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Sec 4 E Maths: Graphs of Functions, practice questions & worked solutions

Graphs of functions questions selected from 2023 to 2025 Singapore school E Maths prelim papers, each with the school’s full worked solution.

12 questionsStep-by-step solutionsFree to access

About this topic & key methods

Graphs in Sec 3 and Sec 4 E Maths cover quadratic, cubic, reciprocal and exponential curves, graphs in real situations such as speed-time graphs, and solving equations with a plotted curve. Most questions here are short Paper 1 items, with two Paper 2 questions that ask you to plot a curve, draw a tangent and add a straight line.

Attempt each question on paper before opening its worked solution. The solutions are the schools’ own marking schemes, so some steps are brief.

Key methods for graphs of functions

Complete the square: $y=(x-h)^2+k$ has turning point $(h,k)$ and line of symmetry $x=h$
For $y=-(x-a)(x-b)$ the curve cuts the $x$ -axis at $a$ and $b$ , opens downwards and turns at $x=\dfrac{a+b}{2}$
For $y=ka^x$ , the $y$ -intercept gives $k$ ; substitute a second point to find $a$
Know the shapes of $y=x^2$ , $y=x^3$ , $y=\dfrac kx$ , $y=\dfrac k{x^2}$ and $y=a^x$ , and how adding a constant shifts them
On a speed-time graph, the gradient is the acceleration and the area underneath is the distance travelled
To solve an equation with a plotted curve, rearrange it so one side is the curve's equation; the other side is the line to draw
For the gradient of a curve, draw a tangent at the point and use two points far apart on it

For lessons that build these methods week by week, see our E Maths tuition for Sec 3 and Sec 4.

Graphs of Functions questions with worked solutions

Unless the question specifies otherwise, give non-exact answers to 3 significant figures and angles in degrees to 1 decimal place.

Q1Dunman Secondary School2025Prelim P1 Q19●○○ Foundational2 marks

Finding the constants of an exponential graph, Dunman Secondary School, 2025 Prelim P1 Q19

The sketch shows the graph of $y=ka^x$ , where $a>0$ .

Dunman Secondary School 2025 P1 Q19: diagram for the question

Find the values of $k$ and $a$ . [2]

Show worked solution▾

$16=ka^0\quad\Rightarrow\quad k=16$

$64=16a^{-2}$

$a^2=\frac14$

$a=\frac12\quad\text{(rej. -ve)}$

Answer: $k=16,\quad a=\dfrac12$

Q2Tanjong Katong Girls’ School2025Prelim P1 Q16●○○ Foundational3 marks

Sketching a quadratic from its factorised form, Tanjong Katong Girls’ School, 2025 Prelim P1 Q16

Sketch the graph $y=-(x+4)(x-6)$ on the axes below. Indicate clearly the equation of the line of symmetry, the coordinates of the turning point and the points where the graph crosses the axes. [3]

Tanjong Katong Girls’ School 2025 P1 Q16: diagram for the question

Show worked solution▾

Tanjong Katong Girls’ School 2025 P1 Q16: worked solution diagram

Eqn of line of symmetry $x=1$ .

Coordinates of the turning point $(1,25)$ ,

$x$ -intercepts $(-4,0)$ , $(6,0)$

and $y$ -intercept $(0,24)$ .

Answer: $x=1;\ (1,25),\ (-4,0),\ (6,0),\ (0,24)$

Q3Methodist Girls’ School2025Prelim P1 Q12●○○ Foundational4 marks

Completing the square to find the turning point, Methodist Girls’ School, 2025 Prelim P1 Q12

Express $x^2-6x+8$ in the form of $a(x+b)^2+c$ . [2]
The curve $y=x^2-6x+8$ is drawn.
Using your answer in (a), write down
1. the coordinates of the turning point, [1]
2. the equation of the line of symmetry of the curve. [1]

Show worked solution▾

(a)

$\begin{aligned}x^2-6x+8&=x^2-6x+(-3)^2+8-(-3)^2\\&=(x-3)^2+8-9\\&=(x-3)^2-1\end{aligned}$

(b)

(i) $(3,-1)$

(ii) $x=3$

Answer: (a) $(x-3)^2-1$ (b)(i) $(3,-1)$ (ii) $x=3$

Q4Fairfield Methodist School (Secondary)2023Prelim P1 Q15●○○ Foundational4 marks

Matching equations to graph shapes, Fairfield Methodist School (Secondary), 2023 Prelim P1 Q15

Write down a possible equation for each of the graph below.
In each case, select one of the equations from the table below.

Fairfield Methodist School (Secondary) 2023 P1 Q15: diagram for the question

Show worked solution▾

(a)

$y=x^2+3$

(b)

$y=x^3+3$

(c)

$y=3^x+3$

(d)

$y=3x^{-2}$

Answer: (a) $y=x^2+3$ (b) $y=x^3+3$ (c) $y=3^x+3$ (d) $y=3x^{-2}$

Q5Anglo-Chinese School (Independent)2024Prelim P1 Q5●●○ Standard3 marks

Quadratic equation from the line of symmetry, Anglo-Chinese School (Independent), 2024 Prelim P1 Q5

Part of the graph of a quadratic function is shown below.

Anglo-Chinese School (Independent) 2024 P1 Q5: diagram for the question

The graph passes through the point $A\ (1,15)$ and it cuts the $x$ -axis at the points $B$ and $C$ .
Given that $C$ is $(6,0)$ and $x=2$ is the line of symmetry of the graph,

write down the equation of the quadratic curve in the form $y=a(x-h)^2+k$ . [2]
find the positive value of $x$ when $y=4$ , leaving your answer in exact form. [1]

Show worked solution▾

(a)

$B\ (-2,0)$
$y=a(x+2)(x-6)$
Substitute $(1,15)$ into eq to find $a$ : $a=-1$
Find maximum coordinate: when $x=2$ , $y=16$ . Equation: $y=-(x-2)^2+16$ .

(b)

$4=-(x-2)^2+16$
$x=2+\sqrt{12}$ or $2+2\sqrt3$ (positive value).

Answer: (a) $y=-(x-2)^2+16$ (b) $2+2\sqrt3$

Q6Bukit Merah Secondary School2025Prelim P1 Q19●●○ Standard5 marks

Overtaking on a speed time graph, Bukit Merah Secondary School, 2025 Prelim P1 Q19

The diagram below shows Car $A$ travelling at a uniform speed of 60 m/s.
Car $B$ accelerated uniformly from rest to reach a speed of 80 m/s in 10s and continued travelling at the same speed thereafter.

Bukit Merah Secondary School 2025 P1 Q19: diagram for the question

Find the acceleration of Car $B$ from 0 to 10s. [1]
Find the time when both cars were travelling at the same speed. [2]
Find the time when Car $B$ overtook Car $A$ . [2]

Show worked solution▾

(a)

$\text{Acceleration}=\frac{80-0}{10-0}=8\text{ m/s}^2$

(b)

Let $t$ be the time when both cars were travelling at the same speed.

$\frac t{10}=\frac{60}{80}$

$t=7.5\text{ s}$

(c)

Let $T$ be the time when Car $B$ overtook Car $A$ .

$\text{Distance travelled}=\frac12\times10\times80+(T-10)(80)=60T$

$80T-400=60T$

$T=20\text{ s}$

Answer: $\text{(a) }8\text{ m/s}^2\quad\text{(b) }7.5\text{ s}\quad\text{(c) }20\text{ s}$

Q7Ngee Ann Secondary School2025Prelim P1 Q18●●○ Standard5 marks

Minimum value from the completed square form, Ngee Ann Secondary School, 2025 Prelim P1 Q18

$y=(x-h)^2+k$

Explain why $y$ has its minimum value when $x=h$ . [1]
The roots of the equation $(x-h)^2+k=0$ are $-2$ and 4.
Explain why the value of $h$ is 1. [1]
Find the minimum value of $y$ . [1]
Sketch the graph of $y$ on the axes below.
Indicate clearly the coordinates where the graph crosses the $x$ -axis and the minimum point on the curve.
[2]

Show worked solution▾

(a)

When $x\ne h$ , $(x-h)^2>0$ and $y=(x-h)^2+k>k$ .

When $x=h$ , $(x-h)^2=0$ and $y=(x-h)^2+k=k$ .

So, $y$ has its minimum value of $k$ when $x=h$ .

(b)

Since the graph of $y=(x-h)^2+k$ is symmetric about the vertical line passing through its minimum point,

$h=\frac{-2+4}{2}$

$h=1\text{ (shown)}$

(c)

Sub $x=4$ , $y=0$ , $h=1$ ,

$0=(4-1)^2+k$

$k=-9$

Min. value of $y$ is $-9$ .

(d)

Ngee Ann Secondary School 2025 P1 Q18: worked solution diagram

Answer: (a) $(x-h)^2\ge0$ , so $y\ge k$ , with $y=k$ only when $x=h$ (b) the line of symmetry is midway between the roots, $h=\dfrac{-2+4}{2}=1$ (c) $-9$ (d) U-shaped curve crossing the $x$ -axis at $(-2,0)$ and $(4,0)$ , minimum point $(1,-9)$

Q8CHIJ St. Nicholas Girls’ School2023Prelim P1 Q22●●○ Standard6 marks

Acceleration and distance from a speed time graph, CHIJ St. Nicholas Girls’ School, 2023 Prelim P1 Q22

The diagram shows the speed-time graph of a car journey.

The car accelerates for 10 seconds reaching a speed of 12 m/s. It then accelerates for another 12 seconds and then travels at a constant speed of 20 m/s.

CHIJ St. Nicholas Girls’ School 2023 P1 Q22: diagram for the question

Calculate the speed of the car when $t=13$ . [2]
Calculate the distance travelled in the first 22 seconds. [2]
A motor cyclist starts 13 seconds later from the same place as the car, and accelerates uniformly for 9 seconds until it reaches a speed of 25 m/s.
Is the motor cyclist behind the car at $t\,=22$ ?
Justify your decision. [2]

Show worked solution▾

(a)

$\begin{aligned}\text{acceleration}&=\dfrac{v-12}{13-10}=\dfrac{20-12}{22-10}\\\dfrac{v-12}3&=\dfrac8{12}\\v-12&=2\\\therefore v&=14\end{aligned}$

(b)

$\begin{aligned}\text{distance}&=\dfrac12\times10\times12+\dfrac12(20+12)\times12\\&=60+192\\&=252\end{aligned}$

(c)

distance by car $=252$ m

distance by motor cyclist $=\dfrac12\times9\times25$

distance by motor cyclist $=112.5$ m $<252$ m

$\therefore$ motor cyclist is behind the car at t $=22$

Answer: (a) 14 m/s (b) 252 m (c) 112.5 m $<$ 252 m, behind the car

Q9Swiss Cottage Secondary School2024Prelim P1 Q24●●○ Standard6 marks

Solving cubic equations by drawing a straight line, Swiss Cottage Secondary School, 2024 Prelim P1 Q24

The graphs of $y=\dfrac{x^3}{2}-4x-1$ and $y=1-x$ are drawn on the grid.

Swiss Cottage Secondary School 2024 P1 Q24: diagram for the question

The points of intersection of the curve and the straight line give the solutions of a cubic equation. Find the cubic equation, giving your answer in the form $ax^3+bx^2+cx+d=0$ , where $a$ , $b$ , $c$ and $d$ are integers. [1]
By drawing a tangent, find the gradient of the curve at $(-2,3)$ . [2]
The equation $x^3-4x=0$ 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 $x^3-4x=0$ . [2]

Show worked solution▾

(a)

$\begin{aligned}\frac{x^3}{2}-4x-1&=1-x\\\frac{x^3}{2}-3x-2&=0\\x^3-6x-4&=0\end{aligned}$

(b)

Draw tangent
Gradient $=2$

(c)(i)

$\begin{aligned}x^3-4x&=0\\\frac{x^3}{2}-2x&=0\\\frac{x^3}{2}-4x-1&=-2x-1\end{aligned}$

Equation of line: $y=-2x-1$

(c)(ii)

Swiss Cottage Secondary School 2024 P1 Q24: worked solution diagram

Draw $y=-2x-1$
$x=-2,0,2$

Answer: (a) $x^3-6x-4=0$ (b) $2$ (c)(i) $y=-2x-1$ (ii) $x=-2,0,2$

Q10Broadrick Secondary School2024Prelim P1 Q26●●● Challenging7 marks

Speed time and distance time graphs with clock times, Broadrick Secondary School, 2024 Prelim P1 Q26

The diagrams show the speed-time graph and the corresponding distance-time graph of a car.
The car travelled from a point $P$ to $Q$ and in the journey, its greatest speed attained was 1.5 km/min.

Broadrick Secondary School 2024 P1 Q26: diagram for the question

Convert 1.5 km/min to m/s. [1]
Show that $v$ , the initial speed, was 0.6 km/min. [2]
Find the speed of the car, in m/s, at 0812. [2]
The total distance travelled by the car is 92.25 km.
Find the time, $T$ , when it comes to a complete stop. [2]

Show worked solution▾

(a)

1.5 km/min
$=\dfrac{1.5\times1000}{1\times60}=25$ m/s

(b)

Area = dist travelled

$\begin{aligned}21&=\frac12(v+1.5)(20)\\21&=10(v+1.5)\\2.1&=v+1.5\\v&=0.6\end{aligned}$

(c)

$\begin{aligned}\frac{\textit{speed}-0.6}{12}&=\frac{1.5-0.6}{20}\\\frac{\textit{speed}-0.6}{12}&=0.045\\\textit{speed}&=1.14\text{ km/min}\end{aligned}$

$=19$ m/s

$\begin{aligned}\textit{acc}&=\frac{1.5-0.6}{20}=0.045\\\textit{speed}&=0.6+12(0.045)=1.14\text{ km/min}=19\text{ m/s}\end{aligned}$

(d)

Area = dist travelled

$\begin{aligned}92.25&=\frac12(0.6+1.5)(20)+1.5(40)+\frac12(t)(1.5)\\92.25&=21+60+0.75t\\11.25&=0.75t\\t&=15\text{ min}\end{aligned}$

$T$ is 0915

Answer: (a) $25$ m/s (c) $19$ m/s (d) $T$ is 0915

Q11Anglo-Chinese School (Barker Road)2023Prelim P2 Q3●●● Challenging9 marks

Exponential growth graph and the gradient as a rate, Anglo-Chinese School (Barker Road), 2023 Prelim P2 Q3

The number of micro-organisms, $y$ , in a test tube after $t$ days is shown in the table below. The relationship between $y$ and $t$ can be modelled by the equation $y=10(2^t)$ .

Find the value of $q$ . [1]
On the grid opposite, draw the graph of $y=10(2^t)$ for $0\le t\le6$ . [3]
1. By drawing a tangent, find the gradient of the curve when $t=3$ . [2]
2. What does this gradient represent? [1]
The number of micro-organisms in another test tube is given by the equation
$y=-80t+600$ .
By drawing the line $y=-80t+600$ , find the number of days for the number of micro-organisms in the two test tubes to be the same. [2]

Show worked solution▾

(a)

$160$

(b)

All 7 points plotted
Smooth curve through plotted points.

Anglo-Chinese School (Barker Road) 2023 P2 Q3: worked solution diagram

(c)(i)

Drawing of tangent at $t=3$
$55.5$

(c)(ii)

The micro organisms are increasing at rate of 55.5 per day after 3 days.

(d)

Drawing of line $y=-80t+600$
$4.6$

Answer: (a) $160$ (c)(i) about $55.5$ (ii) the rate at which the number of micro-organisms is increasing when $t=3$ , about $55.5$ per day (d) $4.6$ days

Q12Greenridge Secondary School2024Prelim P2 Q7●●● Challenging11 marks

Plotting a reciprocal curve, tangent and line intersections, Greenridge Secondary School, 2024 Prelim P2 Q7

Complete the table of values for $y=2x+\dfrac9x-11$ .
Values are given to 1 decimal place where appropriate.
[1]
On the grid, draw the graph of $y=2x+\dfrac9x-11$ for $0\le x\le5$ .
[3]
By drawing a tangent, find the gradient of the curve at $(1.5,-2)$ . [2]
1. On the same axes, draw the graph of the straight line $2y=12-3x$ . [1]
2. Write down the $x$ -coordinates of the points where the graphs meet. [2]
3. These values of $x$ are the solutions of the equation $7x^2+Ax+B=0$ .
  Find the value of $A$ and value of $B$ . [2]