Graphs of quadratic, cubic, reciprocal and exponential functions; turning points – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

(a) Complete the table of values for $y = x^2 - 4$ [2]

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$5$		$-3$			$0$

(b) Write down the equation of the line of symmetry of the graph of $y = x^2 - 4$ [1]

(a) for 2 or 3 correct values from $0$, $-4$, $-3$, $5$B1
(a) (total B2) all four correct: $0$, $-4$, $-3$, $5$B1
(b) $x = 0$ (accept 'the $y$-axis')B1
Answer: (a) $0, -4, -3, 5$ (b) $x = 0$
Examiner tip: $(-2)^2 = +4$ – squaring a negative gives a positive, so $y = 4 - 4 = 0$, not $-8$.

2Grade 4–5Non-calculator5 marks

The diagram shows the graph of $y = x^2 - 2x - 3$

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

(b) Use the graph to write down the solutions of $x^2 - 2x - 3 = 0$ [2]

(a) $(1, -4)$B1
(b) for one of $x = -1$ or $x = 3$B1
(b) for both $x = -1$ and $x = 3$ (and no others)B1
(c) for drawing or using the line $y = 2$, or reading across from $y = 2$M1
(c) $x = -1.4$ to $-1.5$ and $x = 3.4$ to $3.5$A1
Answer: (a) $(1, -4)$ (b) $x = -1$, $x = 3$ (c) $x \approx -1.45$, $x \approx 3.45$
Examiner tip: Solutions of 'curve $= 0$' are where it crosses the $x$-axis; for 'curve $= 2$' read where it crosses the horizontal line $y = 2$.

3Grade 4–5HigherNon-calculator3 marks

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

Match each equation to its graph.

$$y = x^3 \qquad y = \frac{1}{x} \qquad y = 2^x \qquad y = 2 - x^2$$ [3]

for at least one correct matchB1
for at least two correct matchesB1
for all four correct: $y = x^3$ – D, $y = \tfrac1x$ – C, $y = 2^x$ – A, $y = 2 - x^2$ – BB1
Answer: $y = x^3$: D; $y = \tfrac{1}{x}$: C; $y = 2^x$: A; $y = 2 - x^2$: B
Examiner tip: Learn the key features: reciprocal graphs never touch the axes, exponential graphs pass through $(0, 1)$ and never go below the $x$-axis.

4Grade 6–7HigherNon-calculator4 marks

The curve with equation $y = ab^x$, where $b > 0$, passes through the points $(0, 3)$ and $(2, 48)$.

(a) Find the values of $a$ and $b$. [3]

(b) Find the value of $y$ when $x = -1$ [1]

(a) $a = 3$B1
(a) for $3 \times b^2 = 48$ or $b^2 = 16$ (ft their $a$)M1
(a) $b = 4$A1
(b) $\tfrac{3}{4}$ (oe, e.g. $0.75$) (ft their $a$ and $b$)B1
Answer: (a) $a = 3$, $b = 4$ (b) $y = \tfrac34$
Examiner tip: Substitute $x = 0$ first: $b^0 = 1$, so $a$ is just the $y$-intercept.

5Grade 6–7HigherNon-calculator5 marks

The curve $C$ has equation $y = (x + 2)(x - 1)(x - 3)$

(a) Write down the coordinates of the points where $C$ crosses the $x$-axis. [2]

(b) Find the coordinates of the point where $C$ crosses the $y$-axis. [1]

(a) for two correct points from $(-2, 0)$, $(1, 0)$, $(3, 0)$, or for all three $x$-values $-2$, $1$, $3$ given without $y$-coordinatesB1
(a) for all three as coordinates, $(-2, 0)$, $(1, 0)$ and $(3, 0)$, and no othersB1
(b) $(0, 6)$ (accept $y = 6$)B1
(c) for a positive cubic shape: comes up from bottom left, a maximum, then a minimum, then rises to top right (ignore the exact heights of the turning points)B1
(c) for a curve crossing the $x$-axis at $-2$, $1$ and $3$ and the $y$-axis at $6$, with these values marked (ft their values from (a) and (b))B1
Answer: (a) $(-2, 0)$, $(1, 0)$, $(3, 0)$ (b) $(0, 6)$ (c) positive cubic crossing the $x$-axis at $-2$, $1$ and $3$ and the $y$-axis at $6$; its maximum is between $x = -2$ and $x = 1$ (just left of the $y$-axis, near $(-0.8, 8.2)$) and its minimum is between $x = 1$ and $x = 3$ (near $(2.1, -4.1)$)
Examiner tip: Set each bracket equal to zero – $(x + 2) = 0$ gives $x = -2$, not $+2$.

6Grade 8–9HigherNon-calculator4 marks

(a) By completing the square, find the coordinates of the turning point of the curve $y = 5 + 4x - x^2$ [3]

(b) State whether the turning point is a maximum or a minimum. Give a reason for your answer. [1]

(a) for $-(x^2 - 4x) + 5$ or $-(x - 2)^2$ seenM1
(a) for $-(x - 2)^2 + 4 + 5$ or $9 - (x - 2)^2$M1
(a) $(2, 9)$A1
(b) maximum, with a reason, e.g. $-(x - 2)^2 \le 0$ so $y \le 9$, or the coefficient of $x^2$ is negativeC1
Answer: (a) $(2, 9)$ (b) maximum
Examiner tip: Take out the $-1$ carefully: $-(x - 2)^2 = -(x^2 - 4x + 4)$, so you must ADD 4 back on, not subtract it.

7Grade 8–9HigherNon-calculator5 marks

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

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

(b) By drawing a suitable straight line on the grid, find estimates for the solutions of $x^2 - 4x + 1 = 0$ [3]

(a) for one of $-0.2$ to $-0.4$ or $3.2$ to $3.4$B1
(a) for both, $-0.2$ to $-0.4$ and $3.2$ to $3.4$B1
(b) for rearranging to find the line, e.g. $x^2 - 3x - 1 = x - 2$, so $y = x - 2$M1
(b) for the line $y = x - 2$ drawn correctlyM1
(b) $0.2$ to $0.3$ and $3.7$ to $3.8$A1
Answer: (a) $x \approx -0.3$, $x \approx 3.3$ (b) $x \approx 0.3$, $x \approx 3.7$ (line $y = x - 2$)
Examiner tip: Subtract the new equation from the graph's equation to find the line: $(x^2 - 3x - 1) - (x^2 - 4x + 1) = x - 2$.

8Grade 1–3FoundationNon-calculator3 marks

(a) Complete the table of values for $y = x^3 + 1$ [2]

$x$	$-2$	$-1$	$0$	$1$	$2$
$y$	$-7$			$2$

(b) Use your table to write down the coordinates of the point where the graph of $y = x^3 + 1$ crosses the $x$-axis. [1]

(a) at least two of $0$, $1$, $9$ correct (in the correct places)B1
(a) all three correct: $0$, $1$, $9$B1
(b) $(-1, 0)$ cao; allow ft from their table where their $y = 0$. Do not accept $x = -1$ alone or $(0, -1)$B1
Answer: (a) $0,\ 1,\ 9$ (b) $(-1, 0)$
Examiner tip: Cube negatives carefully: $(-1)^3 = -1$, so $y = 0$ at $x = -1$; and the graph crosses the $x$-axis where $y = 0$, so the coordinate is $(-1, 0)$, not $(0, -1)$.

9Grade 4–5Non-calculator4 marks

A ball is kicked straight up into the air from ground level.

Its height, $h$ metres, after $t$ seconds is given by $h = 20t - 5t^2$

The graph of $h$ against $t$ is shown for $0 \le t \le 4$

(a) Use the graph to write down the greatest height the ball reaches. [1]

(b) Use the graph to find the two times at which the ball is $15$ m above the ground. [2]

(a) $20$ (m)B1
(b) for one correct time, $t = 1$ or $t = 3$, or for a horizontal line drawn at $h = 15$M1
(b) $t = 1$ and $t = 3$ (seconds)A1
(c) $2$ (seconds); ft their (b) (larger time minus smaller time)B1
Answer: (a) $20$ m (b) $t = 1$ s and $t = 3$ s (c) $2$ s
Examiner tip: Read the height off the vertical axis at the top of the curve – the answer to (a) is the $h$-value $20$, not the time $2$.

10Grade 4–5Non-calculator3 marks

A rectangle has an area of $12\text{ cm}^2$.

Its width is $x$ cm and its length is $y$ cm, so $y = \dfrac{12}{x}$

(a) Complete the table of values for $y = \dfrac{12}{x}$ [2]

$x$	$1$	$1.5$	$2$	$3$	$4$	$6$	$8$
$y$	$12$		$6$	$4$		$2$

(b) Sam says, “When the width of the rectangle doubles, the length halves.”

Use two pairs of values from your table to show that Sam is correct. [1]

(a) at least two of $8$, $3$, $1.5$ correctB1
(a) all three correct: $8$, $3$, $1.5$ (accept $\tfrac32$ or $1\tfrac12$)B1
(b) a correct pair of table values where $x$ doubles and $y$ halves, e.g. $x = 2,\ y = 6$ and $x = 4,\ y = 3$ (or $1 \to 12$ and $2 \to 6$; $1.5 \to 8$ and $3 \to 4$; $3 \to 4$ and $6 \to 2$; $4 \to 3$ and $8 \to 1.5$). Both pairs must be statedC1
Answer: (a) $8,\ 3,\ 1.5$ (b) e.g. $x = 2 \Rightarrow y = 6$ and $x = 4 \Rightarrow y = 3$
Examiner tip: For (b) you must quote both pairs of values from the table – just saying “it is inverse proportion” does not earn the mark.

11Grade 6–7HigherCalculator4 marks

A scientist models the number of fish, $P$, in a lake $t$ years after a survey by

$$P = 500 \times 0.8^{t}$$

The graph of this model is shown for $0 \le t \le 8$

(a) Use the graph to estimate the value of $t$ when there are $200$ fish in the lake. [1]

(b) Use the model to work out the number of fish in the lake $10$ years after the survey. [2]

(a) answer in the range $3.9$ to $4.3$ (years)B1
(b) $500 \times 0.8^{10}$ oe, or repeated multiplication by $0.8$ ten timesM1
(b) $53$ or $54$ (accept $53.7$ or $53.68\ldots$)A1
(c) e.g. multiplying a positive number by $0.8$ always gives a positive answer / $0.8^t$ is never zero / the curve gets closer to the $t$-axis but never touches it (asymptote). Do not accept “it decreases slowly” aloneC1
Answer: (a) $t \approx 4.1$ (b) $53.7$, i.e. about $54$ fish (c) $0.8^t > 0$ for every $t$, so $P$ is always positive
Examiner tip: In (b) use the formula, not the graph – $t = 10$ is off the grid, and an answer read from a stretched-out curve will not get the accuracy mark.

12Grade 6–7HigherNon-calculator4 marks

The curve with equation $y = x^2 + bx + c$ has a turning point at $(3, -4)$.

(a) Find the value of $b$ and the value of $c$. [2]

(b) Find the coordinates of the points where the curve crosses the $x$-axis. [2]

(a) for $y = (x - 3)^2 - 4$ oe, or for $b = -6$ from the line of symmetry $x = -\tfrac{b}{2} = 3$M1
(a) $b = -6$ and $c = 5$A1
(b) for $(x - 1)(x - 5) = 0$ or $(x - 3)^2 = 4$ or $x - 3 = \pm 2$ (ft their $b$ and $c$ if a 3-term quadratic)M1
(b) $(1, 0)$ and $(5, 0)$A1
Answer: (a) $b = -6$, $c = 5$ (b) $(1, 0)$ and $(5, 0)$
Examiner tip: Work backwards from completed square form: turning point $(p, q)$ means $y = (x - p)^2 + q$, so watch the sign – $(3, -4)$ gives $(x - 3)^2 - 4$, not $(x + 3)^2 - 4$.

13Grade 8–9HigherNon-calculator3 marks

The sketch shows a cubic curve.

The curve touches the $x$-axis at $(-2, 0)$, crosses the $x$-axis at $(3, 0)$ and crosses the $y$-axis at $(0, -6)$.

Find an equation of the curve. [3]

for $y = k(x + 2)^2(x - 3)$ oe, i.e. a repeated factor $(x + 2)^2$ and a factor $(x - 3)$ (with or without $k$)M1
for substituting $(0, -6)$: $-6 = k \times 2^2 \times (-3)$, i.e. $-12k = -6$M1
$y = \tfrac12(x + 2)^2(x - 3)$ oe, e.g. $y = \tfrac12 x^3 + \tfrac12 x^2 - 4x - 6$A1
Answer: $y = \tfrac12(x + 2)^2(x - 3)$, i.e. $y = \tfrac12x^3 + \tfrac12x^2 - 4x - 6$
Examiner tip: A curve that only touches the $x$-axis has a squared factor there; without the scale factor $\tfrac12$ your curve would cross the $y$-axis at $-12$, not $-6$.

14Grade 8–9HigherNon-calculator5 marks

The graph of $y = x^3 - 3x$ is drawn on the grid for $-2.2 \le x \le 2.2$

(a) Write down the coordinates of the two turning points of the graph. [2]

(b) Use the graph to write down the number of solutions of $x^3 - 3x = 1$ [1]

Find all the possible values of $k$. [2]

(a) $(-1, 2)$B1
(a) $(1, -2)$B1
(b) $3$B1
(c) for one correct region, $k > 2$ or $k < -2$, or for identifying $2$ and $-2$ as the critical valuesM1
(c) $k < -2$ or $k > 2$ (both; accept $|k| > 2$). Do not accept $\le$ or $\ge$A1
Answer: (a) $(-1, 2)$ and $(1, -2)$ (b) $3$ (c) $k < -2$ or $k > 2$
Examiner tip: Solutions of $x^3 - 3x = k$ are where the horizontal line $y = k$ meets the curve; at $k = \pm 2$ the line touches a turning point and gives two solutions, so the inequalities must be strict.

15Grade 6–7HigherNon-calculator5 marks

(a) Explain why the graph of $y = \dfrac{4}{x}$ does not cross either axis. [1]

(b) The graph of $y = \dfrac{4}{x}$ and the line $x + y = 5$ meet at two points.

Find, algebraically, the coordinates of both points. [4]

(a) both reasons: $x$ cannot be $0$ (you cannot divide by zero), and $\dfrac{4}{x}$ can never equal $0$ (so $y \ne 0$)C1
(b) for substituting to eliminate one variable, e.g. $\dfrac{4}{x} = 5 - x$ or $x + \dfrac{4}{x} = 5$M1
(b) for rearranging to a 3-term quadratic, e.g. $x^2 - 5x + 4 = 0$ (or $y^2 - 5y + 4 = 0$)M1
(b) for a correct method to solve their quadratic, e.g. $(x - 1)(x - 4) = 0$, giving $x = 1$ and $x = 4$M1
(b) $(1, 4)$ and $(4, 1)$ with coordinates correctly pairedA1
Answer: (a) $x = 0$ is not allowed and $\tfrac4x \ne 0$ for any $x$ (b) $(1, 4)$ and $(4, 1)$
Examiner tip: Multiply every term by $x$ when clearing the fraction – a common slip is writing $4 = 5 - x^2$ instead of $4 = 5x - x^2$.

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