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Sec 1 Maths: Functions & Linear Graphs: practice questions & worked solutions

Functions and graphs questions from real Singapore Secondary 1 examination papers, each with a full, step-by-step worked solution.

12 questionsStep-by-step solutionsFree to access

About this topic & key methods

Practise Cartesian coordinates, drawing and reading straight-line graphs, finding gradients and intercepts, and interpreting linear functions in context. Attempt each complete exam question on paper before opening its worked solution. For guided practice, see our Secondary 1 Maths tuition.

Use graph paper and the scales stated in each question. The on-screen grids resize to fit your device, so their physical centimetre sizes are not fixed.

Key methods

Cartesian coordinates: read the horizontal coordinate first, then the vertical coordinate.
Gradient of a straight line: $m=\dfrac{y_2-y_1}{x_2-x_1}$ . Use coordinate values, especially when the axes have different scales.
Equation of a line: in $y=mx+c$ , $m$ is the gradient and $c$ is the $y$ -intercept.
Drawing a straight-line graph: calculate the table values, plot each point accurately and join them with a ruler.
Horizontal and vertical lines: $y=k$ is horizontal with gradient $0$ ; $x=k$ is vertical and its gradient is undefined.
Interpreting a linear function: the gradient is the rate of change and the vertical intercept is the value when the horizontal variable is zero.

Questions & worked solutions

Q1Temasek Secondary2023EOY-P2 Q8●●○ Standard6 marks

Drawing and reading straight-line graphs

In the grid provided, line $L$ has been drawn.

Grid with line L through (-1, -6) and (1.5, 4).

(a)

The equation of line $L$ is given as $y = 4x + p$ . State the value of $p$ .

(b)

The given table of values for another linear graph, $y = -2x + 1$ .

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

Calculate the value of $q$ .

(c)

By plotting all the points in table, draw the graph of $y = -2x + 1$ in the grid provided in the previous page.

(d)

Using your graph of $y = -2x + 1$ , find the value of $x$ when $y = 2$ .

(e)

The point $R$ lies on both $y = -2x + 1$ and line $L$ . Label point $R$ on the graph and write down the coordinates of the point $R$ .

Show worked solution▾

(a) The vertical intercept

Line $L$ crosses the $y$ -axis at $(0,-2)$ . At $x=0$ , $y=p$ , so $p=-2$ and $L$ has equation $y=4x-2$ .

(b) Complete the table

$\begin{aligned}-3&=-2q+1\\-4&=-2q\\q&=2\end{aligned}$

Plot $(-2,5)$ , $(1,-1)$ and $(2,-3)$ . Join them with a ruler to draw $y=-2x+1$ . The completed graph below shows the line through all three points.

(d) Read the value of x

Read across from $y=2$ to the descending line, then down to the $x$ -axis: $x=-0.5$ . Check:

$2=-2x+1\quad\Rightarrow\quad 1=-2x\quad\Rightarrow\quad x=-0.5.$

(e) Point of intersection

The two lines meet at $R(0.5,0)$ . Check by equating their expressions for $y$ :

$\begin{aligned}4x-2&=-2x+1\\6x&=3\\x&=0.5,\qquad y=0.\end{aligned}$

Completed graph: L is y=4x-2; y=-2x+1 meets L at R(0.5,0), with a reading guide at (-0.5,2).

Q2Tanjong Katong Secondary2021EOY-P2 Q6●●○ Standard7 marks

Finding the equation of a line from a table

Answer the whole of this question on a sheet of graph paper.

The values of $x$ and $y$ shown in the table below are related through a straight line.

$x$	$-1$	$0$	$1$	$2$
$y$	$8$	$5$	$2$	$-1$

(a)

Using a scale of $2$ cm to $1$ unit on the $y$ -axis and $4$ cm to $1$ unit on the $x$ -axis, plot the points given in the table above and join them with a straight line.

(b)

The point $(-0.5, d)$ lies on the graph. Use your graph to find the value of $d$ .

(c)

Using your graph, find the

(i)

gradient of the line,

(ii)

equation of the line.

Show worked solution▾

(a) Plot the graph

Use the stated scales: one $x$ -unit occupies twice the length of one $y$ -unit. Plot $(-1,8)$ , $(0,5)$ , $(1,2)$ and $(2,-1)$ , then join them with a ruler.

(b) Read a value from the graph

At $x=-0.5$ , the point on the line lies halfway between $(-1,8)$ and $(0,5)$ vertically. Hence $d=6.5$ .

(c)(i) Gradient

Use two clear points on the graph, such as $(0,5)$ and $(1,2)$ :

$m=\frac{2-5}{1-0}=-3.$

(c)(ii) Equation

The graph crosses the $y$ -axis at $5$ . Substitute $m=-3$ and $c=5$ into $y=mx+c$ :

$y=-3x+5.$

Check the reading in (b): $-3(-0.5)+5=6.5$ .

Graph through (-1,8), (0,5), (1,2) and (2,-1), with a guide showing y=6.5 when x=-0.5.

Q3Chung Cheng High (Yishun)2023EOY-P2 Q7●●○ Standard8 marks

Interpreting a decreasing linear function

The volume $V$ cm $^3$ , of an ice-cream at time $t$ minutes is given by $V = 50 – 5t$ for $0 \le t \le 10$ . Some corresponding values of $V$ and $t$ are given in the table below.

$t$	$0$	$2$	$4$	$6$	$8$	$10$
$V$	$50$	$40$	$p$	$20$	$10$	$0$

(a)

Calculate the value of $p$ .

(b)

Using a scale of $1$ cm to represent $1$ unit, draw a horizontal $t$ -axis for $0 \le t \le 10$ . Using a scale of $1$ cm to represent $5$ units, draw a vertical $V$ -axis for $0 \le V \le 50$ . On your axes, plot the points given in the table and draw the line $V = 50 – 5t$ for $0 \le t \le 10$ .

Blank grid with t from 0 to 10 and V from 0 to 50.

(c)

State the gradient of the graph.

(d)

Using your graph,

(i)

find the volume of the ice cream after $2.6$ minutes,

(ii)

find the time taken for the volume of the ice cream to be $17$ cm $^3$ .

(e)

Explain the meaning of the constant term $50$ , in the equation $V = 50 – 5t$ .

(f)

Explain why the function $V = 50 – 5t$ is not a good representation of the volume of the ice cream after $10$ minutes.

Show worked solution▾

(a) Missing table value

$p=50-5(4)=30.$

(b) Draw the graph

Use the stated scales and plot $(0,50)$ , $(2,40)$ , $(4,30)$ , $(6,20)$ , $(8,10)$ and $(10,0)$ . Join them with a straight line, stopping at $t=0$ and $t=10$ .

$m=\frac{0-50}{10-0}=-5.$

(d)(i) Volume after 2.6 minutes

Read up from $t=2.6$ to the line, then across to the $V$ -axis: $V=37\text{ cm}^3$ . Check: $50-5(2.6)=37$ .

(d)(ii) Time when the volume is 17 cubic centimetres

Read across from $V=17$ to the line, then down to the $t$ -axis: $t=6.6$ minutes. Check:

$17=50-5t\quad\Rightarrow\quad 5t=33\quad\Rightarrow\quad t=6.6.$

(e) Meaning of 50

At $t=0$ , $V=50$ . The initial volume of the ice-cream is $50\text{ cm}^3$ .

(f) Limitation of the function

At $t=10$ , the volume is zero. For $t>10$ , the formula gives a negative volume, which is physically impossible. The model therefore applies only to $0\le t\le10$ .

V=50-5t from (0,50) to (10,0), with reading guides at (2.6,37) and (6.6,17).

Q4Anglo-Chinese School (Barker Road)2023EOY-P1 Q14●●○ Standard7 marks

Gradient and fixed charge in a linear function

A company charges $\$y$ for $x$ units of electricity used.
The charge $\$y$ can be represented by the function $y=ux+20$ , where $u$ is a constant.
Some corresponding values of $x$ and $y$ are given in the table below.

$x$	$0$	$100$	$200$	$300$	$400$
$y$	$20$	$50$	$80$	$110$	$140$

(a)

On the grid, draw the graph of $y=ux+20$ for $0\le x\le 400$ .

Blank electricity-cost grid with x from 0 to 400 units and y up to 140 dollars.

(b)

Find the total cost if a household uses $140$ units of electricity.

(c)

What does the constant term, $20$ , represent?

(d)

Find the value of $u$ .

(e)

What does the gradient of the graph represent in this context?

Show worked solution▾

(a) Draw the graph

Plot $(0,20)$ , $(100,50)$ , $(200,80)$ , $(300,110)$ and $(400,140)$ . Join the points with a ruler for $0\le x\le400$ .

(b) Cost for 140 units

Read up from $x=140$ to the line and across to the cost axis. The cost is $\$62$ . To check, the cost rises by $\$30$ per $100$ units, so each unit costs $\$0.30$ :

$y=0.30(140)+20=62.$

The $20$ represents the fixed charge of $\$20$ , payable even when no electricity is used.

(d) Value of u

Using the two points $(0,20)$ and $(100,50)$ :

$u=\frac{50-20}{100-0}=\frac{30}{100}=0.3.$

(e) Meaning of the gradient

The gradient is the charge per unit of electricity: $\$0.30$ per unit, or $30$ cents per unit.

Electricity cost y=0.3x+20 through all five table points, with a reading guide at (140,62).

Q5Shuqun Secondary2017SA2-P2 Q9●●○ Standard9 marks

Intersection with a horizontal line

Answer the whole of this question on a sheet of graph paper.

The table below shows the corresponding $x$ and $y$ values for the equation:
$y = -2x + 3.$

$x$	$-2$	$-1$	$0$	$1$	$2$
$y$	$7$	$a$	$3$	$b$	$-1$

(a)

Find the values of $a$ and $b$ .

(b)

Using a scale of $4$ cm to represent $1$ unit on the $x$ -axis and $2$ cm to represent $1$ unit on the $y$ -axis, draw the graph of $y = -2x + 3$ for $-2 \le x \le 2$ .

Blank graph-paper grid for x from -2 to 2, with twice as much space per x-unit as per y-unit.

(c)

From the graph, find the value of $x$ when $y = 6$ .

(d)

Draw the line $y = 2$ and state its gradient.

(e)

Find the coordinates of the point where the two graphs intersect.

Show worked solution▾

(a) Complete the table

$\begin{aligned}a&=-2(-1)+3=5,\\b&=-2(1)+3=1.\end{aligned}$

(b) Draw the graph

Using the stated scales, plot $(-2,7)$ , $(-1,5)$ , $(0,3)$ , $(1,1)$ and $(2,-1)$ . Join these points with a straight line for $-2\le x\le2$ .

Read across from $y=6$ to the line, then down to the $x$ -axis: $x=-1.5$ . Check:

$6=-2x+3\quad\Rightarrow\quad -2x=3\quad\Rightarrow\quad x=-1.5.$

(d) Horizontal line

Draw a horizontal line through $(0,2)$ , labelled $y=2$ . Every point on this line has the same $y$ -coordinate, so the vertical change is zero and the gradient is $0$ .

(e) Point of intersection

The graphs meet at $(0.5,2)$ . Check by substituting $y=2$ :

$2=-2x+3\quad\Rightarrow\quad -2x=-1\quad\Rightarrow\quad x=0.5.$

Completed graph y=-2x+3 and horizontal line y=2, intersecting at (0.5,2), with a guide at (-1.5,6).

Q6Xinmin Secondary2018EOY-1E Q6●○○ Foundational2 marks

Equation of a line y = mx + c

(a)

Find the gradient of the line $L_1$ .

Line L1 crosses the x-axis at -8 and the y-axis at 4.

(b)

Hence state the equation of $L_1$ .

Show worked solution▾

(a) Gradient

The intercepts give two points on $L_1$ : $(-8,0)$ and $(0,4)$ .

$m=\frac{4-0}{0-(-8)}=\frac{4}{8}=\frac12.$

(b) Equation

The $y$ -intercept is $4$ . Substitute $m=\frac12$ and $c=4$ into $y=mx+c$ :

$y=\frac12x+4.$

Check the other intercept: when $x=-8$ , $y=\frac12(-8)+4=0$ .

Q7Chung Cheng High (Yishun)2023EOY-P2 Q4●○○ Foundational3 marks

Vertical lines and a negative gradient

The diagram shows three points $P(2, 6)$ , $Q(2, 4)$ and $R(6, 2)$ .

Triangle with vertices P(2,6), Q(2,4) and R(6,2).

(a)

Write down the equation of line $PQ$ .

(b)

Find the gradient of line $QR$ .

(c)

The line $PR$ has an equation $y + x = 8$ . A point $(4.25, a)$ lies on the line $PR$ . Find the value of $a$ .

Show worked solution▾

(a) Equation of PQ

Both $P(2,6)$ and $Q(2,4)$ have $x$ -coordinate $2$ . Thus $PQ$ is vertical and its equation is $x=2$ .

(b) Gradient of QR

$m=\frac{2-4}{6-2}=\frac{-2}{4}=-\frac12.$

Substitute $(4.25,a)$ into $y+x=8$ :

$a+4.25=8\quad\Rightarrow\quad a=3.75.$

Q8Anglo-Chinese School (Barker Road)2023EOY-P2 Q2●●○ Standard5 marks

Parallel lines and axial intercepts

$A$ is the point $(0,5)$ and the gradient of line $AB$ is $-1.5$ .

(a)

Find the equation of the line $AB$ .

(b)

The equation of line $l$ is $2y+3x=6$ .
Explain why line $l$ does not intersect line $AB$ .

(c)

Line $l$ passes through the $x$ -axis at point $C$ and $y$ -axis at point $D$ .
State the coordinates of

(i)

point $C$ ,

(ii)

point $D$ .

Show worked solution▾

(a) Equation of AB

Since $A=(0,5)$ , the $y$ -intercept is $5$ . With gradient $-1.5$ :

$AB:\quad y=-1.5x+5.$

(b) Why the lines do not intersect

Write the equation of $l$ in the form $y=mx+c$ :

$2y+3x=6\quad\Rightarrow\quad2y=-3x+6\quad\Rightarrow\quad y=-1.5x+3.$

Both lines have gradient $-1.5$ , but their $y$ -intercepts are different ( $5$ and $3$ ). They are distinct parallel lines, so they do not intersect.

(c)(i) x-intercept

On the $x$ -axis, $y=0$ :

$2(0)+3x=6\quad\Rightarrow\quad x=2.\qquad C=(2,0).$

(c)(ii) y-intercept

On the $y$ -axis, $x=0$ :

$2y+3(0)=6\quad\Rightarrow\quad y=3.\qquad D=(0,3).$

Q9Pasir Ris Secondary2021SA2-P1 Q8●○○ Foundational3 marks

Checking whether a point lies on a line

It is given that the equation of a line is $y = 8 – 5x$ .

(a)

State the $y$ -intercept of the line.

(b)

Does the line pass through the point $(-1.5, -20)$ ? Explain your answer.

Show worked solution▾

(a) y-intercept

At the $y$ -axis, $x=0$ . Hence $y=8-5(0)=8$ , so the $y$ -intercept is $8$ .

(b) Test the point

Substitute the proposed $x$ -coordinate, $-1.5$ , into the equation:

$y=8-5(-1.5)=8+7.5=15.5.$

The corresponding $y$ -value is $15.5$ , which is not $-20$ . Therefore the line does not pass through $(-1.5,-20)$ .

Q10CHIJ St. Theresa's Convent2017SA2-P1 Q12●●○ Standard5 marks

Cartesian coordinates and a parallelogram

(a)

Write down the coordinates of point $A$ .

(b)

$B$ is the point $(1, 4)$ and $C$ is the point $(5, 4)$ . Write down the equation of line $BC$ .

(c)

Calculate the area of triangle $ABC$ .

(d)

Find the coordinates of the point $D$ such that $ABCD$ is a parallelogram.

(e)

Calculate the area of parallelogram $ABCD$ .

Show worked solution▾

(a) Coordinates of A

Point $A$ is $2$ units left of the origin and $3$ units below it, so $A=(-2,-3)$ .

(b) Equation of BC

Both $B(1,4)$ and $C(5,4)$ have $y$ -coordinate $4$ . The horizontal line $BC$ has equation $y=4$ .

The base $BC=5-1=4$ units. The perpendicular height from $A$ to the horizontal line $BC$ is $4-(-3)=7$ units.

$\text{Area of }\triangle ABC=\frac12\times4\times7=14\text{ units}^2.$

(d) Coordinates of D

In parallelogram $ABCD$ , the move from $A$ to $D$ is the same as the move from $B$ to $C$ : $4$ units right with no vertical change. Thus

$D=(-2+4,-3)=(2,-3).$

(e) Area of the parallelogram

The base is $4$ units and the perpendicular height is $7$ units:

$\text{Area of }ABCD=4\times7=28\text{ units}^2.$

Completed parallelogram A(-2,-3), B(1,4), C(5,4), D(2,-3), with diagonal AC.

Q11Tanjong Katong Secondary2017SA1-P1 Q9●●● Challenging5 marks

Coordinates, gradient and triangle area

$A$ and $B$ are points on the coordinate plane as shown.

Coordinate plane with A at (0,-2) and B at (3,1).

(a)

Write down the coordinates of point $A$ .

(b)

Find the gradient of line $AB$ , given that $B\,(3,\,1)$ .

(c)

Given that $C\,(k,\,1)$ is a point on the coordinate plane such that the area of triangle $ABC$ is $9$ units $^2$ , find the possible values of $k$ .

Show worked solution▾

(a) Coordinates of A

Point $A$ lies on the $y$ -axis, $2$ units below the origin. Hence $A=(0,-2)$ .

(b) Gradient of AB

$m=\frac{1-(-2)}{3-0}=\frac33=1.$

Both $B(3,1)$ and $C(k,1)$ lie on the horizontal line $y=1$ . Use $BC$ as the base. The perpendicular height from $A$ to this line is $1-(-2)=3$ units.

$\frac12\times BC\times3=9\quad\Rightarrow\quad BC=6.$

Point $C$ can be $6$ units to either side of $B$ . Therefore

$k=3+6=9\quad\text{or}\quad k=3-6=-3.$

Both positions give base $6$ and height $3$ , so both give area $9\text{ units}^2$ .

Q12Temasek JC (IP1)2023IP1-EOY Q10●●● Challenging7 marks

Linear equations and an isosceles triangle

Solutions to this question using accurate drawing will not be accepted.

(i)

State the gradient and the $y$ -intercept of the line $2y=10-2x$ .

(ii)

Show that the point $(1,3)$ does not lie on the line $2y=10-2x$ .

(iii)

The line $2y=10-2x$ intersects the $y$ -axis at $A$ . It is given that the coordinates of $B$ is $(h,k)$ , where $h$ and $k$ are positive real numbers. The area of isosceles triangle $OAB$ , where $OB=AB$ , is $10$ units $^2$ . Find the value of $h$ and of $k$ .

Show worked solution▾

(i) Gradient and y-intercept

Divide every term by $2$ :

$2y=10-2x\quad\Rightarrow\quad y=5-x=-x+5.$

Hence the gradient is $-1$ and the $y$ -intercept is $5$ .

(ii) Test the point (1,3)

When $x=1$ , the line has $y=5-1=4$ . Since $4\ne3$ , the point $(1,3)$ does not lie on the line.

(iii) Find h and k

The intercept is $A=(0,5)$ and $O=(0,0)$ , so $OA=5$ units. Since $OB=AB$ , the perpendicular from $B$ to the base $OA$ bisects $OA$ . Thus $B$ is halfway up from $O$ to $A$ :

$k=\frac{0+5}{2}=\frac52.$

The perpendicular distance from $B(h,k)$ to the vertical base $OA$ is $h$ units, since $h>0$ . Use the triangle’s area:

$\frac12\times5\times h=10\quad\Rightarrow\quad h=4.$

Therefore $h=4$ and $k=\frac52$ . These values were obtained by calculation, as required.

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