GCSE Maths Functions and Advanced Graphs (Higher): check-up and exam-style questions

Key words. A function is a rule that turns each input into one output. f(x) is read “f of x” and means the output when the input is x.

Method. To find f(a), substitute the number a for x everywhere in the rule, then work out the answer.

1. Write down the function. f(x) is a rule that says what to do to the input x. The letter x is a placeholder for whichever number we put in.
   f(x) = 3x² − 2
2. Replace every x with −2, using brackets. The number inside the bracket of f(−2) is the input, so it goes where the x was. Brackets matter with negative numbers.
   f(−2) = 3 × (−2)² − 2
3. Work out each part separately (powers first, then multiplying). BIDMAS: indices before multiplication, then add or subtract.
   = 12 − 2
4. Add and subtract. Work from left to right.
   = 10

Check. Work it out again on a calculator, typing the brackets exactly as shown: the result is 10. ✓

Watch out. Put brackets round negative inputs. For example (−3)² = 9 but −3² = −9.

Key words. The input is the number that goes into the function and the output is the number that comes out.

Method. To solve g(x) = 15, put the rule in for g(x) and solve the equation for x.

1. Write the equation. We are told that the output is 15, so g(x) = 15. Put the rule in for g(x).
   2x + 7 = 15
2. Subtract 7 from both sides. The last thing that happens to the x-part is that 7 is added. The opposite of adding 7 is subtracting 7. Do it to both sides to keep the equation balanced.
   2x = 8
3. Divide both sides by 2. The unknown part is being multiplied by 2. The opposite of multiplying by 2 is dividing by 2.
   x = 4
4. Write the answer. x is now on its own.
   x = 4

Check. Put each solution back into the function: g(4) = 2 × 4 + 7 = 15. ✓

Watch out. Do not confuse the two jobs: finding f(3) means put 3 in and get an output; solving f(x) = 3 means the output is 3 and you must find the input.

Key words. The inverse function f⁻¹ reverses f: if f takes 3 to 11, then f⁻¹ takes 11 back to 3. Subject means the letter on its own on one side.

Method. To find the inverse function f⁻¹, write y = f(x), make x the subject, then swap the letters. The inverse undoes what the function did.

1. Write the function as y = … f(x) is the output, which we call y. We want to turn the rule round so that x is given in terms of y.
   y = x + 52
2. Describe what the function does to x, in order. Think of a function machine. To undo it we must reverse every step, in reverse order.
3. Swap the sides so the x part is on the left. We now rearrange to make x the subject.
   x + 52 = y
4. Multiply both sides by 2. The expression is being divided by 2. The opposite of dividing by 2 is multiplying by 2.
   x + 5 = 2y
5. Subtract 5 from both sides. The last thing that happens to the x-part is that 5 is added. The opposite of adding 5 is subtracting 5. Do it to both sides to keep the equation balanced.
   x = 2y − 5
6. Write x in terms of y. Now x is the subject.
   x = 2y − 5
7. Swap the letters: write x for the input of the inverse function. A function is written with x as its input, so we rename y as x.
   f⁻¹(x) = 2x − 5

Check. Test with a number: f(2) = 3.5. Then f⁻¹(3.5) = 2 × 3.5 − 5 = 2, which takes us back to the input 2. ✓

Watch out. The notation f⁻¹(x) does not mean 1 divided by f(x). It means the inverse function.

Key words. A composite function is two functions joined: the output of the first becomes the input of the second.

Method. fg(x) means: do g first, then f. Replace the x in the outer function by the whole inner function.

1. Remember what fg(x) means. fg(x) is the same as f(g(x)). The function written closest to the x (g) is applied first, then f is applied to the result.
   fg(x) = f(g(x))
2. Write down the two functions.
   g(x) = x²
   f(x) = 2x − 1
3. Put g(x) into f: replace the x in f(x) with x². g(x) is the output of the first function, and it becomes the input of f. Use brackets.
   fg(x) = 2(x²) − 1
4. Expand the brackets. Multiply out each bracket, one term at a time.
   2(x²) = 2x²
5. Simplify. Collect like terms.
   fg(x) = 2x² − 1

Check. Test with x = 2: g(2) = 4, then f(4) = 7. The answer gives 2 × 2² − 1 = 7. They match. ✓

Watch out. Order matters: fg(x) and gf(x) are usually different. The function nearest the x goes first.

Method. Do each part of the transformation to the point: the change inside the bracket moves x, the change outside moves y.

1. Deal with f(x − 3). This moves the graph 3 to the right, so add 3 to the x-coordinate.
   (2, 5) → (5, 5)
2. Deal with the + 1. This moves the graph 1 up, so add 1 to the y-coordinate.
   (5, 5) → (5, 6)

Check. The total move is the vector (3, 1): (2 + 3, 5 + 1) = (5, 6). ✓

Watch out. Inside the bracket the effect is the opposite of the sign.

Key words. The radius is the distance from the centre to the edge. A point is on a circle if its coordinates make the equation true.

Method. A circle with centre (0, 0) and radius r has equation x² + y² = r².

1. Remember the equation of a circle with its centre at the origin. Pick any point (x, y) on the circle. The radius to that point, the horizontal distance x and the vertical distance y make a right-angled triangle. Pythagoras gives x² + y² = (radius)².
   x² + y² = r²
2. Put in the radius. Here the radius is r = 6.
   r² = 6² = 36
3. Write the equation. Keep x² + y² on the left.
   x² + y² = 36

Check. Test the point (6, 0), which is on the circle: 6² + 0² = 36. ✓

Watch out. The right-hand side is r², not r. A circle of radius 5 has equation x² + y² = 25, not = 5.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (4)² + (3)² = 16 + 9 = 25
2. Compare with r². The circle has r² = 25. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   25 = 25
3. Write the conclusion.
   The point (4, 3) is on the circle.

Check. The distance from the origin is √25 ≈ 5.00 and the radius is 5.00: equal. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Method. A tangent is perpendicular to the radius at the point of contact.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   (0)² + (5)² = 0 + 25 = 25 ✓
2. Think about the picture. The radius to (0, 5) lies along the y-axis (vertical). A tangent is perpendicular to the radius, so it is horizontal.
   A horizontal line through y = 5
3. Write the equation. Every point on a horizontal line has the same y-coordinate.
   y = 5

Check. The line y = 5 touches the circle only at (0, 5). ✓

Method. The area under a speed–time graph is the distance travelled. Split it into rectangles, triangles and trapezia, find each area and add.

1. Remember what the area represents. On a graph with time on the horizontal axis and speed on the vertical axis, area = speed × time = distance travelled. So the area under the graph is the distance travelled.
2. Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
3. Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.
   Part 1 (from 0 to 10 s): triangle: ½ × 10 × 20 = 100
   Part 2 (from 10 to 30 s): rectangle: 20 × 20 = 400
4. Add the areas.
   100 + 400 = 500

Check. Total time = 30 s, so the average speed is 500 ÷ 30 = 503 m/s. This lies between the lowest and highest values on the graph. ✓

Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.

Method. Try whole-number powers of 3 until the value matches 19. A power 0 gives 1, a negative power is 1 divided by the positive power.

1. Write the equation. We need the power (the index) that turns the base into the target.
   3^x = 19
2. Work out powers of the base. Start near the answer. Remember that going up one power multiplies by the base.
   3⁻³ = 127
   3⁻² = 19
3. Pick the matching power.
   x = −2

Check. 3⁻² = 19. ✓

Watch out. A negative power means a reciprocal: 5⁻² = 1/25, not −25.

Key words. The inverse function f⁻¹ reverses f: if f takes 3 to 11, then f⁻¹ takes 11 back to 3. Subject means the letter on its own on one side.

Method. To find the inverse function f⁻¹, write y = f(x), make x the subject, then swap the letters. The inverse undoes what the function did.

1. Write the function as y = … f(x) is the output, which we call y. We want to turn the rule round so that x is given in terms of y.
   y = 3x + 12
2. Describe what the function does to x, in order. Think of a function machine. To undo it we must reverse every step, in reverse order.
3. Swap the sides so the x part is on the left. We now rearrange to make x the subject.
   3x + 12 = y
4. Multiply both sides by 2. The expression is being divided by 2. The opposite of dividing by 2 is multiplying by 2.
   3x + 1 = 2y
5. Subtract 1 from both sides. The last thing that happens to the x-part is that 1 is added. The opposite of adding 1 is subtracting 1. Do it to both sides to keep the equation balanced.
   3x = 2y − 1
6. Divide both sides by 3. The unknown part is being multiplied by 3. The opposite of multiplying by 3 is dividing by 3.
   x = 2y − 13
7. Write x in terms of y. Now x is the subject.
   x = 2y − 13
8. Swap the letters: write x for the input of the inverse function. A function is written with x as its input, so we rename y as x.
   f⁻¹(x) = 2x − 13

Check. Test with a number: f(2) = 3.5. Then f⁻¹(3.5) = 2 × 3.53 − 13 = 2, which takes us back to the input 2. ✓

Watch out. The notation f⁻¹(x) does not mean 1 divided by f(x). It means the inverse function.

Marks. 1 mark for multiplying by 2, 1 mark for subtracting 1, 1 mark for dividing by 3.

Key words. The period of a graph is the length of one complete cycle before it repeats. sin and cos repeat every 360°; tan repeats every 180°.

Method. Find the first solution with the inverse function, then use the symmetry and the repeat pattern of the sin graph to find the rest in the range.

1. Write the equation and picture the graph. The solutions of sin x = 0.7 are the x-values where the horizontal line y = 0.7 crosses the graph of y = sin x. Between 0° and 360° it crosses 2 times.
2. Use the inverse button on the calculator for the first (principal) solution. The calculator gives one answer, which may not be in the range we want.
   sin⁻¹(0.7) = 44.4°
3. Use the symmetry of the graph to find the other solutions. The sine graph is symmetrical about x = 90° (and x = 270°). So the second solution in the same cycle is 180° − the first.
   180 − 44.4 = 135.6
4. Write the solutions. There are 2 solutions in the range 0° ≤ x ≤ 360°.
   44.4, 135.6

Check. Put each answer into the calculator: sin 44.4° = 0.70; sin 135.6° = 0.70 (rounding to 2 d.p. should give about 0.70). ✓

Watch out. The calculator only gives one answer. Always ask yourself where the graph crosses the line again.

Marks. 1 mark for 44.4, 1 mark for the method 180 − 44.4, 1 mark for 135.6.

Method. Read the amplitude, the period and the middle line from the graph, then match them to y = a sin(bx) + d (or cos).

1. Read the maximum and minimum values. The graph oscillates between a highest and lowest value.
   maximum = 3, minimum = −3
2. Find the amplitude and the middle line. The amplitude is half the distance between the maximum and the minimum. The middle line is halfway between them.
   amplitude = (3 − (−3)) ÷ 2 = 3
   middle line: y = (3 + (−3)) ÷ 2 = 0
3. Read the period (the length of one full cycle). The standard sin graph has period 360°. If one cycle takes 180° then the angle is multiplied by 2.
   period = 180° ⇒ b = 360 ÷ 180 = 2
4. Write the equation. Use y = a sin (bx) + d with a = 3, b = 2, d = 0.
   y = 3 sin 2x

Check. Test one point: when x = 0, the equation gives 0.00, and the graph passes through that value. ✓

Watch out. The period is the length of one whole cycle, not half of it.

Marks. 1 mark for a = 3 (the amplitude), 1 mark for the period 180°, 1 mark for b = 2.

Key words. The multiplier is the number you multiply by to apply a percentage change in one step. Compound means each change is applied to the new amount, not the original.

Method. Growth and decay problems use a multiplier. Find the multiplier, raise it to the power of the number of years, and multiply by the starting amount.

1. Find the growth multiplier. An increase of 2.5% means we add 2.5% of the amount. So we are left with 100% + 2.5% = 102.5% of the amount, and as a decimal that is the multiplier.
   102.5% = 102.5 ÷ 100 = 1.025
2. Write the formula. The amount is multiplied by 1.025 once for each year. After n years we multiply n times, which is the multiplier to the power n.
   V = (starting amount) × (multiplier)ⁿ
3. Put in the numbers: start = £6000, multiplier = 1.025, n = 4. Replace each letter by its value.
   V = £6000 × 1.025⁴
4. Work it out on a calculator. Do the power 1.025⁴ first, then multiply by 6000. Keep all the digits until the final rounding.
   1.025⁴ = 1.10381
   V = £6000 × 1.10381 = 6622.88
5. Round sensibly. Money is rounded to the nearest penny (2 decimal places).
   V = £6622.88

Check. Check the direction: the amount increases, so the answer should be bigger than £6000. It is £6622.88. ✓

Watch out. Do not work out the percentage once and multiply by the number of years. That gives simple growth. Compound growth multiplies by the multiplier every period, so use a power.

Marks. 1 mark for the multiplier 1.025, 1 mark for 6000 × 1.025⁴, 1 mark for the answer.

Key words. A tangent is a straight line that touches a circle at exactly one point. It meets the radius at that point at a right angle.

Method. The tangent is perpendicular to the radius. Find the gradient of the radius, flip it and change the sign, then use the point to find the equation.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   ((−5))² + (12)² = 25 + 144 = 169 ✓
2. Find the gradient of the radius. The radius joins the centre (0, 0) to the point. Gradient = change in y ÷ change in x.
   gradient of radius = (12 − 0) ÷ ((−5) − 0) = −125
3. Find the gradient of the tangent. A tangent is perpendicular to the radius at the point of contact (Chapter 26). For perpendicular lines the gradients multiply to give −1: flip the fraction and change the sign.
   gradient of tangent = −1 ÷ −125 = 512
4. Use y − y₁ = m(x − x₁) with the point. The tangent has gradient m and passes through (a, b).
   y − 12 = 512(x − (−5))
5. Expand and rearrange to y = mx + c. Multiply out the bracket: 512 × (−5) = −2512. Then add 12 to both sides.
   y = 512x + 2512 + 12
   y = 5x12 + 16912
6. (Optional) A neater form. Multiplying through by the bottom of the fraction removes the fractions. The result always has this pattern: ax + by = r².
   −5x + 12y = 169

Check. Put the point back in: y = 512 × (−5) + 16912 = 12, and the point has y = 12. ✓

Watch out. The tangent gradient is the negative reciprocal of the radius gradient. A common slip is to flip the fraction but forget the sign change.

Marks. 1 mark for the gradient of OP, 1 mark for the perpendicular gradient, 1 mark for the equation of the line, 1 mark for the correct final form.

Method. Substitute the equation of the line into the circle equation. A tangent touches at one point only, so the resulting quadratic has exactly one solution (a repeated root).

1. Rearrange the line to make y the subject. We will substitute this into the circle equation.
   3x + 4y = 25 ⇒ y = −3x + 254
2. Substitute into the circle equation. Where the line meets the circle, both equations are true at once. Replace y in x² + y² = 25.
   x² + (−3x + 254)² = 25
3. Expand, clear the fractions and simplify to a quadratic equal to 0. Multiply every term by 16 to clear the fractions, then move everything to one side.
   x² − 6x + 9 = 0
4. Factorise. Look for a perfect square.
   (x − 3)² = 0
5. Count the solutions. A perfect square gives only one value of x, so the line meets the circle at exactly one point. That is what a tangent does.
   x = 3 only. (Discriminant b² − 4ac = 0.)

Check. The point of contact is (3, 4); it satisfies both equations. ✓

Watch out. If the discriminant were positive the line would cut the circle twice (a secant); if negative it would miss the circle.

Marks. 1 mark for rearranging to y = …, 1 mark for substituting, 1 mark for the quadratic, 1 mark for the repeated root and a conclusion.

Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.

Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.

1. Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at t = 4 and has the curve balanced evenly on both sides of that point; draw along the ruler.
2. Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.
   (1, −2) and (7, 10)
3. Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.
   run = 7 − 1 = 6
   rise = 10 − (−2) = 12
4. Work out the gradient. Gradient = rise ÷ run.
   gradient = 12 ÷ 6 = 2
5. Interpret the answer. The gradient of the tangent is the rate of change at that instant.
   The gradient of a distance–time graph is speed, so the speed after 4 seconds is 2 m/s.

Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 1.8 and 2.2). The exact value for this curve is 2. ✓

Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.

Marks. 1 mark for a tangent drawn at t = 4, 1 mark for a rise and run from the tangent, 1 mark for an answer between 1.6 and 2.4 m/s with units.

Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).

Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.

1. Split the area into strips of equal width. The region from time = 0 to 4 is 4 wide. With 4 strips, each strip has width 4 ÷ 4 = 1.
2. Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).

   time	0	1	2	3	4
   speed	16	15	12	7	0

3. Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.
   Strip 1: ½ × (16 + 15) × 1 = 15.5
   Strip 2: ½ × (15 + 12) × 1 = 13.5
   Strip 3: ½ × (12 + 7) × 1 = 9.5
   Strip 4: ½ × (7 + 0) × 1 = 3.5
4. Add the strip areas to get the estimate.
   15.5 + 13.5 + 9.5 + 3.5 = 42
5. Comment on the accuracy. The curve bends downwards (concave), so each trapezium top lies below the curve. So the estimate is an under-estimate. More strips would give a better estimate.
   (The exact area is 42.67 m, so the estimate is smaller.)

Check. Use the trapezium rule formula: ½ × 1 × [16 + 0 + 2 × (15 + 12 + 7)] = 42.00. ✓

Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.

Marks. 1 mark for the speeds 16, 15, 12, 7, 0, 1 mark for the trapezium areas, 1 mark for 42 m, 1 mark for under-estimate with the reason (the curve bends downwards).

Method. Use the multiplier repeatedly, building a table, until the amount reaches the target.

1. Find the multiplier. Each year the amount is multiplied by the same number.
   105% = 1.05
2. Work out the amount at the end of each year. Multiply by 1.05 each time. Stop as soon as the amount reaches or passes 4000.

   After n years	Amount
   0	2000
   1	2100
   2	2205
   12	3592
   13	3771
   14	3960
   15	4158

3. Read off the answer. The first row that reaches or passes 4000 is after 15 years.
   n = 15

Check. Use the formula: 2000 × 1.05¹⁵ = 4158 (above 4000) but after 14 years it is 3960 (not yet there). ✓

Watch out. Count the years carefully. The row marked 0 is the start, before any year has passed.

Marks. 1 mark for the multiplier 1.05, 1 mark for 2000 × 1.05¹⁴ < 4000, 1 mark for 2000 × 1.05¹⁵ > 4000, 1 mark for the conclusion.

Method. Find fg(x), invert it, then compose the two separate inverses in the reverse order.

1. Find fg(x). g acts first.
   fg(x) = (3x − 2) + 5 = 3x + 3
2. Invert it. Write y = 3x + 3, subtract 3, divide by 3.
   x = y − 33
3. Write (fg)⁻¹(x). Swap the letters.
   (fg)⁻¹(x) = x − 33
4. Find the separate inverses. Reverse each function.
   f⁻¹(x) = x − 5 and g⁻¹(x) = (x + 2) ÷ 3
5. Compose g⁻¹f⁻¹(x). Undo f first, then g.
   g⁻¹(x − 5) = ((x − 5) + 2) ÷ 3 = (x − 3) ÷ 3
6. Compare. Both are (x − 3)/3.
   Shown. ✓

Check. fg(2) = 3 × 2 + 3 = 9 and (fg)⁻¹(9) = (9 − 3) ÷ 3 = 2. ✓

Watch out. Marks: 1 for fg(x), 1 for the method of inverting, 1 for (x − 3)/3, 1 for each inverse (2 marks), total 5.