Functions & graphs — Mathematics Grade 12

8 exam-style question sets on Functions & graphs (Paper 1), each with a hint and a fully worked answer. The app holds 40 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Exponential function on a restricted domain, straight line & inverseParabola & hyperbola: product inequality, discriminant & tangencyExponential functions & graphsHyperbola: asymptotes, symmetry & transformationParabola, straight line & tangencyLogarithmic function, its inverse & translationParabola & hyperbola system: interpretation, intercepts & tangentHyperbola & straight line: determining a, p and q

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Question 1

Exponential function on a restricted domain, straight line & inverseRoutine14 marksPaper 1

The sketch shows the graph of $f (x) = 3^{x} - 9$ , drawn for $- 1 \leq x < 3$ ONLY. The dashed line is the asymptote of $f$ .

The graph starts at $P$ , where $x = - 1$ , and ends at $Q$ , where $x = 3$ . $P$ is a point of the graph, but $Q$ is NOT.
$A$ and $B$ are the $y$ -intercept and the $x$ -intercept of $f$ respectively.

aWrite down the equation of the asymptote of $f$ . (1)
bDetermine the coordinates of $B$ . (2)
cDetermine the equation of the straight line that passes through $A$ and $B$ , in the form $y = m x + c$ . (3)
dCalculate the vertical distance between the straight line in QUESTION (c) and the graph of $f$ at $x = 1$ . (3)
eThe graph of $g (x) = f (x) + c$ has the $x$ -axis as its asymptote. Write down the equation of $g$ . (1)
fWrite down the domain of $g^{- 1}$ . (2)
gDetermine the equation of $g^{- 1}$ in the form $y = \dots$ (2)

This question has a diagram, shown in the app.

Hint

For $B$ solve $3^{x} = 9$ by writing $9$ as a power of $3$ . The domain of $g^{- 1}$ is the RANGE of $g$ : because $g$ is increasing, substitute the two endpoints $x = - 1$ and $x = 3$ into $g$ and keep the inclusive/exclusive symbols in the same order as the given domain.

Worked answer

a) $y = - 9$ ✓

b) $3^{x} - 9 = 0 \Rightarrow 3^{x} = 9 = 3^{2}$ , so $x = 2$ ✓ $B (2; 0)$ ✓

c) $A (0; f (0)) = A (0; - 8)$ ✓ Gradient $= \frac{0 - ( - 8 )}{2 - 0} = 4$ ✓ The $y$ -intercept is $- 8$ , so $y = 4 x - 8$ ✓

d) Line at $x = 1$ : $y = 4 (1) - 8 = - 4$ ✓ Curve at $x = 1$ : $f (1) = 3 - 9 = - 6$ ✓ Distance $= - 4 - (- 6) = 2$ units ✓

e) The asymptote $y = - 9$ must move up to $y = 0$ , so $c = 9$ : $g (x) = 3^{x}$ ✓

f) Domain of $g^{- 1}$ $=$ range of $g$ on $- 1 \leq x < 3$ . $g$ is increasing: $g (- 1) = 3^{- 1} = \frac{1}{3}$ (included, since $x = - 1$ is in the domain) and $g (3) = 27$ (NOT included) ✓ So the domain of $g^{- 1}$ is $\frac{1}{3} \leq x < 27$ , i.e. $x \in [\frac{1}{3}; 27)$ ✓

g) $g$ : $y = 3^{x}$ . Interchange $x$ and $y$ : $x = 3^{y}$ ✓ Hence $g^{- 1}$ : $y = lo g_{3} x$ ✓ (for $\frac{1}{9} \leq x < 27$ )

Modelled on Nov 2023 Paper 1, Question 4 — same skills, new scenario.

Question 2

Parabola & hyperbola: product inequality, discriminant & tangencyComplex18 marksPaper 1

The sketch shows the graphs of $f (x) = - 2 (x - 1)^{2} + 18$ and $g (x) = \frac{d}{x}$ .

$B$ is the turning point of $f$ , and the graph of $g$ passes through $B$ .
The graph of $f$ cuts the $x$ -axis at $A (- 2; 0)$ and $E (4; 0)$ , and cuts the $y$ -axis at $C$ .

aWrite down the coordinates of $B$ . (2)
bCalculate the coordinates of $C$ . (2)
cDetermine the value of $d$ . (1)
dWrite down the range of $g$ . (1)
eUse the graphs to solve for $x$ if $f (x) \cdot g (x) \leq 0$ . (3)
fConsider the straight line $h (x) = - 8 x + k$ , where $k$ is a real number. Determine the values of $k$ for which the graph of $h$ will NOT intersect the graph of $g$ . (5)
gFor one value of $k$ , the graph of $h$ touches the first-quadrant branch of $g$ at exactly one point, $R$ . Determine the value of $t$ for which the graph of $y = f (x) + t$ passes through $R$ . (4)

This question has a diagram, shown in the app.

Hint

Read $B$ straight from the turning-point form and substitute $x = 0$ for $C$ . For the product inequality, $g$ is negative for $x < 0$ , positive for $x > 0$ and never zero, so look for the intervals where $f$ has the OPPOSITE sign to $g$ (or is zero). For $h$ and $g$ : equate, multiply by $x$ , and use $Δ < 0$ for no intersection and $Δ = 0$ for a tangent; the sign of $k$ decides on which branch the point of contact lies.

Worked answer

a) $f (x) = - 2 (x - 1)^{2} + 18$ has the form $a (x - p)^{2} + q$ with $p = 1$ and $q = 18$ : $B (1; 18)$ ✓✓

b) $f (0) = - 2 (0 - 1)^{2} + 18$ ✓ $= 16$ , so $C (0; 16)$ ✓

c) $B (1; 18)$ lies on $g$ : $18 = \frac{d}{1}$ , so $d = 18$ ✓

d) $y \in R$ , $y \neq = 0$ ✓

e) $g (x) < 0$ for $x < 0$ and $g (x) > 0$ for $x > 0$ ; $g$ is never $0$ and is undefined at $x = 0$ . So $f (x) \cdot g (x) \leq 0$ where $f (x) \geq 0$ with $x < 0$ , or where $f (x) \leq 0$ with $x > 0$ : $- 2 \leq x < 0$ ✓✓ or $x \geq 4$ ✓

f) $- 8 x + k = \frac{18}{x}$ ✓ $\Rightarrow - 8 x^{2} + k x = 18 \Rightarrow 8 x^{2} - k x + 18 = 0$ ✓ No point of intersection: $Δ < 0$ ✓ $k^{2} - 4 (8) (18) < 0 \Rightarrow k^{2} - 576 < 0$ ✓ $\Rightarrow (k - 24) (k + 24) < 0 \Rightarrow - 24 < k < 24$ ✓

g) One point of contact: $Δ = 0$ , so $k = \pm 24$ . The equal roots are $x = \frac{k}{16}$ , which is positive (first quadrant) only when $k = 24$ ✓ $8 x^{2} - 24 x + 18 = 0 \Rightarrow 2 (2 x - 3)^{2} = 0 \Rightarrow x = \frac{3}{2} = 1, 5$ ✓ and $y = g (1, 5) = \frac{18}{1 , 5} = 12$ : $R (1, 5; 12)$ ✓ $f (1, 5) + t = 12 \Rightarrow 17, 5 + t = 12 \Rightarrow t = - 5, 5$ ✓

Modelled on Nov 2023 Paper 1, Question 5 — same skills, new scenario.

Question 3

Exponential functions & graphsRoutine9 marksPaper 1

The graph of $f (x) = a^{x} - 4$ , where $a > 0$ and $a \neq = 1$ , passes through the point $T (2; - 3, 75)$ .

aDetermine the value of $a$ . (2)
bWrite down the range of $f$ . (1)
cSketch the graph of $f$ . Clearly show the horizontal asymptote and ALL intercepts with the axes. (3)
dThe point $C$ on the graph of $f$ has a $y$ -coordinate of $4$ . The point $M$ is the image of $C$ after reflection about the line $y = x$ . Determine the coordinates of $M$ . (3)

Hint

Substitute the coordinates of the given point into the equation to find $a$ ; the horizontal asymptote $y = q$ controls the range, and reflection about $y = x$ swaps the coordinates of a point.

Worked answer

a) Substitute $T (2; - 3, 75)$ : $a^{2} - 4 = - 3, 75$ ✓ so $a^{2} = 0, 25$ and, since $a > 0$ , $a = 0, 5$ ✓

b) $y > - 4$ , $y \in R$ ✓

c) Decreasing exponential shape with horizontal asymptote $y = - 4$ ✓; $y$ -intercept: $f (0) = 1 - 4 = - 3$ , i.e. $(0; - 3)$ ✓; $x$ -intercept: $(\frac{1}{2})^{x} = 4 = (\frac{1}{2})^{- 2}$ , so $x = - 2$ , i.e. $(- 2; 0)$ ✓

d) $(\frac{1}{2})^{x} - 4 = 4$ ✓ $\Rightarrow (\frac{1}{2})^{x} = 8 = (\frac{1}{2})^{- 3} \Rightarrow x = - 3$ , so $C (- 3; 4)$ ✓ Reflection about $y = x$ swaps the coordinates: $M (4; - 3)$ ✓

Modelled on Nov 2024 Paper 1, Question 4 — same skills, new scenario.

Question 4

Hyperbola: asymptotes, symmetry & transformationComplex10 marksPaper 1

In the diagram, the graph of $f (x) = \frac{a}{x + p} + q$ is drawn. The domain of $f$ is $x \in R$ , $x \neq = 2$ . The line $g$ , defined by $g (x) = x - 5$ , is an axis of symmetry of $f$ . The graph of $f$ cuts the $y$ -axis at $A (0; 3)$ . The asymptotes of $f$ are shown as dashed lines.

aWrite down the value of $p$ . (1)
bDetermine the value of $q$ . (2)
cDetermine the value of $a$ . (2)
dFor which values of $x$ is $f (x) \geq 0$ ? (3)
eThe graph of $f$ is transformed to a graph $h$ that has the SAME domain and range as $f$ , but with $h^{'} (x) < 0$ for all $x \neq = 2$ . Describe ONE transformation of $f$ that produces $h$ . (2)

This question has a diagram, shown in the app.

Hint

Every axis of symmetry of a hyperbola passes through the intersection of its asymptotes, so evaluate $g$ at the vertical asymptote to find $q$ ; use the $x$ -intercept and the asymptote to read off where $f$ is non-negative.

Worked answer

a) The vertical asymptote is $x = 2$ , so $p = - 2$ ✓

b) The axes of symmetry of a hyperbola pass through the point where the asymptotes intersect ✓, so $q = g (2) = 2 - 5 = - 3$ ✓

c) Substitute $A (0; 3)$ : $\frac{a}{0 - 2} - 3 = 3$ ✓ so $\frac{a}{- 2} = 6$ , giving $a = - 12$ ✓

d) $x$ -intercept: $\frac{- 12}{x - 2} - 3 = 0 \Rightarrow x - 2 = - 4 \Rightarrow x = - 2$ ✓ On the branch left of the asymptote, $f$ rises from its $x$ -intercept towards the vertical asymptote; right of the asymptote $f (x) < - 3$ ✓ So $f (x) \geq 0$ for $- 2 \leq x < 2$ ✓

e) Reflect $f$ about its horizontal asymptote, the line $y = - 3$ ✓ (or equivalently about its vertical asymptote $x = 2$ ). This gives $h (x) = \frac{12}{x - 2} - 3$ , which has the same asymptotes, domain and range, and $h^{'} (x) < 0$ for all $x \neq = 2$ ✓

Modelled on Nov 2024 Paper 1, Question 5 — same skills, new scenario.

Question 5

Parabola, straight line & tangencyComplex15 marksPaper 1

In the diagram, the graphs of $f (x) = - x^{2} + 2 x + 8$ and a straight line $g$ are drawn. The graph of $f$ cuts the $x$ -axis at $A$ and $B$ . The line $g$ passes through $A$ and cuts the graph of $f$ again at $C (2; 8)$ . $E$ is a point on the graph of $f$ and $H$ is a point on the graph of $g$ such that $E H$ is parallel to the $y$ -axis, with $E$ and $H$ between $A$ and $C$ .

aDetermine the coordinates of $T$ , the turning point of $f$ . (3)
bShow that the equation of $g$ is given by $g (x) = 2 x + 4$ . (3)
cDetermine the maximum length of $E H$ . (4)
dThe graph of $f$ is shifted horizontally to form the graph of $k$ , where $k (x) = f (x + m)$ . Determine the value of $m$ for which the line $g$ is a tangent to the graph of $k$ . (5)

This question has a diagram, shown in the app.

Hint

Use $x = \frac{- b}{2 a}$ for the turning point; the length $E H$ is the vertical difference $f (x) - g (x)$ , itself a parabola, and for tangency set $k (x) = g (x)$ and make the discriminant zero.

Worked answer

a) $x = \frac{- b}{2 a} = \frac{- 2}{2 ( - 1 )} = 1$ ✓✓ and $f (1) = - 1 + 2 + 8 = 9$ , so $T (1; 9)$ ✓

b) $x$ -intercepts of $f$ : $- (x - 4) (x + 2) = 0$ , so $x = - 2$ or $x = 4$ , giving $A (- 2; 0)$ ✓ Gradient of $g$ : $\frac{8 - 0}{2 - ( - 2 )} = 2$ ✓ Then $y = 2 (x + 2)$ , so $g (x) = 2 x + 4$ ✓

c) $E H = f (x) - g (x) = (- x^{2} + 2 x + 8) - (2 x + 4)$ ✓ $= - x^{2} + 4$ ✓ This is a maximum at $x = 0$ ✓, so the maximum length of $E H$ is $4$ units ✓

d) $k (x) = - (x + m)^{2} + 2 (x + m) + 8$ ✓ Equate to $g$ : $- (x + m)^{2} + 2 (x + m) + 8 = 2 x + 4$ , which simplifies to $x^{2} + 2 m x + (m^{2} - 2 m - 4) = 0$ ✓ For a tangent, $Δ = 0$ : $(2 m)^{2} - 4 (m^{2} - 2 m - 4) = 0$ ✓ $\Rightarrow 8 m + 16 = 0$ ✓ $\Rightarrow m = - 2$ ✓

Modelled on Nov 2024 Paper 1, Question 6 — same skills, new scenario.

Question 6

Logarithmic function, its inverse & translationRoutine10 marksPaper 1

In the diagram, the graph of $f (x) = lo g_{\frac{1}{5}} x$ is drawn. The point $A (25; t)$ lies on the graph of $f$ .

aWrite down the value of $t$ . (1)
bWrite down the coordinates of the point at which $f$ cuts the $x$ -axis. (1)
cDetermine, in the form $y = \dots$ , the equation of the inverse $f^{- 1}$ . (2)
d $f^{- 1}$ has ONE asymptote. Write down its equation. (1)
eDraw a sketch graph of $f^{- 1}$ . Indicate the $y$ -intercept and ONE other point on this graph. (3)
fThe graph of $h$ is formed when the graph of $f^{- 1}$ is shifted ONE unit to the LEFT. Determine the range of $h$ for $x \geq - 3$ . (2)

This question has a diagram, shown in the app.

Hint

For a base between $0$ and $1$ the log curve falls, and so does its inverse $y = (\frac{1}{5})^{x}$ ; a horizontal shift leaves the asymptote $y = 0$ unchanged, so evaluate the shifted graph at the boundary $x$ -value to find the top of the range.

Worked answer

a) $t = lo g_{\frac{1}{5}} 25 = - 2$ , because $(\frac{1}{5})^{- 2} = 25$ ✓

b) $f (1) = 0$ , so the graph cuts the $x$ -axis at $(1; 0)$ ✓

c) Interchange $x$ and $y$ : $x = lo g_{\frac{1}{5}} y$ ✓ $\Rightarrow f^{- 1} : y = (\frac{1}{5})^{x}$ ✓

d) $y = 0$ (the $x$ -axis) ✓

e) Decreasing exponential shape lying above the $x$ -axis and approaching $y = 0$ to the right ✓; $y$ -intercept $(0; 1)$ ✓; one further point, e.g. $(- 1; 5)$ ✓

f) $h (x) = (\frac{1}{5})^{x + 1}$ ✓ $h$ is decreasing with asymptote $y = 0$ , so on $x \geq - 3$ the greatest value is $h (- 3) = (\frac{1}{5})^{- 2} = 25$ and it is attained: $0 < y \leq 25$ ✓

Modelled on Nov 2025 Paper 1, Question 4 — same skills, new scenario.

Question 7

Parabola & hyperbola system: interpretation, intercepts & tangentComplex18 marksPaper 1

The sketch shows the graphs of a parabola $f$ and of $g (x) = \frac{8}{x - 2} + 8$ . The dashed lines are the asymptotes of $g$ ; they cross at $P$ , which is also the maximum turning point of $f$ . The two graphs meet at the point $D (0; 4)$ . $M$ is the smaller $x$ -intercept of $f$ and $T$ is the $x$ -intercept of $g$ .

aWrite down the domain of $g$ . (1)
bWrite down the range of $f$ . (1)
cFor which values of $x$ is $g (x) \leq f (x)$ ? (2)
dFor which values of $x$ is $f (x) < 4$ ? (2)
eShow that $f (x) = - x^{2} + 4 x + 4$ . (3)
fCalculate the length of $M T$ . Give your answer correct to TWO decimal places. (6)
gDetermine the equation of the tangent to $f$ at $D$ . (3)

This question has a diagram, shown in the app.

Hint

The equation of $g$ hands you both asymptotes, and therefore the turning point of $f$ ; write $f$ in vertex form through $D$ , then solve each $x$ -intercept from its own equation before subtracting.

Worked answer

a) $g$ is undefined only at its vertical asymptote: $x \in R$ , $x \neq = 2$ ✓

b) The asymptotes of $g$ cross at $(2; 8)$ , so $P (2; 8)$ is the maximum turning point of $f$ : $y \leq 8$ ✓

c) From the sketch, the hyperbola lies on or below the parabola between $D$ and the vertical asymptote ✓: $g (x) \leq f (x)$ for $0 \leq x < 2$ ✓

d) $f (0) = 4$ at $D$ , and by symmetry about $x = 2$ the parabola returns to a height of $4$ at $x = 4$ ✓, so $f (x) < 4$ for $x < 0$ or $x > 4$ ✓

e) $f (x) = a (x - 2)^{2} + 8$ ✓ Substitute $D (0; 4)$ : $4 a + 8 = 4$ , so $a = - 1$ ✓ Hence $f (x) = - (x - 2)^{2} + 8 = - x^{2} + 4 x + 4$ ✓

f) At $M$ : $- (x - 2)^{2} + 8 = 0 \Rightarrow (x - 2)^{2} = 8$ ✓ $\Rightarrow x = 2 \pm 22$ ✓ $M$ is the smaller intercept: $x_{M} = 2 - 22 \approx - 0, 83$ ✓ At $T$ : $\frac{8}{x - 2} = - 8$ ✓ $\Rightarrow x - 2 = - 1$ , so $x_{T} = 1$ ✓ $M T = 1 - (2 - 22) = 22 - 1 \approx 1, 83$ units ✓

g) $f^{'} (x) = - 2 x + 4$ ✓, so the gradient of the tangent is $f^{'} (0) = 4$ ✓ Through $D (0; 4)$ : $y = 4 x + 4$ ✓

Modelled on Nov 2025 Paper 1, Question 5 — same skills, new scenario.

Question 8

Hyperbola & straight line: determining a, p and qProblem-solving8 marksPaper 1

In the sketch, the hyperbola $f (x) = \frac{a}{x + p} + q$ and the straight line $g (x) = x + c$ are drawn. The dashed lines are the asymptotes of $f$ .

$g$ cuts the $x$ -axis at the point where the vertical asymptote of $f$ meets the $x$ -axis.
$g$ crosses the horizontal asymptote of $f$ at $K$ , the point on $g$ where $x = 2$ .
$f$ and $g$ cut the $y$ -axis at the SAME point.
$R (3; 4)$ lies on the graph of $f$ .

aWrite down, in terms of $p$ , the coordinates of the $x$ -intercept of $g$ . (1)
bDetermine the equation of $f$ . (5)
cThe graph of $g$ is translated so that its image is the axis of symmetry of $f$ that has a POSITIVE gradient. Describe this translation. (2)

This question has a diagram, shown in the app.

Hint

Because $g$ passes through $(- p; 0)$ we get $c = p$ ; use $K$ to write $q$ in terms of $p$ , the common $y$ -intercept to write $a$ in terms of $p$ , then substitute $R$ into $f$ to obtain a quadratic in $p$ .

Worked answer

a) $g$ passes through the point where the vertical asymptote $x = - p$ meets the $x$ -axis, so the $x$ -intercept of $g$ is $(- p; 0)$ ✓

b) $g$ cuts the $x$ -axis at $(- p; 0)$ : $0 = - p + c$ , so $c = p$ and $g (x) = x + p$ ✓ $K$ lies on $g$ and on the asymptote $y = q$ : $q = g (2) = 2 + p$ ✓ Equal $y$ -intercepts: $f (0) = g (0)$ : $\frac{a}{p} + 2 + p = p$ , so $a = - 2 p$ ✓ Substitute $R (3; 4)$ : $\frac{- 2 p}{3 + p} + 2 + p = 4 \Rightarrow - 2 p + (2 + p) (3 + p) = 4 (3 + p) \Rightarrow p^{2} - p - 6 = 0 \Rightarrow (p - 3) (p + 2) = 0$ ✓ The sketch places the vertical asymptote LEFT of the $y$ -axis, so $p > 0$ : $p = 3$ , $q = 5$ , $a = - 6$ and $f (x) = \frac{- 6}{x + 3} + 5$ ✓

c) The axes of symmetry of $f$ pass through the intersection of the asymptotes, $(- 3; 5)$ ; the increasing one is $y = (x + 3) + 5 = x + 8$ ✓ Since $g$ : $y = x + 3$ , the required translation is $5$ units UP (or, equivalently, $5$ units to the LEFT) ✓

Modelled on Nov 2025 Paper 1, Question 6 — same skills, new scenario.

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