + q. (Higher)
Surd
An exact root such as √5 that is not a whole number or fraction. (Higher)

Exercise 13Z  Check-up (non-calculator)

Non-calculator
1
Complete the table of values for y = x2 − 3.
x−2−1012
y               
2
Here is the graph of y = x² − 4x. Write down (a) the roots, (b) the coordinates of the turning point.
xy-112345-5-4-3-2-112345
(a)the roots
(b)the turning point
3
Here is the graph of y = 8 + 2x − x². Write down (a) the y-intercept, (b) the equation of the line of symmetry, (c) the maximum value of y.
xy-3-2-112345-4-3-2-112345678910
(a)the y-intercept
(b)the line of symmetry
(c)the maximum value of y
4
Use the graph of y = x² − 4x (above) to solve x² − 4x = 5.
xy-112345-5-4-3-2-112345
5
Name the type of graph for each equation and say what its shape is.
(a)y = x3 + 1
(b)y = −x2
(c)y = 3x
6
Complete the table of values for y = x3.
x−2−1012
y               
7
Work out the value of y when x = −3 for y = 12 ÷ x.
8
Use the car journey graph to work out the speed on the way back (from 3 hours to 4 hours).
123451015202530354045505560Distance (km)Time (hours)
9
Use the train’s speed-time graph to work out its acceleration in the first 10 seconds.
5101520253035402468101214161820Speed (m/s)Time (seconds)
10
Use the lorry’s speed-time graph to work out the total distance it travelled in 10 seconds.
12345678910123456789101112Speed (m/s)Time (seconds)

Total 52 marks. Allow about 52 minutes.

Exercise 13X  Exam-style questions

Non-calculator
1
F(a) Complete the table of values for y = x² − 4x + 1. (b) On a grid, draw the graph of y = x² − 4x + 1 for values of x from −1 to 5. (c) Write down the coordinates of the turning point of the graph.[4 marks]
x−1012345
y   1   −3   1   
2
FThe graph of y = x² − 4x is shown. (a) Write down the roots of x² − 4x = 0. (b) Write down the equation of the line of symmetry. (c) Write down the coordinates of the turning point.[3 marks]
xy-112345-5-4-3-2-112345
(a)the roots
(b)the line of symmetry
(c)the turning point
3
FThe graph of y = x² − 2x − 3 is shown. Use it to find estimates of the solutions of x² − 2x − 3 = 2. Give your answers to 1 decimal place.[3 marks]
xy-2-11234-5-4-3-2-112345
4
FThe graphs of y = x² − 1 and y = x + 1 are drawn on the grid. Use the graphs to solve x² − 1 = x + 1.[3 marks]
xy-2-1123-2-1123456789
5
FHere are four equations: A: y = 4 − x, B: y = x² + 2, C: y = x³, D: y = −4 ÷ x. Write down the type of graph for each equation and describe its shape.[4 marks]
(a)A: y = −x + 4
(b)B: y = x2 + 2
(c)C: y = x3
(d)D: y = −4x
6
F(a) Complete the table of values for y = x³ − x. (b) Draw the graph of y = x³ − x for x from −2 to 2.[4 marks]
x−2−1.5−1−0.500.511.52
y   −1.875   0.375   −0.375   1.875   
7
FThe graph of y = 8 ÷ x is shown. Use it to find the value of x when y = 3.2.[2 marks]
xy-8-6-4-22468-8-6-4-22468
8
FThe graph shows a cycle ride. (a) Work out the speed in the first hour. (b) For how long did the cyclist stop? (c) Work out the average speed for the whole ride.[4 marks]
1235101520253035Distance (km)Time (hours)
(a)the speed in the first hour
(b)how long the cyclist stopped
(c)the average speed
9
PSThe speed-time graph shows a car. (a) Work out the acceleration in the first 8 seconds. (b) Work out the total distance travelled by the car.[5 marks]
246810121416182022246810121416Speed (m/s)Time (seconds)
(a)the acceleration
(b)the total distance
10
PSThe graph shows the height of a ball thrown upwards, h = 20t − 5t². (a) Write down the greatest height of the ball. (b) Find the times when the ball is 15 m above the ground. (c) For how long is the ball in the air?[3 marks]
1234510152025Height (m)Time (seconds)
(a)the greatest height
(b)the times at 15 m
(c)time in the air
11
PSPriya cycles from A to B, 80 km, at a steady speed. Quinn leaves B one hour later and travels towards A. Use the graph to find (a) when they meet, (b) how far from A they meet.[3 marks]
12341020304050607080Distance (km)Time (hours)
(a)when they meet
(b)how far from A
12
FSketch the graph of y = x² − 2x − 8. Show the coordinates of the points where the graph crosses the axes and of the turning point.[4 marks]
13
F(a) Write x² + 8x + 3 in the form (x + p)² + q. (b) Write down the coordinates of the turning point of the graph of y = x² + 8x + 3.[3 marks]
14
PSThe graph of y = x² − 2x − 3 is shown. By drawing a suitable straight line on the grid, find estimates of the solutions of x² − x − 5 = 0.[4 marks]
xy-2-11234-5-4-3-2-112345
15
FSolve x² − 4x + 1 = 0 by completing the square. Give your answers in the form a ± √b.[3 marks]
← Chapter 12: Straight-line graphs and coordinatesChapter 14: Quadratics and simultaneous equations →

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