Graphs of Functions & Graphical Solution

`n`	Function	Name	Shape in one line
3	`y` = `ax`³	cubic	through the origin, flat in the middle, opposite behaviour at the two ends
2	`y` = `ax`²	quadratic	a parabola with its turning point at the origin ⓘ See ch 2
1	`y` = `ax`	linear	a straight line through the origin
0	`y` = `a`	constant	a horizontal straight line
−1	`y` = `a`/`x`	reciprocal	two separate branches that hug both axes
−2	`y` = `a`/`x`²	inverse square	two branches on the same side of the `x`-axis, mirror images in the `y`-axis

The sign of `a` decides the shape

For every power function y = axⁿ, the value of n chooses the family and the sign of a chooses which way up it sits.

The graph of y = axⁿ for a < 0 is the graph for the matching positive value of a reflected in the x-axis. Every one of these graphs passes through the origin except y = a (n = 0), which is a horizontal line, and the two reciprocal ones, which never reach it at all.

Fig. 4.1

The two shapes of y = ax³. Both pass through the origin and flatten there; the sign of a decides whether the curve rises or falls as x increases.

More detail

Those two choices are the only ones there are: n fixes the family and the sign of a fixes which way up it sits, and nothing else changes the shape. Changing a from 1 to 5 makes the curve steeper; changing it from 5 to −5 turns the whole picture upside down.

Past-paper questions

G3 N2016 · Paper 1 · Q11 3 marks

Express 9 − 8x + x² in the form p + (x + q)². [2]
Write down the coordinates of the minimum point of the graph of y = 9 − 8x + x². [1]

Worked solution

(a) Write the terms in the usual order first: 9 − 8x + x² = x² − 8x + 9. Half of −8 is −4, so x² − 8x = (x − 4)² − 16, and x² − 8x + 9 = (x − 4)² − 16 + 9 = −7 + (x − 4)², so p = −7 and q = −4. (b) A square is never negative, so y is smallest when (x − 4)² = 0, that is when x = 4, and then y = −7. Minimum point (4, −7). Why this works. Completing the square in (a) hands you (b): the bracket is the only place x appears, and the coefficient of x² is positive, so the curve is U-shaped and y is least when the bracket is zero. The mark is the pair (4, −7), read off −7 + (x − 4)². Keep the two signs apart: q = −4, turning point at x = +4.

G3 N2017 · Paper 1 · Q4 2 marks

Sketch the graph of y = −(x − 8)(x + 3) on the axes below. Indicate clearly the values where the graph crosses the x- and y- axes. [2]

Worked solution

Multiplying out would give −x² + …, so the coefficient of x² is negative and the curve is ∩-shaped. x-axis: y = 0 when −(x − 8)(x + 3) = 0, so x = 8 or x = −3. y-axis: put x = 0: y = −(0 − 8)(0 + 3) = −(−8)(3) = 24. The maximum is halfway between the two roots, at x = ½(−3 + 8) = 2.5, where y = −(2.5 − 8)(2.5 + 3) = 30.25 — not asked for here, but it is what keeps the sketch the right shape.

Why this works. The brackets give the roots, x = 8 and x = −3, and the leading minus gives the ∩ shape. The two marks are for that shape and for the three numbers on the axes, the y-intercept being −(−8)(3) = +24. A ∩ cutting the y-axis above the origin stays above the axis between its roots, which checks the sketch against its own numbers.

G3 N2019 · Paper 1 · Q15 3 marks

Sketch the graph of y = −(x − 2)² + 9 on the axes below. Indicate clearly the coordinates of the points where the graph crosses the axes and the maximum point on the curve. [3]

Worked solution

The equation is already in completed-square form, and the squared bracket carries a minus sign, so the curve is ∩-shaped with a maximum. Maximum: (x − 2)² is smallest, namely 0, when x = 2, and then y = 9. Maximum point (2, 9). y-axis: put x = 0: y = −(−2)² + 9 = −4 + 9 = 5, so (0, 5). x-axis: put y = 0: (x − 2)² = 9, so x − 2 = ±3, giving (−1, 0) and (5, 0).

Why this works. Completed-square form hands over the turning point with no working: (2, 9), a maximum because of the leading minus. The paper asks for coordinates, so each answer is a pair — (0, 5) on the y-axis, (−1, 0) and (5, 0) on the x-axis. Square before applying the minus, since −(−2)² = −4, and expect two roots, equally spaced about x = 2.

G3 N2020 · Paper 1 · Q10 3 marks

Express x² + 9x − 4 in the form (x + q)² + p. [2]
Write down the coordinates of the minimum point of the graph of y = x² + 9x − 4. [1]

Worked solution

(a) Half of 9 is

\frac{9}{2}

, so

x^{2} + 9 x = {(x + \frac{9}{2})}^{2} - \frac{81}{4}

.

Therefore $x^{2} + 9 x - 4 = {(x + \frac{9}{2})}^{2} - \frac{81}{4} - 4$ , and $\frac{81}{4} + 4 = \frac{97}{4}$ , so

$x^{2} + 9 x - 4 = {(x + \frac{9}{2})}^{2} - \frac{97}{4}$ (so q = 4.5 and p = −24.25).

(b) The bracket is zero when $x = - \frac{9}{2}$ , and then y is at its least value $- \frac{97}{4}$ .

Minimum point $(- \frac{9}{2}, - \frac{97}{4})$ , that is (−4.5, −24.25).

Why this works. Halving 9 gives a fraction, so keep $\frac{9}{2}$ and $\frac{81}{4}$ exact (4.5 and 20.25 as decimals work too) — nothing here asks for rounding. The mark in (b) is the pair (−4.5, −24.25), read off the completed square. Squaring back checks it: ${(x + 4.5)}^{2}$ is 24.25 more than the given expression, the amount part (a) subtracts.

G3 N2023 · Paper 1 · Q6 3 marks

The expression x² − 14x + b is equivalent to (x + a)² − 25.

Find the value of a and the value of b. [2]
The curve y = x² − 14x + b is drawn. Write down the equation of the line of symmetry of the curve. [1]

Worked solution

(a) Complete the square on the left-hand expression: half of −14 is −7, so

x² − 14x + b = (x − 7)² − 49 + b.

Comparing with (x + a)² − 25 gives a = −7 and −49 + b = −25, so b = 24.

(b) The curve is y = (x − 7)² − 25, whose turning point is at x = 7, so its line of symmetry is x = 7.

Why this works. "Is equivalent to" means the two expressions agree for every x, so completing the square on one side and comparing the pieces settles both letters; the x-terms give 2a = −14 in one line. Part (b) asks for a line, so the answer is an equation, x = 7. The signs differ: a = −7, line of symmetry x = +7.

G3 N2023 · Paper 1 · Q19 2 marks

Sketch the graph of y = −(x − 3)(x + 7) on the axes below. Indicate clearly the points where the graph crosses the axes. [2]

Worked solution

The leading coefficient is negative (expanding gives −x² − 4x + 21), so the curve is ∩-shaped.

x-axis: y = 0 gives (3, 0) and (−7, 0).

y-axis: x = 0 gives y = −(−3)(7) = 21, that is (0, 21).

The maximum lies halfway between the roots, at x = ½(−7 + 3) = −2, where y = 25; the y-intercept 21 must therefore sit a little below the top of the curve, which is a quick check that the sketch is consistent.

Why this works. Two marks, and they are the shape and the crossings: the leading minus makes the curve ∩-shaped, and the brackets give (3, 0), (−7, 0) and (0, 21) without any expanding. The peak lies halfway between the roots, at x = −2, so it sits to the left of the y-axis. Expanding in your head is the quick way to confirm the leading sign is negative.

Cubic functions

A cubic function is a function of the form y = ax³ + bx² + cx + d, where a, b, c and d are constants and a ≠ 0.

Fig. 4.2

The four possible shapes of y = ax³ + bx² + cx + d: a cubic has two turning points or none, never exactly one.

Where does a cubic cut the y-axis? At (0, d) — put x = 0 and every term with an x in it vanishes. So the graph passes through the origin exactly when d = 0.

Walkthrough 1 — table of values, then draw the curve Basic

The variables x and y are connected by y = x³ − 4. (a) Copy and complete the table of values. (b) Using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 5 units on the y-axis, draw the graph for −3 ≤ x ≤ 3. (c) Use your graph to find the value of y when x = 1.5, and the value of x when y = 10.

`x`	−3	−2	−1	0	1	2	3
`y`	−31		−5	−4	−3		23

When x = −2: y = (−2)³ − 4 = −8 − 4 = −12

Substitute into the equation you were given, and keep the negative number in brackets. (−2)³ is −8, not +8 — an odd power keeps the sign of the number.
When x = 2: y = 2³ − 4 = 8 − 4 = 4

The second missing entry, worked the same way. The table is not worth marks on its own, but one wrong value distorts the curve that is.
Draw axes to the given scales, plot the seven points, and join them with a single smooth curve.

Use the scales the question states, or the curve comes out distorted and every reading is wrong. Join the points freehand in one continuous sweep — no ruled segments and no hairy double lines.
From the graph, when x = 1.5, y = −0.6.

Read the value off the curve, not from the equation: this part is testing the graph. Draw a faint vertical line up from x = 1.5 and read across, to the nearest half a small square.
Fig. 4.3

The seven plotted points joined by one smooth curve; the readings are taken off the curve, not calculated.

From the graph, when y = 10, x = 2.4.

The second reading, taken the other way round: across from y = 10, then down. Give both answers to 1 decimal place, because half a small square is about all the accuracy a drawn graph carries.

Walkthrough 2 — match each graph to its equation Basic

Each of the four sketches below is the graph of one of these five functions: y = 3x², y = −3x², y = 2x³, y = −2x, y = 4. Match each graph to its function.

Fig. 4.4

Four sketches of power functions — one of the five listed equations is not used.

Sort the five functions by n: two with n = 2, one with n = 3, one with n = 1 and one with n = 0.

The power decides the family, so this one line already splits five candidates into four different shapes. Only the two quadratics share a shape, and they are told apart by the sign of a.
Graph A is a U-shaped parabola with its lowest point at the origin, so a > 0: Graph A is y = 3x².

A parabola means n = 2. It opens upwards, so the coefficient of x² is positive, which rules out y = −3x².
Graph B is a straight line through the origin sloping downwards, so n = 1 with a < 0: Graph B is y = −2x.

Straight and through the origin means y = ax. Sloping down from left to right means a is negative.
Graph C passes through the origin, is flat there, and goes down on the left and up on the right: Graph C is y = 2x³.

Opposite behaviour at the two ends is the signature of an odd power. A parabola would come back up on both sides; only the cubic dives on one side and climbs on the other.
Graph D is a horizontal line above the x-axis, so n = 0: Graph D is y = 4. The unused function is y = −3x².

A horizontal line is y = a, and this one lies above the axis, so a is positive. Name the leftover function too: five equations were offered for four graphs, and saying which is unused completes the matching.

Walkthrough 3 — draw the graph, hence solve an equation Intermediate

The variables x and y are connected by y = 5 − x³, and some corresponding values are given in the table. (a) Find the value of p. (b) Draw the graph for −3 ≤ x ≤ 3. (c) Explain from your graph why −x³ = 8 has only one solution. (d) On the same axes draw the line y = 3x + 1 for −2 ≤ x ≤ 2, write down the x-coordinate of the point where it meets the curve, and use it to find A and B in x³ + Ax + B = 0.

`x`	−3	−2	−1	0	1	2	3
`y`	`p`	13	6	5	4	−3	−22

When x = −3: y = 5 − (−3)³ = 5 + 27 = 32, so p = 32.

Subtracting a negative cube adds: −(−27) = +27. Keeping the bracket on (−3)³ is what makes that sign visible.
−x³ = 8 → 5 − x³ = 13, so draw the line y = 13.

Add 5 to both sides so the left-hand side becomes exactly the expression the curve already draws. Now the equation reads "curve = 13", and a horizontal line is all that is needed.
The line y = 13 meets the curve at one point only, at x = −2, so −x³ = 8 has one solution.

Solutions of the equation are the x-coordinates of the intersections, so counting the crossings counts the solutions. This curve never turns back, so a horizontal line can only ever cut it once.
For y = 3x + 1: (−2, −5), (0, 1) and (2, 7). Plot them and join with a ruled line.

Three points, not two: the third is a free check that the line is right. A straight line is the one graph on the page that should be drawn with a ruler.
From the graph, the line meets the curve at x = 1.

Read only the x-coordinate — that is what the question asked for, and it is the solution of the equation the two graphs represent together.
At the intersection, 3x + 1 = 5 − x³, so x³ + 3x − 4 = 0. Comparing with x³ + Ax + B = 0 gives A = 3 and B = −4.

Fig. 4.5

One smooth curve, a ruled line, and the two crossings marked: the line y = 13 cuts the curve once, at x = −2, and y = 3x + 1 cuts it once, at x = 1.

At a point of intersection both equations hold, so the two right-hand sides are equal. Rearrange to make the x³ term positive and everything equal to zero, then read off A and B by matching term with term.

Walkthrough 4 — a sum of two power terms, and a line you have to choose Advanced

(a) Draw the graph of y = x³ − 4x for −3 ≤ x ≤ 3. (b) Write down the x-coordinates of the three points where the curve crosses the x-axis. (c) By drawing a suitable straight line on the same axes, solve x³ − 6x + 2 = 0, giving your answers correct to 1 decimal place.

Table of values, at half-unit steps near the turning points: x = −3, −2, −1, 0, 1, 2, 3 give y = −15, 0, 3, 0, −3, 0, 15.

A sum of two power terms (n = 3 and n = 1) is still a cubic, so it still needs one smooth curve. The half-unit values pay off near the two turning points, which integer points alone leave too flat.
x³ − 4x = x(x² − 4) = x(x + 2)(x − 2), so the curve crosses the x-axis at x = −2, 0 and 2.

The crossings are where y = 0, and factorising gives them exactly — far better than reading them off. Three crossings is also a check on the plot: if your curve only cuts the axis twice, a table value is wrong.
x³ − 6x + 2 = 0 → x³ − 4x = 2x − 2, so the line to draw is y = 2x − 2.

Split the equation so that one side is exactly the expression already drawn. Move the leftover terms across: −6x becomes −4x on the left and 2x on the right, and +2 becomes −2. Whatever is left on the right is the line.
For y = 2x − 2: (−3, −8), (0, −2) and (3, 4). Draw it with a ruler across the full range.

Draw the line right across the given range, not just the middle: two of the three crossings are near the ends, and a short line would miss them entirely.
Fig. 4.6

The line y = 2x − 2 cuts the curve three times; the three x-coordinates are the three solutions.

From the graph, x = −2.6, x = 0.3 or x = 2.3 (1 d.p.).

Give all three roots — a cubic can have three, and the question said "solve", not "find a solution". Read each crossing to the nearest half a small square and round to the 1 decimal place the question specified.

Where this comes from

Adding the lower-power terms bends the middle of the curve, but it cannot change what happens at the two ends: that is fixed by ax³, the term that grows fastest. So a cubic graph has only two possible sets of ends, and in between it either wiggles through two turning points or climbs straight through with none — which is why the four pictures above are the whole story.

Check yourself

Describe the shape of the graph of y = ax³ when a > 0 and when a < 0, and write down the coordinates of the one point every such graph passes through.
Answer
For a > 0 the curve rises from the bottom left to the top right; for a < 0 it falls from the top left to the bottom right. Both flatten out as they pass through the origin, (0, 0).
How many turning points can the graph of a cubic function have?
Answer
Two, or none — never exactly one. With two it wiggles (up, down, up, or the mirror image); with none it climbs or falls all the way.
The curve y = x³ − 9x cuts the x-axis at three points. Find their coordinates.
Answer
Put y = 0: x(x² − 9) = 0, so x(x + 3)(x − 3) = 0.
The points are (−3, 0), (0, 0) and (3, 0).
A student has drawn the graph of y = x³ − 9x and now wants to solve x³ − 9x = 4. What line should they draw, and where on the diagram are the answers?
Answer
Draw the horizontal line y = 4. The solutions are the x-coordinates of the points where that line meets the curve. (Read each to 1 decimal place.)
Which one of y = 5x², y = 5x³, y = 5x and y = 5 has a graph that is a straight line not passing through the origin?
Answer
y = 5. It is the case n = 0, a horizontal line 5 units above the x-axis. y = 5x is also a straight line, but it does pass through the origin.
Write down the coordinates of the point where the curve y = 2x³ − x² + 7 cuts the y-axis.
Answer
Put x = 0: every term with an x disappears, leaving y = 7. The point is (0, 7) — the constant term d.

Past-paper questions

G3 N2017 · Paper 1 · Q21 6 marks

The graphs of y = 2x² − 3x − 7 and y = 5 − 2x are drawn on the grid.

Explain why the equation 2x² − 3x − 7 = k does not have solutions for some values of k. [1]
The points of intersection of the curve and the straight line give the solutions of a quadratic equation. Find the quadratic equation, giving your answer in the form ax² + bx + c = 0. [1]
The equation 2x² − 6x − 3 = 0 can be solved by drawing a suitable straight line on the grid.
1. Find the equation of the straight line. [1]
2. By drawing this straight line, solve the equation 2x² − 6x − 3 = 0. [3]

Worked solution

(a) y = k is a horizontal line, and the solutions of 2x² − 3x − 7 = k are where that line meets the curve.

The curve has a lowest point, at x = ¾, where y = 2(¾)² − 3(¾) − 7 = −8.125. For any k below that minimum value the line y = k lies entirely below the curve and never cuts it, so the equation has no solutions.

(b) At a point of intersection the two y-values are equal:

2x² − 3x − 7 = 5 − 2x ⇒ 2x² − x − 12 = 0

(c)(i) The curve's equation is fixed, so subtract the equation you want from it:

(2x² − 3x − 7) − (2x² − 6x − 3) = 3x − 4.

So where the curve meets y = 3x − 4, the difference between the two expressions is zero, which is exactly 2x² − 6x − 3 = 0.

(c)(ii) Draw y = 3x − 4 through (0, −4) and (3, 5). It cuts the curve twice, and reading the two x-coordinates gives

x = −0.45 or x = 3.4 (from the graph).

Exactly, $x = \frac{3 \pm \sqrt{15}}{2}$ = −0.4365… or 3.4365…, so the graph readings are right to the accuracy a drawn graph can give.

Why this works. Every part rests on one idea: a solution is a crossing point. In (a) a ∪-shaped curve has nothing below its minimum for y = k to meet; in (b) and (c) the line comes from subtracting the equation you want from the curve's expression. In (c)(ii) one of the three marks is for the drawn line itself, so run it right across the grid.

G3 N2019 · Paper 2 · Q4 6 marks

The variables x and y are connected by the equation $y = \frac{x^{3}}{5} - 2 x + 1$ . Some corresponding values of x and y are given in the table below.

`x`	−4	−3	−2	−1	0	1	2	3	4
`y`	`p`	1.6	3.4	2.8	1	−0.8	−1.4	0.4	5.8

Find the value of p. [1]
Use your graph to write down an inequality in x to describe the range of values where y > 4. [1]
1. Show that the points of intersection of the line [5y + x = 10] and the curve give the solutions of the equation x³ − 9x − 5 = 0. [2]
2. Use your graphs to solve the equation x³ − 9x − 5 = 0. [2]

Worked solution

(a) Put x = −4: $y = \frac{{(−4)}^{3}}{5} - 2 (−4) + 1 = −12.8 + 8 + 1$

p = −3.8

(c) Draw the horizontal line y = 4 across the grid. The curve's local maximum is only about 3.4, so the line is above the whole left-hand hump; the curve rises past y = 4 only at the far right.

Reading the one crossing gives 3.75 < x ≤ 4. (The crossing is exactly the root of x³ − 10x − 15 = 0, which is x = 3.7427…, so 3.74 or 3.75 both match the drawn graph; the upper limit 4 is the end of the range the graph is drawn over.)

(d)(ii) Rearranging the line, $y = \frac{10 - x}{5}$ . At a point of intersection this equals the curve's y:

$\frac{x^{3}}{5} - 2 x + 1 = \frac{10 - x}{5}$

Multiplying every term by 5: x³ − 10x + 5 = 10 − x, so

x³ − 9x − 5 = 0 (shown).

(d)(iii) The line cuts the curve three times, and the three x-coordinates are the three solutions:

x = −2.65, x = −0.6 or x = 3.25 (from the graphs).

The exact roots are −2.6697, −0.5769 and 3.2466 (to 4 d.p.), so each graph reading is correct to the half-square a graph can be read to.

Why this works. Part (d)(ii) is a "show that", so the marks are for the working: substitute the line, multiply every term by 5 to clear the fifths, and arrive at the printed equation. Part (c) wants an inequality, and since the left-hand hump peaks below 4 only one stretch of the curve rises above y = 4. Counting crossings against the picture checks both.

G3 N2022 · Paper 2 · Q4 7 marks

Complete the table of values for $y = \frac{x^{3}}{5} - 2 x + 1$ . [1]

x −4 −3 −2 −1 0 1 2 3 4

y 1.6 3.4 2.8 1 −0.8 −1.4 0.4 5.8
The equation $\frac{x^{3}}{5} - 2 x + 1 = k$ has two solutions. Use your graph to find the two possible values of k. [2]
By drawing a suitable straight line on the grid, solve the equation 2x³ − 25x + 20 = 0. [4]

Worked solution

(a) At x = −4: $y = \frac{−64}{5} + 8 + 1 = −12.8 + 9$ , so the missing value is −3.8.

(c) y = k is a horizontal line. A cubic with a maximum and a minimum is cut by such a line three times between the two turning values and once outside them; the only way to get exactly two solutions is for the line to pass through a turning point, where two of the crossings merge into one.

From the graph the maximum is about 3.4 (near x = −1.8) and the minimum about −1.4 (near x = 1.8), so

k = 3.4 or k = −1.4.

(Exactly, the turning points are at $x = \pm \sqrt{\frac{10}{3}}$ = ±1.8257…, where y = 3.4343… and −1.4343…)

(d) The grid carries $y = \frac{x^{3}}{5} - 2 x + 1$ , so divide the target equation by 10 to get the same $\frac{x^{3}}{5}$ term:

$\frac{x^{3}}{5} - \frac{5}{2} x + 2 = 0$

Subtracting this from the curve's expression leaves $\frac{1}{2} x - 1$ , so the line to draw is $y = \frac{1}{2} x - 1$ — through (−4, −3), (0, −1) and (4, 1).

It cuts the curve three times, giving

x = −3.9, x = 0.85 or x = 3 (from the graph; the exact roots are −3.8827, 0.8489 and 3.0338).

Why this works. Part (c) is really about turning points: slide y = k down the picture and the crossings go 1 → 3, hitting exactly 2 as the line passes a turning point. Read those turning values off the graph: k = 3.4 or −1.4. In (d), divide by 10 so the cubic terms match — dividing by 2 leaves an x³ in the line.

G3 N2024 · Paper 2 · Q4(a) 5 marks

Complete the table of values for $y = 2 x^{2} - \frac{x^{3}}{2} - 4$ . [1]

x −2 −1 0 1 2 3 4

y −1.5 −4 −2.5 0 0.5 −4
Use your graph to explain why the equation $2 x^{2} - \frac{x^{3}}{2} - 4 = x$ has only one solution. [1]
By drawing a suitable straight line on the grid, solve the equation 4x² − x³ − 4 = 0. [3]

`x`	−2	−1	0	1	2	3	4
`y`		−1.5	−4	−2.5	0	0.5	−4

Worked solution

(i) At x = −2: $y = 2 (4) - \frac{−8}{2} - 4 = 8 + 4 - 4$ , so the missing value is 8.

(iii) Draw the line y = x on the same grid. It meets the curve at exactly one point (a little to the left of x = −1), and each meeting point is a solution of $2 x^{2} - \frac{x^{3}}{2} - 4 = x$ , so the equation has only one solution.

(Exactly, that root is x = −1.0739…, and x³ − 4x² + 2x + 8 = 0 has no other real root.)

(iv) Halve the target equation so that the x² and x³ terms match the curve:

$2 x^{2} - \frac{x^{3}}{2} - 2 = 0$

Now subtract 2 from both sides so the left-hand side is the curve's own expression:

$2 x^{2} - \frac{x^{3}}{2} - 4 = −2$

So draw the horizontal line y = −2. It cuts the curve three times, giving

x = −0.9, x = 1.15 or x = 3.7 (from the graph). The exact roots of x³ − 4x² + 4 = 0 are −0.9032, 1.1939 and 3.7093, so a reading anywhere within about half a small square of each is credited.

Why this works. Both graphical parts use one move: get the curve's exact expression onto one side, and whatever is left is the line to draw. In (iv) the leftover is a constant, so the line is horizontal — the easiest kind to draw accurately. In (iii) the word "explain" puts the mark on the reason: name the line you drew and say it cuts the curve once.

Graphs of `y` = `a`/`x`

A reciprocal function is what you get when two quantities multiply to a constant: a rectangle of fixed area, a journey of fixed distance, a gas at fixed temperature. In every case one variable is the constant divided by the other.

A reciprocal function is a function of the form

y = \frac{a}{x}

where a is a constant and x ≠ 0.

a > 0 → the branches lie in the 1st and 3rd quadrants, because x and y then always have the same sign.
a < 0 → the branches lie in the 2nd and 4th quadrants, because x and y then always have opposite signs.

A line that a curve approaches as closely as you like, without ever touching it, is called an asymptote. For y = a/x the asymptotes are the two axes: x = 0 and y = 0.

Fig. 4.7

The two shapes of y = a/x. The dashed lines are the two asymptotes, x = 0 and y = 0 — the branches get arbitrarily close to them but never touch.

Along a branch, as x increases, y decreases. As x approaches zero, y becomes very large; as x becomes very large, y approaches zero. So the curve gets very close to the x-axis and the y-axis without touching them.

Walkthrough 5 — draw y = a/x and read off values Basic

Using a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to represent 4 units on the y-axis, draw the graph of y = 8/x for −8 ≤ x ≤ 8, x ≠ 0. Use your graph to find the value of y when x = 3, and the value of x when y = −5.

x −8 −4 −2 −1 −0.5 0.5 1 2 4 8

y −1 −2 −4 −8 −16 16 8 4 2 1

Choose the x-values yourself, and include some close to zero — that is where the curve does everything interesting. Whole-number values alone would leave the two steep parts of the graph unplotted. Note there is no entry for x = 0: 8/0 is undefined.
Plot the five points with x > 0 and join them with one smooth curve; do the same, separately, for the five points with x < 0.

Two branches means two curves. Joining them across the y-axis would draw a line through x = 0, where the function has no value at all.
From the graph, when x = 3, y = 2.7.

Read off the curve, to the nearest half a small square. (By calculation 8 ÷ 3 = 2.666…, so 2.7 is comfortably inside the tolerance a graphical answer is allowed.)
Fig. 4.8

Two separate branches, each drawn as one smooth curve, with a clean break at x = 0.

From the graph, when y = −5, x = −1.6.

A negative y puts the reading on the third-quadrant branch, so the answer is negative. A quick sign check confirms the right branch was used.

`x`	−8	−4	−2	−1	−0.5	0.5	1	2	4	8
`y`	−1	−2	−4	−8	−16	16	8	4	2	1

Walkthrough 6 — find the asymptotes and explain them Basic

A curve has equation y = 12/x². (a) Explain why the curve meets neither the x-axis nor the y-axis. (b) Write down the equations of its two asymptotes and of its line of symmetry.

When x = 0, 12/x² is undefined, so there is no point on the curve with x = 0 and the curve never meets the y-axis.

Meeting the y-axis would mean having a point with x = 0, and division by zero has no value. The break in the graph is exactly this fact, drawn.
y = 0 would need 12/x² = 0, i.e. 12 = 0, which is impossible — so the curve never meets the x-axis either.

A fraction is zero only when its numerator is zero, and here the numerator is the fixed number 12. Saying "the curve gets close to the axis" is not the same as proving it never reaches it; this line does prove it.
As x becomes very large, y approaches 0: when x = 1000, y = 12/1000² = 0.000 012.

One worked value makes the claim concrete: the curve gets arbitrarily close to the x-axis while, by the line above, never actually reaching it.
Asymptotes: x = 0 (the y-axis) and y = 0 (the x-axis). Line of symmetry: x = 0.

An asymptote is a line, so it is quoted as an equation. The line of symmetry is the y-axis because (−x)² = x², so x and −x give the same y.

Where this comes from

The condition x ≠ 0 is not a technicality — it is the whole shape of the graph. Dividing by zero is undefined, so there is no point on the curve at x = 0: the graph comes in two separate branches with a break down the middle.

Follow the first-quadrant branch from left to right and the three statements in the box above are what you read off it. The numbers make the two limits concrete: for y = 1/x, when x = 0.000 001 then y = 1 000 000, and when x = 1 000 000 then y = 0.000 001. The graph also has rotational symmetry of order 2 about the origin.

Past-paper questions

G3 N2020 · Paper 2 · Q3 8 marks

Complete the table of values for $y = 2 x + \frac{6}{x} - 9$ . Give your answer correct to 1 decimal place. [1]

x 0.5 1 1.5 2 3 4 5 6 7

y 4 −1 −2 −2 −1 0.5 2.2 4
Use your graph to find the solutions of the equation $2 x + \frac{6}{x} = 10$ in the range 0.5 ≤ x ≤ 7. [2]
1. Write down the x-coordinates of the points where this line [3y = 2x − 3] intersects the curve. [2]
2. These values of x are the solutions of the equation 2x² + Ax + B = 0. Find the value of A and the value of B. [3]

`x`	0.5	1	1.5	2	3	4	5	6	7
`y`	4	−1	−2	−2	−1	0.5	2.2	4

Worked solution

(a) At x = 7: $y = 14 + \frac{6}{7} - 9 = 5.857…$ = 5.9 (1 d.p.).

(c) The curve drawn is $y = 2 x + \frac{6}{x} - 9$ , so subtract 9 from both sides of the equation to be solved:

$2 x + \frac{6}{x} - 9 = 10 - 9 = 1$

Draw y = 1 and read the two crossings: x = 0.7 or x = 4.3.

(Exactly, 2x² − 10x + 6 = 0 gives $x = \frac{5 \pm \sqrt{13}}{2}$ = 0.6972… or 4.3028…)

(d)(ii) The line $y = \frac{2 x - 3}{3}$ passes through (0, −1) and (6, 3). It cuts the curve twice, at

x = 0.9 and x = 5.1 (from the graph).

(d)(iii) At an intersection the two y-values are equal:

$2 x + \frac{6}{x} - 9 = \frac{2 x - 3}{3}$

Multiplying every term by 3x (allowed, since x ≠ 0):

6x² + 18 − 27x = 2x² − 3x ⇒ 4x² − 24x + 18 = 0 ⇒ 2x² − 12x + 9 = 0

A = −12 and B = 9.

Check: those roots are $x = \frac{6 \pm 3 \sqrt{2}}{2}$ = 0.8787… and 5.1213…, which agree with the readings in (d)(ii).

Why this works. One habit runs through the question: make the left-hand side the curve already drawn, and the right-hand side is the line. In (c) that is one subtraction and a horizontal line. In (d)(iii) both graphs are known, so equate them and multiply every term by 3x — including the 9, which supplies the −27x — then divide by 2 to match the printed form.

G3 N2021 · Paper 2 · Q3(c)–(f) 3 marks

Parts (a) and (b) of the paper's question set up a shape whose shaded area is y cm², and part (d) asks for the curve to be drawn on the printed grid; none of the three can be reproduced here, so the paper's own letters (c), (e) and (f) are kept for the three parts that follow.

Complete the table of values for $y = \frac{320}{x} + 4 x - 16$ . Values are given to 1 decimal place where appropriate. [1]

x 4 6 8 10 12 14 16 18 20

y 80 61.3 56 56 58.7 62.9 68 80
Use your graph to find the greatest value of x when the shaded area is equal to 60 cm². [1]
Explain how the graph shows that the shaded area cannot be equal to 50 cm². [1]

`x`	4	6	8	10	12	14	16	18	20
`y`	80	61.3	56	56	58.7	62.9	68		80

Worked solution

(c) At x = 18: $y = \frac{320}{18} + 72 - 16 = 17.77… + 56$ = 73.8 (1 d.p.).

(e) Draw the horizontal line y = 60. It cuts the curve twice, because the curve falls to a minimum and then rises again; the question asks for the greater of the two x-values.

From the graph, x = 12.7 (the other crossing is at about x = 6.3).

Exactly: $\frac{320}{x} + 4 x - 16 = 60$ gives x² − 19x + 80 = 0, so $x = \frac{19 \pm \sqrt{41}}{2}$ = 6.2984… or 12.7016…

(f) The lowest point of the curve is at about y = 55.5 (near x = 9), so the whole curve lies above the horizontal line y = 50.

The line y = 50 never meets the curve, so there is no value of x for which the shaded area is 50 cm².

(The true minimum is at $x = \sqrt{80}$ = 8.944…, where y = 55.554…)

Why this works. A $\frac{320}{x}$ term plus a 4x term gives the minimum-area shape: one part shrinks as x grows, the other grows, so the curve dips to a lowest point. In (e) greatest signals that the horizontal line cuts twice, so give the larger crossing. In (f) the mark is for reading the graph: the minimum, 55.5, is above 50, so y = 50 misses it.

Graphs of `y` = `a`/`x`²

Squaring the denominator changes one thing, and that one thing changes the picture completely: x² is never negative, so y keeps the sign of a whatever x does.

a > 0 → every value of y is positive, so the whole graph lies above the x-axis (branches in the 1st and 2nd quadrants).
a < 0 → every value of y is negative, so the whole graph lies below the x-axis (branches in the 3rd and 4th quadrants).
Either way the graph is symmetrical about the y-axis.

Fig. 4.9

The two shapes of y = a/x². Both branches sit on the same side of the x-axis and are mirror images of each other in the y-axis — the difference from y = a/x, whose branches sit on opposite sides.

Walkthrough 7 — draw y = a/x² Intermediate

Using 1 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of y = 9/x² for −6 ≤ x ≤ 6, x ≠ 0. Use your graph to find the value of y when x = 2.5, and the values of x when y = 6.

x ±1 ±1.5 ±2 ±3 ±4 ±6

y 9 4 2.25 1 0.56 0.25

One row of working does both branches, because x and −x give the same y. Values are given to 2 decimal places so that they can be plotted to the nearest half square on this scale.
Plot the six points on the right and join them smoothly; repeat on the left. Both branches lie above the x-axis.

Because 9 > 0 and x² > 0, every y is positive — so a branch dipping below the axis is a plotting error, not a feature.
From the graph, when x = 2.5, y = 1.4.

Read up from x = 2.5 and across. (By calculation 9 ÷ 6.25 = 1.44, so the graphical reading is well within tolerance.)
Fig. 4.10

Both branches above the x-axis and symmetrical about the y-axis.

From the graph, when y = 6, x = −1.2 or x = 1.2.

A horizontal line at y = 6 crosses both branches, so there are two answers, equal and opposite. Give both: the symmetry of the graph is the point of the question.

`x`	±1	±1.5	±2	±3	±4	±6
`y`	9	4	2.25	1	0.56	0.25

Walkthrough 8 — a sum of two power terms, and its minimum Advanced

(a) Draw the graph of y = x + 4/x for 0.5 ≤ x ≤ 6. (b) State the minimum value of y and the value of x at which it occurs. (c) By drawing a suitable straight line, solve x + 4/x = 5.

x 0.5 1 1.5 2 3 4 5 6

y 8.5 5 4.2 4 4.3 5 5.8 6.7

Add the two parts: with n = 1 and n = −1 this is a sum of power functions. At x = 1.5, 1.5 + 2.667 = 4.167, and the values fall then rise, so a minimum lies between x = 1 and 3.
Plot the eight points and join them with one smooth curve: no break is needed, because x = 0 is outside the given range.

The function is still undefined at x = 0, but the question starts at x = 0.5, so only the first-quadrant branch is drawn. Read the stated range before deciding how many pieces to draw.
From the graph, the minimum value of y is 4, and it occurs at x = 2.

The lowest point of the curve is where it stops falling and starts rising. Give both numbers, and say which is which: the minimum value is the y-coordinate; the x-coordinate is where it happens.
The left-hand side is already the curve, so the line to draw is y = 5.

When the equation to solve is "curve = number", the line is simply that horizontal number line. No rearranging is needed here.
Fig. 4.11

The dashed line y = 5 cuts the curve twice; the two x-coordinates are the two solutions.

From the graph, x = 1 or x = 4.

Two crossings, so two solutions — and both are needed. (Checking algebraically: x + 4/x = 5 gives x² − 5x + 4 = 0, i.e. (x − 1)(x − 4) = 0, so the readings are exact.)

`x`	0.5	1	1.5	2	3	4	5	6
`y`	8.5	5	4.2	4	4.3	5	5.8	6.7

More detail

This is the inverse square law. The brightness of a light at distance d, and the gravitational pull between two masses at distance r, both follow y = a/x² — so a question about light intensity or about gravity is a question about this graph.

Check yourself

For y = 12/x, find y when x = 8, and find x when y = −3.
Answer
y = 12 ÷ 8 = 1.5.
−3 = 12/x gives −3x = 12, so x = −4.
Write down the equations of the two asymptotes of y = 7/x, and state which quadrants its two branches lie in.
Answer
Asymptotes x = 0 and y = 0. Since 7 > 0, x and y always have the same sign, so the branches lie in the 1st and 3rd quadrants.
Explain why the graph of y = −5/x² lies entirely below the x-axis.
Answer
For every x ≠ 0, x² > 0, so 5/x² > 0 and therefore −5/x² < 0. Every value of y is negative, so no part of the curve can reach or cross the axis.
The graph of y = a/x passes through (4, −3). Find a, and state which quadrants the branches lie in.
Answer
−3 = a ÷ 4, so a = −12. As a < 0, x and y have opposite signs, so the branches lie in the 2nd and 4th quadrants.
The table shows some values of y = x + 9/x. Copy and complete it, then write down the least value of y for 1 ≤ x ≤ 6.

x 1 2 3 4 6

y 10 6 7.5

Answer
At x = 2: 2 + 4.5 = 6.5. At x = 4: 4 + 2.25 = 6.25.
The values fall to 6 and then rise, so the least value of y is 6, at x = 3.
Why can a table of values for y = 6/x never include x = 0, and what does that do to the graph?
Answer
6 ÷ 0 is undefined, so there is no y-value to write. On the graph it produces a break at x = 0: the curve is drawn as two separate branches which must never be joined across the y-axis.

`x`	1	2	3	4	6
`y`	10		6		7.5

Past-paper questions

G3 N2020 · Paper 1 · Q2 1 mark

y is inversely proportional to x². Which of these diagrams represents the graph of y against x? [1]

Worked solution

"y is inversely proportional to x²" means $y = \frac{k}{x^{2}}$ for some constant k, so the graph is the reciprocal-squared shape:

as x increases, y decreases; as x gets close to zero, y becomes very large; and as x becomes very large, y approaches zero without ever reaching it. The curve therefore hugs both axes.

Diagram 1 is a straight line with a negative gradient (y = c − mx); Diagram 2 and Diagram 4 both increase with x, so neither can show an inverse relationship (Diagram 4 is direct proportion, y = kx).

Diagram 3

Why this works. The mark rests on knowing the shape: near x = 0 the denominator is tiny, so y shoots up, and for large x the curve flattens onto the x-axis. Those two extremes rule out three diagrams. Diagram 1 decreases, but a straight line reaching the x-axis would need y = 0 at a finite x, and $\frac{k}{x^{2}}$ is never zero.

G3 N2023 · Paper 2 · Q4 6 marks

Complete the table of values for $y = \frac{2}{x^{2}} + 3 x - 1$ . Values are given to one decimal place where appropriate. [1]

x −3 −2 −1 −0.5 0.5 1 1.5 2 3

y −6.5 −2 5.5 8.5 4 4.4 5.5 8.2
1. Write down the x-coordinates of the points where the line [2y − 5x = 7] intersects the curve. [2]
2. These values of x are solutions of the equation x³ + Ax² + B = 0. Find the value of A and the value of B. [3]

`x`	−3	−2	−1	−0.5	0.5	1	1.5	2	3
`y`		−6.5	−2	5.5	8.5	4	4.4	5.5	8.2

Worked solution

(a) At x = −3: $y = \frac{2}{9} - 9 - 1 = −9.777…$ = −9.8 (1 d.p.).

(Note (−3)² = 9, so the first term is positive even though x is negative.)

(c)(ii) The line is $y = \frac{5 x + 7}{2}$ , through (−3, −4), (−1, 1) and (3, 11). It cuts the left branch once and the right branch once:

x = −0.65 and x = 0.7 (from the graph; exactly −0.6440 and 0.6940).

(c)(iii) At an intersection the two y-values are equal:

$\frac{2}{x^{2}} + 3 x - 1 = \frac{5 x + 7}{2}$

Multiply every term by 2x² (allowed, since x ≠ 0):

4 + 6x³ − 2x² = 5x³ + 7x²

x³ − 9x² + 4 = 0, so A = −9 and B = 4.

Check: the real roots of that cubic are −0.6440, 0.6940 and 8.9501. Only the first two lie in the range −3 ≤ x ≤ 3 that the graph covers, which is exactly the two crossings found in (c)(ii).

Why this works. The $\frac{2}{x^{2}}$ term makes this curve unusual: positive on both sides of the y-axis and shooting upwards as x approaches 0, so the graph breaks into two branches at x = 0. In (c)(iii) multiply by 2x², not x², so the 2 in the line's denominator clears too. No x term appears in the printed form, which checks the rearrangement.

Graphs of `y` = `ka`^x

Bacteria multiplying, a rumour spreading, a video going viral: all start small and then climb astonishingly fast. What they share is a multiplier applied over and over, which puts the variable in the index.

An exponential function is a function of the form y = ka^x, where k is a constant and a is a positive integer with a > 1.

Fig. 4.12

The two shapes of y = ka^x. The curve cuts the y-axis at (0, k) and flattens towards the dashed asymptote y = 0 on the left, without ever reaching it.

Three facts. The curve cuts the y-axis at (0, k), since a⁰ = 1. a^x is always positive, so the whole curve keeps the sign of k. And y = 0 is a horizontal asymptote, never touched or crossed.

Walkthrough 9 — draw y = a^x and solve from it Basic

Using a scale of 4 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of y = 3^x for −2 ≤ x ≤ 2. Use your graph to find (a) the value of y when x = 1.2, and (b) the solution of 3^x = 6.

x −2 −1.5 −1 −0.5 0 0.5 1 1.5 2

y 0.11 0.19 0.33 0.58 1 1.73 3 5.20 9

Half-unit steps, because the curve turns upwards quickly and whole-number values alone would leave the steep part unplotted. Negative indices give small positive values — 3⁻² = 1/9 = 0.11, never a negative answer.
Plot the nine points and join them with one smooth curve, cutting the y-axis at (0, 1).

The y-intercept is a free accuracy check: every graph y = a^x passes through (0, 1), because a⁰ = 1. If your curve misses it, a plotted point is wrong.
From the graph, when x = 1.2, y = 3.7.

Read up from x = 1.2 and across, to the nearest half a small square. On the steep part of an exponential curve a small slip along the x-axis costs a lot of y, so line the ruler up carefully.
Fig. 4.13

The dashed line y = 6 meets the curve once, just past x = 1.5.

Draw y = 6. It meets the curve once, so x = 1.6.

Solving 3^x = 6 means finding where the curve reaches height 6, so a horizontal line at y = 6 is all that is needed. There is one answer, because a rising curve passes any given height once.

`x`	−2	−1.5	−1	−0.5	0	0.5	1	1.5	2
`y`	0.11	0.19	0.33	0.58	1	1.73	3	5.20	9

Walkthrough 10 — find k and a from two points Intermediate

The graph of y = ka^x passes through the points (0, 5) and (3, 40). (a) Find the value of k and of a. (b) Hence find the value of y when x = 5, and write down the equation of the asymptote.

Substituting (0, 5): 5 = ka⁰ = k × 1, so k = 5.

Start with the point on the y-axis. Any number to the power 0 is 1, so that substitution removes a and hands you k in one line.
Substituting (3, 40) with k = 5: 40 = 5a³, so a³ = 8.

With k already known there is only one unknown left, so the second point gives a single equation in a. Divide by the 5 before doing anything with the index.
a = ∛8 = 2, so the function is y = 5(2^x).

Take the cube root of both sides. The answer must be a positive integer greater than 1 — that is the syllabus condition on a — and 2 is, so the working is consistent.
When x = 5: y = 5 × 2⁵ = 5 × 32 = 160. The asymptote is y = 0.

Now that k and a are known, substitute rather than read off a graph — the answer is exact. The asymptote of every y = ka^x is the x-axis, because a^x shrinks towards zero but never reaches it.

Walkthrough 11 — exponential growth in context Intermediate

A colony of bacteria triples in number every hour. At the start of an experiment there are 40 bacteria. The number of bacteria after t hours is N. (a) Write down a formula for N in terms of t, and complete the table for 0 ≤ t ≤ 3. (b) Draw the graph. (c) Use your graph to estimate the time taken for the colony to reach 800.

`t` (hours)	0	1	2	3
`N`	40		360

N = 40(3^t)

"Triples every hour" is a multiplier of 3 applied t times, so the base is 3 and the index is t. The starting number 40 is the value when t = 0, so it is the k in y = ka^x.
When t = 1, N = 40 × 3 = 120; when t = 3, N = 40 × 27 = 1080.

Fill the gaps straight from the formula. The given value at t = 2 (360 = 40 × 9) confirms the formula is the right one before any plotting is done.
Choose 4 cm to represent 1 hour on the horizontal axis and 1 cm to represent 100 bacteria on the vertical axis; plot the four points and join them with a smooth curve.

The vertical scale has to reach 1080, so a scale in hundreds is the only workable choice; a scale of 1 cm to 10 would need a metre of paper. Choosing a sensible scale is part of the mark for drawing.
Fig. 4.14

The dashed line N = 800 meets the curve a little before t = 2.75.

Draw N = 800. From the graph, t = 2.7 hours.

Draw the horizontal line at the number the question gives, and read the t-coordinate where it crosses. Answer in the units of the question — hours — because a bare "2.7" does not answer "how long".

Where this comes from

Suppose one person tells five others in the first hour, and each of those tells five more in the next hour:

Hour	0	1	2	3	4	5	6
New people who hear it	1	5	25	125	625	3125	15 625

Each entry is the one before it multiplied by 5, so after N hours the count is 5^N. The variable is in the index, not the base — that is what makes it an exponential function rather than a power function.

Exponential growth outruns every power function eventually. Between x = 0 and x = 2 the graph of y = 3^x looks gentle; by x = 10 it has reached 59 049. That is why a table of values for an exponential rarely needs more than a small range of x — and why the y-axis scale is the thing to think hardest about.

Check yourself

Write down the coordinates of the point where the graph of y = 4^x cuts the y-axis, and find y when x = 2.5.
Answer
4⁰ = 1, so the graph cuts the y-axis at (0, 1).
4^2.5 = 32.
The graph of y = ka^x passes through (0, 7) and (2, 63). Find k and a.
Answer
From (0, 7): k × 1 = 7, so k = 7.
From (2, 63): 7a² = 63, so a² = 9 and a = 3 (taking the positive integer root).
State the equation of the asymptote of y = 5(2^x) and explain why the curve never meets it.
Answer
The asymptote is y = 0, the x-axis. 2^x is positive for every x, so 5(2^x) is always positive: it can be made as small as you like but never becomes 0.
A population of 60 insects doubles every day. Write a formula for the number N after t days, and find N when t = 5.
Answer
N = 60(2^t).
When t = 5, N = 60 × 32 = 1920.
Explain why y = 1^x is not an exponential graph of the kind described in the syllabus.
Answer
The syllabus requires a to be a positive integer with a > 1. Here a = 1, and 1^x = 1 for every x, so the "curve" is just the horizontal line y = 1 — it does not grow at all.

Past-paper questions

G3 N2016 · Paper 1 · Q17 4 marks

The sketch shows the graph of y = ka^−x. The points A(−2, 100) and B(0, 4) lie on the graph.

Find the values of k and a. [2]
A straight line is drawn from A to B. Find the equation of the line AB. [2]

Worked solution

(a) Use B(0, 4) first, because a zero index kills the power:

4 = ka⁰ = k × 1, so k = 4.

Now use A(−2, 100): 100 = 4a⁻⁽⁻²⁾ = 4a², so a² = 25 and a = ±5.

A base must be positive, so −5 is rejected: a = 5.

(b) Gradient of AB = $\frac{4 - 100}{0 - (−2)} = \frac{−96}{2} = −48$

B is on the y-axis, so the intercept is 4 and

y = −48x + 4

Why this works. Two points, two unknowns, and the order decides the algebra: take the point on the y-axis first and k falls out in one line, since a⁰ = 1. Watch the double negative: a^−x at x = −2 is a⁺², not $\frac{1}{a^{2}}$ . Part (b) is a straight line through two known points, with gradient −48.

G3 N2024 · Paper 2 · Q4(b) 3 marks

The sketch shows the graph of y = ka^x, where a > 0. The graph passes through the points A (0, 5), B (3, 135) and C (4, p).

Find the value of p. [3]

Worked solution

Substitute A(0, 5): 5 = ka⁰ = k, so k = 5.

Substitute B(3, 135) into y = 5a^x:

135 = 5a³ ⇒ a³ = 27 ⇒ a = 3

Substitute C(4, p) into y = 5(3)^x:

p = 5 × 3⁴ = 5 × 81 = 405

Why this works. Three marks, three substitutions, in that order: each one supplies the constant the next one needs. The middle step carries them — undo a³ = 27 with a cube root, which gives exactly 3. Check the answer by the shape: one step of 1 in x multiplies y by a = 3, so C is three times B, and 3 × 135 = 405.

The tangent, and the gradient at a point

A straight line has one gradient everywhere. A curve does not, so "the gradient of the curve" only makes sense at a stated point.

When a straight line touches a curve at a single point A, the line is called the tangent to the curve at A.

The gradient of a curve at a point is defined as the gradient of the tangent to the curve at that point.

Fig. 4.15

The tangent touches the curve at P and nowhere near it. Two points on the tangent give a right-angled triangle, and the gradient of the curve at P is the vertical change divided by the horizontal change.

gradient of the curve at P = \frac{vertical change}{horizontal change}

Drawing the tangent. Put a ruler against the curve at the point and turn it until it just touches, with the curve falling away equally on both sides. Draw the line long: a short stub gives a tiny triangle and a rough reading.

Reading the gradient off it. Take two points on the tangent, as far apart as the page allows and with coordinates that are easy to read, then divide the vertical change by the horizontal change.

Both points come off the tangent, not the curve. The only point the two share is the point of contact; anywhere else the curve has already bent away, so a point taken from the curve gives a different gradient.

At a maximum or minimum the tangent is horizontal, so the gradient is 0. It works backwards too: a point where the gradient is zero is a turning point, so draw a horizontal tangent and read (h, k) off the graph.

Walkthrough 12 — find a gradient from a tangent already drawn Basic

The diagram shows part of a curve. The straight line drawn on it is the tangent to the curve at the point P(2, 3). Find the gradient of the curve at P.

Fig. 4.16

The tangent at P, with two of its points marked.

Two points on the tangent whose coordinates can be read exactly: (0, −1) and (5, 9).

Choose points where the tangent passes through a grid intersection, and choose them far apart — a long triangle makes a small reading error much less damaging than a short one does.
Vertical change = 9 − (−1) = 10; horizontal change = 5 − 0 = 5.

Subtract in the same order for both: second point minus first point, every time. Reversing one of them flips the sign of the gradient.
$gradient = \frac{10}{5} = 2$ The gradient of the curve at P is 2.

The tangent has gradient 2 and touches the curve at P, so the curve has gradient 2 there too — that is the definition. Give a number and say it is the gradient at P; elsewhere the curve has a different gradient.

Walkthrough 13 — draw the tangent yourself, and find the turning point Intermediate

The variables x and y are connected by y = 6x − x². (a) Draw the graph for 0 ≤ x ≤ 6. (b) By drawing a tangent, find the gradient of the curve at the point where x = 1. (c) The gradient of the curve at the point (h, k) is zero. Find h and k.

x 0 1 2 3 4 5 6

y 0 5 8 9 8 5 0

Substitute each x into 6x − x². The values read the same forwards and backwards, which is the free check that a symmetrical curve gives you: if yours is not symmetrical, a value is wrong.
Plot the seven points and join them with one smooth curve, cutting the x-axis at (0, 0) and (6, 0).

Two of the plotted points sit on the x-axis itself, so mark them with crosses like the rest — they are on the curve, not just on the axis.
At x = 1, y = 5. Lay a ruler so that it touches the curve only at (1, 5), and draw a long tangent there.

Find the point of contact from the table first, so the ruler goes in the right place. Turn the ruler until the curve falls away on both sides by the same amount — that is what "touching at a single point" looks like in practice.
Two points on the tangent: (1, 5) and (3, 13). Vertical change = 8, horizontal change = 2.

Both points are read off the tangent. The point of contact is a legitimate choice for one of them, because it lies on the tangent as well as on the curve.
$gradient = \frac{8}{2} = 4$ So the gradient of the curve at x = 1 is 4.

Divide the vertical change by the horizontal change. It is positive, and it should be: at x = 1 the curve is still climbing towards its maximum, so a negative answer would mean the tangent was drawn the wrong way.
Fig. 4.17

The tangent at (1, 5) with its right-angled triangle, and the horizontal tangent at the maximum point.

A gradient of zero means a horizontal tangent, which touches this curve only at its maximum point: h = 3 and k = 9.

Draw the horizontal tangent, see where it touches, and read the coordinates off the graph. Because the curve is symmetrical about x = 3, the table already agrees: y = 8 at both x = 2 and x = 4.

`x`	0	1	2	3	4	5	6
`y`	0	5	8	9	8	5	0

More detail

A curve is steep in some places and nearly flat in others, so measuring its gradient means replacing the curve, just at that point, by the straight line that matches it. Everything the tangent does elsewhere is beside the point: it is the contact that carries the meaning.

The reading is an estimate, and the syllabus says so. The outcome is worded "estimation of the gradient of a curve by drawing a tangent". Two slightly different tangents give noticeably different gradients, and both can be right — which is why the two points you used and the division are shown as the working.

Past-paper questions

G3 N2015 · Paper 2 · Q4 8 marks

The variables x and y are connected by the equation $y = \frac{x^{2}}{5} + \frac{4}{x} - 3$ . Some corresponding values of x and y, correct to two decimal places, are given in the table below.

`x`	0.5	1	1.5	2.0	2.5	3	4	5	6
`y`	`p`	1.20	0.12	−0.20	−0.15	0.13	1.20	2.80	4.87

Find the value of p. [1]
Use your graph to find the solutions to the equation $\frac{x^{2}}{5} + \frac{4}{x} = 3$ in the range 0 ≤ x ≤ 6. [2]
By drawing a tangent, find the gradient of the curve at (4, 1.20). [2]
1. Write down the equation of this line [the line with gradient −1.5 that passes through the point with coordinates (2, 1)]. [1]
2. Write down the coordinates of the points where the line intersects the curve. [2]

Worked solution

(a) Put x = 0.5: $y = \frac{{0.5}^{2}}{5} + \frac{4}{0.5} - 3 = 0.05 + 8 - 3$

p = 5.05

(c) The curve drawn is $y = \frac{x^{2}}{5} + \frac{4}{x} - 3$ , so the equation to be solved is exactly y = 0 — read where the curve crosses the x-axis.

x = 1.6 or x = 2.8.

(Exactly, multiplying by 5x gives x³ − 15x + 20 = 0, whose roots in the range are 1.6132… and 2.8056…)

(d) Draw a long tangent touching the curve at (4, 1.20) and take two widely spaced points on it — for example (3.2, 0.1) and (4.9, 2.4):

gradient = $\frac{2.4 - 0.1}{4.9 - 3.2} = \frac{2.3}{1.7}$ = 1.35 (3 s.f.)

(The exact gradient there is $\frac{2 x}{5} - \frac{4}{x^{2}}$ at x = 4, which is 1.6 − 0.25 = 1.35, so a well-drawn tangent lands on the exact value.)

(e)(ii) Gradient −1.5 through (2, 1): 1 = −1.5(2) + c, so c = 4.

y = −1.5x + 4

(e)(iii) Reading the two crossings of that line with the curve:

(0.7, 2.95) and (2.7, −0.05).

(Exactly, the crossings are at x = 0.6792… and 2.7052…, where y = 2.9811… and −0.0578…, so the readings are right to the nearest half-square.)

Why this works. The grid's curve is $\frac{x^{2}}{5} + \frac{4}{x} - 3$ , so "= 3" becomes "y = 0" and (c)'s answers are the x-intercepts. Part (d) wants a long tangent, both points off the ruler, and the changes divided the right way round; the curve rises at (4, 1.20), so the gradient is positive. In (e)(iii) "write down" wants pairs, not algebra.

G3 N2016 · Paper 2 · Q5 8 marks

The variables x and y are connected by the equation $y = \frac{x^{3}}{2} - 5 x - 2$ . Some corresponding values of x and y are given in the table below.

`x`	−3	−2	−1	0	1	2	3	4
`y`	`p`	4	2.5	−2	−6.5	−8	−3.5	10

Find the value of p. [1]
The equation $\frac{x^{3}}{2} - 5 x = 8$ has only one solution. Explain how this can be seen from your graph. [2]
By drawing a tangent, find the gradient of the curve at (1, −6.5). [2]
1. Write down the x-coordinate of the point where this line [y = 4 − 3x] intersects the curve. [1]
2. This value of x is a solution of the equation x³ + Ax + B = 0. Find the value of A and the value of B. [2]

Worked solution

(a) Put x = −3: $y = \frac{−27}{2} + 15 - 2 = −13.5 + 13$

p = −0.5

(c) Subtract 2 from both sides so the left-hand side becomes the curve's own expression:

$\frac{x^{3}}{2} - 5 x - 2 = 8 - 2 = 6$

Draw the line y = 6. It cuts the curve at only one point, so the equation has only one solution.

(That crossing is at x = 3.774 to 3 d.p.; the curve's local maximum, near x = −1.8, is well below 6, which is why the line misses the left-hand part of the curve altogether.)

(d) Draw a long tangent touching the curve at (1, −6.5) and read two points on the ruler — for example (−1, 0.8) and (2, −10):

gradient = $\frac{0.8 - (−10)}{−1 - 2} = \frac{10.8}{−3}$ = −3.6

The exact gradient at x = 1 is $\frac{3 x^{2}}{2} - 5 = 1.5 - 5 = −3.5$ , so any drawn tangent giving about −3.5 to −3.6 is a good one.

(e)(ii) The line y = 4 − 3x passes through (−1, 7), (2, −2) and (4, −8); it cuts the curve once, at

x = 2.85 (from the graph; exactly 2.8623…).

(e)(iii) At that point the two y-values are equal:

$\frac{x^{3}}{2} - 5 x - 2 = 4 - 3 x$

$\frac{x^{3}}{2} - 2 x - 6 = 0$ , and doubling every term gives x³ − 4x − 12 = 0.

A = −4 and B = −12.

Why this works. Part (c) wants two things: which line you draw (y = 6, from matching the curve) and how many times it cuts the curve. In (d) the curve is falling at (1, −6.5), so the gradient is negative, and a steep tangent wants well-separated reading points. In (e)(iii) the printed form has leading coefficient 1 and no x² term, so clear the halves.

G3 N2018 · Paper 1 · Q23 5 marks

The graph of $y = \frac{1}{2} (x - x^{2})$ is drawn on the grid.

Write down the equation of the line of symmetry of the curve. [1]
Use the graph to solve the equation x − x² = −4. [2]
The point P has coordinates (−2, 5). A tangent to the curve can be drawn so that the tangent passes through P.
1. Find the equation of this tangent. [2]

Worked solution

(a) The curve cuts the x-axis where x − x² = x(1 − x) = 0, that is at x = 0 and x = 1. A parabola is symmetrical about the midpoint of its roots, so the line of symmetry is

x = 0.5

(b) The curve drawn is $y = \frac{1}{2} (x - x^{2})$ , so halve both sides of the equation to be solved:

$\frac{1}{2} (x - x^{2}) = −2$

Draw y = −2 and read the two crossings:

x = −1.55 or x = 2.55 (from the graph).

(Exactly, x² − x − 4 = 0 gives $x = \frac{1 \pm \sqrt{17}}{2}$ = −1.5616… or 2.5616…)

(c)(ii) Lay a ruler through P(−2, 5) and rotate it until it just touches the curve; the point of contact on this grid is (2, −1). Two points on the tangent are therefore (−2, 5) and (2, −1):

gradient = $\frac{−1 - 5}{2 - (−2)} = \frac{−6}{4} = −1.5$

Through P: 5 = −1.5(−2) + c, so c = 2 and

y = −1.5x + 2

Check: the gradient of the curve at x = 2 is $\frac{1}{2} (1 - 2 x) = \frac{1}{2} (1 - 4) = −1.5$ , which matches, and the curve passes through (2, −1). A hand-drawn tangent that gives something close, such as y = −1.48x + 2.04, is credited too.

Why this works. This is a tangent question run backwards: you are given a point the tangent passes through, and you find where it touches. Pivot the ruler about P until the line grazes the curve rather than cutting it: only one line through P does that inside the drawn range. The answer is an equation, so it contains y and x.

G3 N2021 · Paper 1 · Q23 6 marks

The graph of $y = \frac{1}{2} (x^{3} + x^{2}) - 4 x + 3$ is drawn on the grid.

By drawing a tangent, find the gradient of the curve at (2, 1). [2]
The equation x³ + x² − 6x = 0 can be solved by drawing a suitable straight line on the grid.
1. Find the equation of the straight line. [2]
2. By drawing this straight line, solve the equation x³ + x² − 6x = 0. [2]

Worked solution

(a) First check the point of contact: at x = 2, y = ½(8 + 4) − 8 + 3 = 1, so (2, 1) really is on the curve.

Draw a long tangent touching there and read two points on the ruler — for example (0.5, −5) and (3, 4.9):

gradient = $\frac{4.9 - (−5)}{3 - 0.5} = \frac{9.9}{2.5}$ = 3.96 (3 s.f.)

The exact gradient there is $\frac{1}{2} (3 x^{2} + 2 x) - 4$ at x = 2, which is ½(12 + 4) − 4 = 4, so any tangent reading close to 4 is credited.

(b)(i) Halve the equation to match the curve's cubic and square terms:

$\frac{1}{2} (x^{3} + x^{2}) - 3 x = 0$

Subtracting this from the curve's expression leaves (−4x + 3) − (−3x) = −x + 3, so the line to draw is

y = 3 − x — through (0, 3) and (3, 0).

(b)(ii) The line cuts the curve three times, and the three x-coordinates are the solutions:

x = −3, x = 0 or x = 2

Check by factorising: x³ + x² − 6x = x(x + 3)(x − 2), so the roots really are −3, 0 and 2 — the graph readings are exact here.

Why this works. The two halves pull in opposite directions: in (a) a drawn tangent is approximate, so a band around the true value 4 is accepted, and a long ruler line with well-separated points earns it. In (b) nothing is approximate: x³ + x² − 6x factorises, so the roots are whole numbers. Halve the target equation first, to match the curve's $\frac{x^{3}}{2}$ .

Reading a rate off a real-world curve ⓘ Curves only

On a real-world graph the gradient is not just a number: it is a rate of change of one quantity with respect to the other, and it carries units. If the graph is a curve, that rate is different at every instant — so finding it means drawing a tangent.

Graph	Gradient of the tangent means	Its units
distance against time	the speed at that instant	distance unit ÷ time unit, e.g. m/s or km/h
speed against time	the acceleration at that instant	speed unit ÷ time unit, e.g. m/s²
volume against time	the rate of filling or draining	volume unit ÷ time unit, e.g. litres per second

The units come straight off the axes. Gradient is (change in the vertical variable) ÷ (change in the horizontal variable), so its unit is the vertical axis's unit divided by the horizontal axis's unit.

A steepening curve means the rate is increasing. On a distance–time graph, a curve that gets steeper shows the object speeding up; one that flattens shows it slowing down.

Walkthrough 14 — a distance–time curve, with units Intermediate

A cyclist starts from rest. The table shows the distance, d metres, travelled t seconds after starting. (a) Plot the points and join them with a smooth curve. (b) Use your graph to estimate the time taken to travel the first 20 m. (c) By drawing a tangent, find the speed of the cyclist 8 seconds after starting. (d) Describe how the cyclist's speed changes over the 12 seconds.

`t` (s)	0	2	4	6	8	10	12
`d` (m)	0	1	4	9	16	25	36

Take 1 cm to represent 1 second horizontally and 1 cm to represent 5 metres vertically; plot the seven points and join them with one smooth curve.

Label the axes with the quantity and its unit, as the table does. On a real-world graph the units are half the meaning, and the gradient later depends on having them written down.
From the graph, the cyclist travels 20 m after about 8.9 seconds.

Read across from d = 20 to the curve, then down to the t-axis. The answer is a time, so it needs the unit "seconds".
At t = 8, d = 16. Draw a long tangent touching the curve at (8, 16).

Take the point of contact from the table, so the ruler starts in exactly the right place. Draw the tangent right across the grid: the longer it is, the more accurately its two reading points can be taken.
Two points on the tangent: (6, 8) and (12, 32). Vertical change = 24 m; horizontal change = 6 s.

Keep the units attached to the two changes. That is what turns an abstract gradient into a speed on the next line — and it is what stops metres and seconds being divided the wrong way round.
Fig. 4.18

The tangent at (8, 16) with its right-angled triangle; the dashed lines show the reading for the first 20 m.

$speed = \frac{24 m}{6 s} = 4 m/s$ The curve gets steeper throughout, so the cyclist is speeding up the whole time.

Dividing metres by seconds gives metres per second, so the unit is part of the answer. The description follows from the shape alone: a steepening distance–time curve means a rising gradient, and the gradient is the speed.

Walkthrough 15 — a speed–time curve, and an acceleration Advanced

A particle moves in a straight line so that, t seconds after passing a point P, its speed v m/s is given by v = 2t² − 12t + 25. (a) Copy and complete the table and draw the graph for 0 ≤ t ≤ 6. (b) Use your graph to find the two times at which the speed is 20 m/s. (c) By drawing a tangent, find the gradient of the curve at t = 5, and state what it represents. (d) Write down the time at which the acceleration is zero.

`t` (s)	0	1	2	3	4	5	6
`v` (m/s)	25		9	7		15	25

When t = 1: v = 2 − 12 + 25 = 15. When t = 4: v = 32 − 48 + 25 = 9.

Substitute into the given expression, one term at a time. The completed row reads 25, 15, 9, 7, 9, 15, 25 — symmetrical about t = 3, which is a free check that both new values are right.
Plot the seven points and join them with one smooth curve; it falls to a lowest point at t = 3 and rises again.

Draw one continuous curve through the minimum, not two straight pieces meeting in a corner. A speed–time curve is what makes this question ours: its gradient changes from instant to instant.
Draw v = 20. From the graph, t = 0.5 s or t = 5.5 s.

A horizontal line at the stated speed cuts the curve twice, because the particle slows down and then speeds up again — so give both times, each with its unit.
At t = 5, v = 15. Draw a long tangent touching the curve at (5, 15).

The point of contact comes from the table, so the ruler can be placed exactly. The curve is steep here, so a long tangent matters even more than usual.
Two points on the tangent: (4.25, 9) and (6.25, 25). Vertical change = 16 m/s; horizontal change = 2 s.

Read both points off the tangent, not the curve — at t = 6.25 the curve is already well above the tangent. Carry the units through so the division produces them automatically.
$gradient = \frac{16 m/s}{2 s} = 8 m/s^{2}$ This is the acceleration of the particle at t = 5 s.

Metres per second divided by seconds gives metres per second squared, so the unit itself says the answer is an acceleration. On a speed–time graph the gradient is always the acceleration, and here it is positive, so the particle is speeding up.
Fig. 4.19

The tangent at (5, 15) with its right-angled triangle, and the horizontal tangent at the minimum point (3, 7).

The acceleration is zero when the tangent is horizontal, which happens at the lowest point: t = 3 s.

Zero acceleration means zero gradient, and a zero gradient means a horizontal tangent — which touches this curve only at its turning point. At that instant the particle has stopped slowing down and has not yet begun to speed up.

More detail

Read the units off the axis labels, and convert if the question asks for different ones — a speed of 1.5 km per minute asked for in km/h, for instance.

The shape of the curve already says which way the rate is going, so a comparison like "the cyclist is speeding up here and slowing down there" can be made without drawing a single tangent; questions often ask for exactly that.

Check yourself

The tangent to a curve at the point P passes through (1, 2) and (5, 14). Find the gradient of the curve at P.
Answer
Vertical change = 14 − 2 = 12; horizontal change = 5 − 1 = 4.
Gradient = 12 ÷ 4 = 3.
What is the gradient of a curve at its maximum point, and what does the tangent there look like?
Answer
The gradient is 0, and the tangent is a horizontal line. The same is true at a minimum point.
On a graph of distance (km) against time (hours), what does the gradient of the tangent represent, and in what units?
Answer
The speed at that instant, in km/h — the vertical axis's unit divided by the horizontal axis's unit.
Two students draw a tangent at the same point on the same curve and get gradients of 2.8 and 3.1. Explain how both can be acceptable.
Answer
The tangent is positioned by eye, so the graphical method gives only an estimate: the syllabus itself says "estimation of the gradient of a curve by drawing a tangent". A slight difference in the tilt of the ruler changes the reading, so answers close to the true value are accepted.
On a speed–time curve with v in m/s and t in s, the tangent at t = 4 has gradient −3. What does this tell you about the motion?
Answer
The acceleration at that instant is −3 m/s². The gradient is negative, so the object is slowing down — decelerating at 3 m/s².
Why should you read the two points for the gradient off the tangent rather than off the curve?
Answer
The tangent and the curve meet at only one point. Anywhere else the curve has bent away from the line, so a point taken from the curve gives the gradient of a different line — not the tangent, and so not the gradient of the curve at that point.

Past-paper questions

G3 N2017 · Paper 2 · Q3 6 marks

A number of bacteria are introduced to a culture. The number of bacteria, y, in the culture t hours after they are first introduced is given by the formula y = 50 × 2^t. The table shows some corresponding values of t and y, correct to 3 significant figures.

`t`	0.5	1	1.5	2	2.5	3	3.5	4
`y`	`p`	100	141	200	283	400	566	800

How many bacteria are introduced to the culture at the start? [1]
Calculate the value of p. [1]
Use your graph to find how many hours it takes for the number of bacteria to reach 500. [1]
1. By drawing a tangent, find the gradient of the curve at (2, 200). [2]
2. What does this gradient represent? [1]

Worked solution

(a) "At the start" means t = 0: y = 50 × 2⁰ = 50 × 1 = 50 bacteria.

(b) Put t = 0.5: y = 50 × 2^0.5 = 50√2 = 70.7106…

p = 70.7 (3 s.f., the accuracy the table is given to; as a count of bacteria this is 71).

(d) Read across from y = 500 to the curve and down to the t-axis:

t = 3.32 hours (about 3.3 hours off a hand-drawn graph).

(Exactly, 50 × 2^t = 500 gives 2^t = 10, so t = 3.3219…)

(e)(i) Draw a long tangent touching the curve at (2, 200) and read two points on the ruler — for example (1.5, 130) and (2.5, 265):

gradient = $\frac{265 - 130}{2.5 - 1.5}$ = 135

(The true gradient at t = 2 is 138.6, so any reading of roughly 130 to 145 from a well-drawn tangent is credited.)

(e)(ii) The vertical axis is a number of bacteria and the horizontal axis is a time in hours, so the gradient is bacteria ÷ hours:

It is the rate at which the number of bacteria is increasing, two hours after they were introduced — about 135 bacteria per hour.

Why this works. A gradient on a real-world graph is "vertical quantity per horizontal quantity" — here bacteria per hour, and naming both quantities is the whole of (e)(ii). Part (b) turns on 2^0.5 = √2, with the table's own "3 significant figures" fixing 70.7. The gradient at t = 2 is 135 per hour and four times that by t = 4, so the model cannot last.

G3 N2019 · Paper 1 · Q17 4 marks

The distance–time graph shows the journeys for two people, Nur and Lim, between two villages, A and B.

Find Nur's speed for the first hour of her journey. [1]
1. Find the time when Nur and Lim pass each other. [1]
2. How far are they from village B when they pass each other? [1]
Lim jogs back to village B at 8 km/h. Change 8 km/h into metres per second. [1]

Worked solution

(a) Nur leaves A at 10 am and her line is straight until she stops, so read the distance she has covered by 11 am: 3.6 km in 1 hour.

Nur's speed = 3.6 km/h

(The same line reaches 5.4 km at 11.30 a.m., and $\frac{5.4}{1.5} = 3.6$ too, which is a useful check on the reading.)

(b)(i) They pass each other where the two lines cross — Nur going up, Lim coming down. Reading straight down from the crossing to the time axis:

12.24 p.m.

(b)(ii) Reading across from the same crossing to the distance axis gives 6.8 km from A. Village B is 14.8 km from A, so

distance from B = 14.8 − 6.8 = 8 km

(c) 8 km/h means 8000 m in 3600 s:

$\frac{8000 m}{3600 s} = 2 \frac{2}{9}$ m/s (2.22 m/s to 3 s.f.)

Why this works. On a distance–time graph the gradient is the speed and a flat section means stopped, and the vertical axis is measured from A — so (b)(ii) needs a subtraction to turn a distance from A into a distance from B. Converting km/h to m/s is ×1000 then ÷3600, or ÷3.6 in one step. A jogger at a few metres a second is the common-sense check.

G3 N2021 · Paper 1 · Q24 6 marks

The diagram shows speed–time graphs for two cyclists, A and B, travelling along the same road in the same direction.

A accelerates for 20 seconds and then travels at a constant speed of 10 m/s.
B travels at a constant speed of 8 m/s.

Calculate the acceleration of A for the first 20 seconds. [1]
The area beneath each speed–time graph represents the distance travelled by each cyclist.
1. Find an expression, in terms of T, for the distance travelled by B in the T seconds. [1]
2. When t = 0, A is 90 m behind B. When t = T, A passes B. Find the value of T. [4]

Worked solution

(a) On a speed–time graph the acceleration is the gradient. A's speed rises from 5 m/s to 10 m/s in 20 s:

acceleration = $\frac{10 - 5}{20}$ = 0.25 m/s²

(b)(i) B's graph is a rectangle of height 8 and width T:

distance = 8T m

(b)(ii) A's graph is a trapezium (parallel sides 5 and 10, width 20) followed by a rectangle (height 10, width T − 20):

distance travelled by A = $\frac{1}{2} (5 + 10) (20) + 10 (T - 20) = 150 + 10 T - 200$

= (10T − 50) m.

A starts 90 m behind and finishes level, so A must travel 90 m more than B in the T seconds:

10T − 50 = 8T + 90

2T = 140, so T = 70

Check: A covers 10(70) − 50 = 650 m and B covers 8(70) = 560 m, a difference of exactly 90 m.

Why this works. On a speed–time graph the gradient is the acceleration and the area is the distance. A does not start from rest — the line begins at 5 m/s — so the first 20 seconds is a trapezium of area 150, not a triangle of 100. A starts 90 m behind B and later passes it, so A's distance is the larger.

The shape atlas

Twelve pictures cover the whole of outcome N6.9. If you can draw these from memory, you can answer any "sketch the graph" or "match the equation" question in the topic.

Fig. 4.20

The graphs of y = axⁿ for the six values of n in the syllabus, each with a positive and a negative. A negative a always reflects the picture in the x-axis.

The whole chapter in one table

Equation	Shape	Key features to state
`y` = `ax`³ (`n` = 3)	through the origin, flat there, opposite ends	passes through (0, 0); `a` > 0 rises left to right, `a` < 0 falls
`y` = `ax`³ + `bx`² + `cx` + `d`	cubic: two turning points, or none	`y`-intercept (0, `d`); the sign of `a` fixes the two ends
`y` = `ax`² (`n` = 2)	parabola, turning point at the origin ⓘ See ch 2	`a` > 0 a minimum, `a` < 0 a maximum; symmetrical about the `y`-axis
`y` = `ax` (`n` = 1)	straight line through the origin	gradient `a`; passes through (0, 0)
`y` = `a` (`n` = 0)	horizontal straight line	gradient 0; cuts the `y`-axis at (0, `a`)
`y` = `a`/`x` (`n` = −1)	two branches, one in each of two opposite quadrants	asymptotes `x` = 0 and `y` = 0; `a` > 0 → 1st and 3rd quadrants, `a` < 0 → 2nd and 4th; rotational symmetry of order 2 about the origin
`y` = `a`/`x`² (`n` = −2)	two branches on the same side of the `x`-axis	asymptotes `x` = 0 and `y` = 0; entirely above the axis if `a` > 0, entirely below if `a` < 0; symmetrical about the `y`-axis
`y` = `ka`^x (`a` > 1)	exponential growth curve	cuts the `y`-axis at (0, `k`); horizontal asymptote `y` = 0; entirely above the axis if `k` > 0, entirely below if `k` < 0
Sums, e.g. `y` = `x` + 4/`x`	add the two graphs value by value	up to three power terms; the term that grows fastest controls the ends, the others bend the middle

The rules to know by heart

Asymptotes. A line the curve approaches as closely as you like without touching it. Both reciprocal graphs have x = 0 and y = 0; the exponential has one, y = 0. Quote an asymptote as an equation.

Why the reciprocal graphs break. Division by zero is undefined, so neither has a point at x = 0: draw two separate branches. They do not meet the x-axis either, because a fraction is zero only when its numerator is.

Exponentials. y = ka^x cuts the y-axis at (0, k), since a⁰ = 1, and a^x is positive, so the curve keeps the sign of k. Given two points, start with the one on the y-axis.

Gradient of a curve at a point. Draw the tangent there and find its gradient, taking both reading points off the tangent, as far apart as the page allows. At a maximum or minimum the tangent is horizontal and the gradient is 0.

gradient = \frac{vertical change}{horizontal change}

Units of a gradient. Vertical axis unit ÷ horizontal axis unit. On a distance–time curve the tangent's gradient is the speed at that instant; on a speed–time curve it is the acceleration. A negative gradient means the quantity is decreasing.

"Hence solve" by drawing a straight line. Rearrange the equation so one side is exactly the expression already drawn; whatever is left is the line. The solutions are the x-coordinates where the line meets the curve, and the crossings count them.

The habits that earn the marks

1. Use the scale the question states, and label both axes with the variable and its unit.
2. Include x-values close to the interesting part of the curve — near a turning point, or near an asymptote.
3. Join the points with one smooth freehand curve; ruled segments and hairy double lines lose the drawing mark.
4. Never draw through a value the function does not have — a reciprocal graph is two curves.
5. Read answers to the nearest half a small square and give them to 1 decimal place.
6. For a gradient, show the two points you used and the division, not just the answer.
7. Answer the question that was asked, in context, with units.

Syllabus at a glance

The three G3 Mathematics (K310) outcomes this chapter covers, quoted from the Secondary Three/Four section of the syllabus.

G3 Mathematics (K310, first examination 2027) — these notes are written to it.

LO	What you must be able to do	Covered in
6.9	graphs of power functions `y` = `ax`ⁿ, where `n` = −2, −1, 0, 1, 2, 3, and simple sums of not more than three of these	4.1 Power graphs, 4.2 Reciprocals
6.10	graphs of exponential functions `y` = `ka`^x, where `a` is a positive integer, `a` > 1	4.3 Exponentials
6.11	estimation of the gradient of a curve by drawing a tangent	4.4 Tangents, 4.4 Rates

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus. The quadratic case n = 2 appears in the shape atlas for completeness, but sketching a parabola from a given algebraic form is outcome N6.8, which is taught in chapter 2. The G2 syllabus contains these same three outcomes word for word, numbered 6.8, 6.9 and 6.10, so nothing on this page is marked as G3-only and the G2 view shows this chapter in full; a G2-specific outcome list for this chapter is authored in a later phase. Straight-line distance–time and speed–time graphs are Secondary 1/2 carry-over and are listed in neither Secondary Three/Four syllabus — only curved real-world graphs, where a tangent has to be drawn, belong here.

📈 Graphs of Functions & Graphical Solution — Revision Notes

The family of graphs — and the four things a question can ask

The sign of `a` decides the shape

Past-paper questions

Cubic functions

Check yourself

Past-paper questions

Graphs of `y` = `a`/`x`

Past-paper questions

Graphs of `y` = `a`/`x`²

Check yourself

Past-paper questions

Graphs of `y` = `ka`^x

Check yourself

Past-paper questions

The tangent, and the gradient at a point

Past-paper questions

Reading a rate off a real-world curve ⓘ Curves only

Check yourself

Past-paper questions

The shape atlas

The whole chapter in one table

The rules to know by heart

The habits that earn the marks

Syllabus at a glance

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📈 Graphs of Functions & Graphical Solution — Revision Notes

The family of graphs — and the four things a question can ask

The sign of a decides the shape

Past-paper questions

Cubic functions

Check yourself

Past-paper questions

Graphs of y = a/x

Past-paper questions

Graphs of y = a/x²

Check yourself

Past-paper questions

Graphs of y = kax

Check yourself

Past-paper questions

The tangent, and the gradient at a point

Past-paper questions

Reading a rate off a real-world curve ⓘ Curves only

Check yourself

Past-paper questions

The shape atlas

The whole chapter in one table

The rules to know by heart

The habits that earn the marks

Syllabus at a glance

The sign of `a` decides the shape

Graphs of `y` = `a`/`x`

Graphs of `y` = `a`/`x`²

Graphs of `y` = `ka`^x