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GCSE Maths · EdexcelChapter 13 · Quadratic and other graphs
GCSE Maths · Quadratic and Other Graphs · Higher · Lesson 13.4

GCSE Maths: Cubic and reciprocal graphs (Higher)

Part of the GCSE Maths quadratic and other graphs chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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13.4

Cubic and reciprocal graphs

Edexcel 1MA1: A12
Grade 4-6

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

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