Cubic, reciprocal & trig graphs; transformations of graphs – IGCSE Maths exam questions with mark scheme

1Grade 4–5HigherCalculator5 marks

Here is an incomplete table of values for $y = x^3 - 3x + 1$

$x$	$-2$	$-1.5$	$-1$	$-0.5$	$0$	$0.5$	$1$	$1.5$	$2$
$y$	$-1$	$2.125$		$2.375$	$1$		$-1$		$3$

(a) Complete the table of values. [2]

The graph of $y = x^3 - 3x + 1$ for $-2 \le x \le 2$ is drawn on the grid.

(b) Use the graph to find estimates of the solutions of $x^3 - 3x + 1 = 0$ [2]

(a) at least two of the three missing values correctB1
all three values correct: $x=-1 \to 3$, $x=0.5 \to -0.375$, $x=1.5 \to -0.125$ (accept $-0.4$ and $-0.1$; ignore further rounding)B1
(b) for reading where the curve crosses the $x$-axis ($y = 0$), shown by at least one correct estimateM1
all three: $-1.9$ (accept $-1.95$ to $-1.85$), $0.35$ (accept $0.3$ to $0.4$), $1.5$ (accept $1.45$ to $1.6$)A1
(c) $(-1,\,3)$B1
Answer: (a) $3,\ -0.375,\ -0.125$ (b) $x \approx -1.9,\ 0.35,\ 1.5$ (c) $(-1,\ 3)$
Examiner tip: Use brackets when you key negative values into the calculator: $(-1)^3$ and $1.5^3 - 4.5$ are where most table errors come from.

2Grade 4–5HigherCalculator3 marks

Here are four graphs, A, B, C and D.

Here are six equations.

$y = x^3 + 2 \qquad y = -\dfrac{2}{x} \qquad y = \dfrac{2}{x} \qquad y = -x^3 \qquad y = \tan x^\circ \qquad y = x^2 - 2$

Match each graph to its equation. [3]

at least one graph matched correctlyB1
at least two graphs matched correctlyB1
all four correct: A $y=-\dfrac{2}{x}$, B $y=x^3+2$, C $y=\tan x^\circ$, D $y=-x^3$B1
Answer: A: $y=-\dfrac{2}{x}$; B: $y=x^3+2$; C: $y=\tan x^\circ$; D: $y=-x^3$
Examiner tip: Check which quadrants a reciprocal graph sits in: $y=\dfrac{2}{x}$ is in the 1st and 3rd quadrants, so the curve in the 2nd and 4th must have a negative numerator.

3Grade 4–5HigherCalculator4 marks

The diagram shows the graph of $y = \sin x^\circ$ for $0 \le x \le 360$

(a) Write down the coordinates of the minimum point of the graph. [1]

$\sin 30^\circ = 0.5$

(b) Find the other value of $x$, between $0$ and $360$, for which $\sin x^\circ = 0.5$ [1]

(a) $(270,\,-1)$B1
(b) $150$B1
(c) one correct value, $210$ or $330$, or a correct method using symmetry, e.g. $180 + 30$ or $360 - 30$M1
$210$ and $330$ and no other values in the rangeA1
Answer: (a) $(270,\ -1)$ (b) $150$ (c) $210,\ 330$
Examiner tip: Use the symmetry of the sine curve: it is symmetrical about $x = 90$ and $x = 270$, so values come in pairs equally spaced either side of those lines.

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4Grade 6–7HigherCalculator3 marks

The diagram shows the graph of the quadratic curve $y = \mathrm{f}(x)$

The minimum point of the curve is $(2,\,-3)$

Write down the coordinates of the turning point of the curve with equation

(a) $y = \mathrm{f}(x) + 4$ [1]

(b) $y = \mathrm{f}(x + 3)$ [1]

(a) $(2,\,1)$B1
(b) $(-1,\,-3)$B1
(c) $(2,\,3)$B1
Answer: (a) $(2,\ 1)$ (b) $(-1,\ -3)$ (c) $(2,\ 3)$
Examiner tip: $\mathrm{f}(x + 3)$ moves the graph 3 to the left, not the right: the change inside the bracket does the opposite of what it looks like.

5Grade 6–7HigherCalculator4 marks

(a) Describe fully the single transformation that maps the graph of $y = \cos x^\circ$ onto the graph of $y = \cos(x + 60)^\circ$ [2]

(b) Explain why the equation $3\cos x^\circ = 4$ has no solutions. [1]

(a) translationB1
by vector $\begin{pmatrix} -60 \\ 0 \end{pmatrix}$ (accept '60 units to the left' or '60 in the negative $x$-direction'); B0 for any second transformation statedB1
(b) e.g. the maximum value of $3\cos x^\circ$ is $3$ (so it can never equal $4$), or $\cos x^\circ$ would have to be $\tfrac{4}{3}$, which is greater than $1$C1
(c) $4$B1
Answer: (a) Translation by $\begin{pmatrix} -60 \\ 0 \end{pmatrix}$ (b) $\cos x^\circ \le 1$, so $3\cos x^\circ \le 3 < 4$ (c) $4$
Examiner tip: 'Describe fully' needs both the word 'translation' and the vector; giving two transformations (e.g. 'translation and stretch') scores no marks.

6Grade 6–7HigherCalculator5 marks

The grid shows the graph of $y = \dfrac{8}{x}$ for $0.8 \le x \le 8$

A tangent to the curve has been drawn at the point $P(2,\,4)$.

(a) Use the tangent to find an estimate for the gradient of the curve at $P$. [2]

(b) By drawing a suitable straight line on the grid, find the solutions of the equation $\dfrac{8}{x} = 6 - x$ [3]

(a) for a correct method using two points on the tangent, e.g. $\dfrac{8 - 0}{0 - 4}$ (difference in $y$ ÷ difference in $x$)M1
$-2$ (accept $-2.2$ to $-1.8$)A1
(b) for the line $y = 6 - x$ identifiedM1
for $y = 6 - x$ drawn correctly (through, e.g., $(0,6)$ and $(6,0)$)M1
$x = 2$ and $x = 4$ (accept values in the ranges $1.9$ to $2.1$ and $3.9$ to $4.1$)A1
Answer: (a) $-2$ (b) $x = 2,\ x = 4$
Examiner tip: A curve sloping downwards has a negative gradient: losing the minus sign is the most common way to drop the accuracy mark in (a).

7Grade 6–7HigherCalculator4 marks

The grid shows the graph of $y = x^3 - 4x$ for $-3 \le x \le 3$

(a) Write down the solutions of $x^3 - 4x = 0$ [1]

(b) By drawing a suitable straight line on the grid, find estimates of the solutions of the equation $x^3 - 6x + 2 = 0$ [3]

(a) $-2,\ 0,\ 2$B1
(b) for rearranging to $x^3 - 4x = 2x - 2$, i.e. the line $y = 2x - 2$ identifiedM1
for the line $y = 2x - 2$ drawn correctly for at least the range where it meets the curveM1
all three solutions: $-2.6$ (accept $-2.7$ to $-2.5$), $0.3$ (accept $0.2$ to $0.4$), $2.3$ (accept $2.2$ to $2.4$)A1
Answer: (a) $-2,\ 0,\ 2$ (b) $x \approx -2.6,\ 0.3,\ 2.3$
Examiner tip: Make the left-hand side match the curve first: $x^3 - 6x + 2 = 0$ becomes $x^3 - 4x = 2x - 2$, and the right-hand side is the line to draw.

8Grade 8–9HigherCalculator4 marks

The diagram shows the graph of $y = a\sin(bx)^\circ + c$ for $0 \le x \le 360$

$a$, $b$ and $c$ are positive integers.

Find the values of $a$, $b$ and $c$. [3]

Hence write down the number of solutions of the equation $a\sin(bx)^\circ + c = 0$ for $0 \le x \le 360$ [1]

for $c = 2$ (midline halfway between maximum $5$ and minimum $-1$) or $a = 3$ (half of $5 - (-1)$)M1
$a = 3$ and $c = 2$A1
$b = 2$ (two complete cycles in $360^\circ$, or period $180^\circ$)B1
$4$ (the curve crosses the $x$-axis four times)B1
Answer: $a = 3,\ b = 2,\ c = 2$; $4$ solutions
Examiner tip: Read the maximum and minimum carefully: the amplitude $a$ is half the distance between them, and the midline gives $c$; do not just take the maximum as $a$.

9Grade 8–9HigherCalculator4 marks

The curve $C$ has equation $y = \mathrm{f}(x)$ where $\mathrm{f}(x) = x^2 - 6x + 11$

(a) The curve $C$ is translated by the vector $\begin{pmatrix} -5 \\ 1 \end{pmatrix}$ to give the curve $D$.

Find an equation for $D$. Give your answer in the form $y = x^2 + px + q$, where $p$ and $q$ are integers. [3]

(b) Write down the coordinates of the turning point of the curve with equation $y = \mathrm{f}(2x)$ [1]

(a) for $(x - 3)^2 + 2$ or turning point of $C$ at $(3,\,2)$M1
for turning point of $D$ at $(-2,\,3)$, or $(x + 2)^2 + 3$, or $\mathrm{f}(x + 5) + 1$ written correctly, e.g. $(x + 5)^2 - 6(x + 5) + 11 + 1$M1
$y = x^2 + 4x + 7$A1
(b) $(1.5,\,2)$B1
Answer: (a) $y = x^2 + 4x + 7$ (b) $(1.5,\ 2)$
Examiner tip: A stretch $\mathrm{f}(2x)$ halves every $x$-coordinate and leaves the $y$-coordinates alone, so the turning point moves from $(3, 2)$ to $(1.5, 2)$.

10Grade 8–9HigherCalculator6 marks

Here is a table of values for $y = x^2 + \dfrac{4}{x}$ (values given to 2 decimal places where necessary).

$x$	$0.5$	$1$	$1.5$	$2$	$2.5$	$3$	$3.5$	$4$
$y$	$8.25$		$4.92$		$7.85$	$10.33$	$13.39$	$17$

(a) Complete the table. [1]

The graph of $y = x^2 + \dfrac{4}{x}$ for $0.5 \le x \le 4$ is drawn on the grid.

(b) Use the graph to find estimates of the solutions of $x^2 + \dfrac{4}{x} = 7$ [2]

(c) Show that the $x$-coordinates of the points where the line $y = 2x + 3$ meets the curve $y = x^2 + \dfrac{4}{x}$ satisfy the equation $x^3 - 2x^2 - 3x + 4 = 0$ [2]

(d) Show that $x = 1$ is a solution of $x^3 - 2x^2 - 3x + 4 = 0$ [1]

(a) $5$ and $6$ both correctB1
(b) for drawing (or using) the line $y = 7$ to meet the curve, shown by at least one correct estimateM1
$0.6$ (accept $0.55$ to $0.65$) and $2.3$ (accept $2.25$ to $2.35$)A1
(c) for setting $x^2 + \dfrac{4}{x} = 2x + 3$ and multiplying every term by $x$, e.g. $x^3 + 4 = 2x^2 + 3x$M1
for a fully correct rearrangement to $x^3 - 2x^2 - 3x + 4 = 0$ with no errors seenA1
(d) for substituting $x = 1$: $1 - 2 - 3 + 4 = 0$ (or $1 + 4 = 5$ and $2 + 3 = 5$)B1
Answer: (a) $5,\ 6$ (b) $x \approx 0.6,\ 2.3$ (c) shown (d) $1 - 2 - 3 + 4 = 0$
Examiner tip: In (c) every term must be multiplied by $x$, including the $3$ on the right-hand side; forgetting one term is the usual slip.

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