Sec 2 Maths: Linear & Quadratic Graphs, practice questions & worked solutions

Q: How do I find the line of symmetry of a quadratic graph?

Find two points on the curve with the same y-value, such as the two x-intercepts, and take the x-value halfway between them. For y = (x+5)(3-x) the intercepts are -5 and 3, so the line of symmetry is x = -1.

Q: How do I tell whether a parabola has a maximum or a minimum point?

Look at the sign of the x² term. If it is positive the curve is U-shaped with a minimum point; if it is negative the curve is n-shaped with a maximum point.

Q: What do the points where a line meets a curve tell me?

Their x-coordinates are the solutions of the equation formed by setting the two expressions equal. For example, y = x + 1 meets y = x² - 6x + 7 at x = 1 and x = 6, which solve x² - 7x + 6 = 0.

Linear and quadratic graph questions from 2025 Singapore Secondary 2 End-of-Year examination papers. They cover the gradient and equation of a straight line, intercepts and the line of symmetry of a parabola, and plotting quadratic graphs from a table of values to read off maximum points and solutions, and every question has a full step-by-step worked solution.

About this topic & key methods

In Secondary 2 Mathematics, graphs connect algebra to pictures. A linear graph $y=mx+c$ is a straight line with gradient $m$ and $y$ -intercept $c$ . A quadratic graph $y=ax^2+bx+c$ is a parabola: it opens upwards with a minimum point when $a\gt 0$ , and downwards with a maximum point when $a\lt 0$ . Every parabola has a vertical line of symmetry through its turning point.

The questions below start with straight lines, move on to reading intercepts and the line of symmetry of a sketched parabola, then ask you to complete a table of values, plot the curve, draw a straight line on the same grid and read off where they meet. The last group sets the curve in a real context: a ball, an arrow, a bird in a game and a garden. For a structured programme covering this topic, see our Sec 2 Maths tuition.

Key methods

Gradient: $m=\dfrac{y_2-y_1}{x_2-x_1}$ ; the equation of the line is $y=mx+c$ , where $c$ is the $y$ -intercept.
Intercepts: put $y=0$ to find where a graph cuts the $x$ -axis, and $x=0$ to find where it cuts the $y$ -axis.
Line of symmetry: for a parabola it is the vertical line halfway between the two $x$ -intercepts, or halfway between any two points with the same $y$ -value.
Table of values: substitute each $x$ carefully, with brackets around negative values, e.g. $-(-5)^2=-25$ .
Plotting: plot every point from the table and join them with one smooth curve, not straight segments; label the graph.
Reading a graph: draw a horizontal or vertical line to the curve and read the other coordinate; the $x$ -coordinates where a line meets the curve solve the equation formed by setting the two expressions equal.

Questions & worked solutions

Q1Woodgrove Secondary2025EOY P1 Q7●○○ Foundational3 marks

Table of values and plotting a straight line

(a)

Complete the table of values for the linear function $y=2x+1$ .

$x$	$-0.5$	$0$	$1$	$2.5$
$y$		$1$	$3$	$6$

(b)

Plot the graph of $y=2x+1$ on the grid below and label the line.

Blank axes with x from -2 to 4 and y from -2 to 8

Show worked solution▾

(a)

Substitute $x=-0.5$ :

$\begin{aligned}y&=2(-0.5)+1\\ &=-1+1\\ &=0\end{aligned}$

(b)

Plot the four points $(-0.5,0)$ , $(0,1)$ , $(1,3)$ and $(2.5,6)$ , join them with a ruler into one straight line, and label it $y=2x+1$ . The line crosses the $y$ -axis at $1$ and rises $2$ units for every $1$ unit across.

Straight line y = 2x + 1 crossing the x-axis at -0.5 and the y-axis at 1

Q2St Margaret’s Secondary2025EOY Q7●○○ Foundational3 marks

Gradient and equation of a straight line

The line $L$ passes through the point $A(0,2)$ and $B(7,-2)$ .

Line L on a grid passing through A(0, 2) on the y-axis and B(7, -2)

(a)

Find the gradient of the line $L$ .

(b)

Write down the equation of line $L$ .

Show worked solution▾

(a)

$\begin{aligned}\text{Gradient}&=\frac{y_2-y_1}{x_2-x_1}\\ &=\frac{-2-2}{7-0}\\ &=-\frac47\end{aligned}$

(b)

The line passes through $A(0,2)$ , so the $y$ -intercept is $c=2$ . Using $y=mx+c$ :

$\begin{aligned}y&=-\frac47x+2\end{aligned}$

This can also be written as $7y=-4x+14$ .

Q3Queensway Secondary2025EOY P1 Q10●●○ Standard5 marks

Intersection of two lines and the area of a triangle

In the diagram, the equation of line $p$ is $2y=x+5$ and the equation of line $q$ is $y+2x=0$ . Line $p$ and line $q$ intersect at point $C$ .

Line p, 2y = x + 5, meeting the x-axis at A, and line q, y + 2x = 0, through the origin O, crossing at C

(a)

By solving the simultaneous equations $y+2x=0$ and $2y=x+5$ , show that the coordinates of $C$ is $(-1,2)$ .

(b)

It is also given that line $p$ intersects the $x$ -axis at point $A$ and line $q$ passes through $(0,0)$ . Calculate the area of triangle $OAC$ .

Show worked solution▾

(a)

From line $q$ , $y=-2x$ . Substitute into $2y=x+5$ :

$\begin{aligned}2(-2x)&=x+5\\ -4x&=x+5\\ -5x&=5\\ x&=-1\end{aligned}$
$\begin{aligned}y&=-2(-1)\\ &=2\end{aligned}$

So $C$ is $(-1,2)$ (shown).

(b)

At $A$ , $y=0$ on line $p$ :

$\begin{aligned}2(0)&=x+5\\ x&=-5\end{aligned}$

So $OA=5$ units along the $x$ -axis. The height of triangle $OAC$ above $OA$ is the $y$ -coordinate of $C$ , which is $2$ :

$\begin{aligned}\text{Area of }\triangle OAC&=\frac12\times5\times2\\ &=5\text{ units}^2\end{aligned}$

Q4St Margaret’s Secondary2025EOY Q12●●○ Standard4 marks

Intercepts and line of symmetry of a parabola

The graph of $y=(x+5)(3-x)$ is given below.

Sketch of the n-shaped curve y = (x + 5)(3 - x) cutting the x-axis at M and N and the y-axis at P

(a)

State the values of the two $x$ -intercepts.

(b)

Write down the coordinates of $P$ .

(c)

Find the equation of the line of symmetry of the graph.

Show worked solution▾

(a)

At the $x$ -intercepts, $y=0$ :

$\begin{aligned}(x+5)(3-x)&=0\\ x=-5\quad&\text{or}\quad x=3\end{aligned}$

(b)

$P$ is where the curve cuts the $y$ -axis, so put $x=0$ :

$\begin{aligned}y&=(0+5)(3-0)\\ &=15\end{aligned}$

$P$ is $(0,15)$ .

(c)

The line of symmetry is halfway between the $x$ -intercepts:

$\begin{aligned}x&=\frac{-5+3}{2}\\ &=-1\end{aligned}$

The line of symmetry is $x=-1$ .

Q5Queensway Secondary2025EOY P1 Q5●●○ Standard5 marks

Intercepts, line of symmetry and minimum value

In the diagram, the curve $y=2x^2+5x+3$ cuts the $x$ -axis at two points $A$ and $B$ , and the $y$ -axis at point $C$ .

U-shaped curve cutting the negative x-axis at A and B and the y-axis at C

(a)

Calculate the coordinates of $A$ , $B$ and $C$ .

(b)

Write down the line of symmetry of the curve.

(c)

Jane claims that the minimum $y$ -value of the curve is $-0.1$ . Do you agree with her? Show your working clearly.

Show worked solution▾

(a)

For $A$ and $B$ , put $y=0$ and factorise:

$\begin{aligned}2x^2+5x+3&=0\\ (2x+3)(x+1)&=0\\ x=-1.5\quad&\text{or}\quad x=-1\end{aligned}$

$A$ is further from the origin, so $A=(-1.5,0)$ and $B=(-1,0)$ .

For $C$ , put $x=0$ : $y=2(0)^2+5(0)+3=3$ , so $C=(0,3)$ .

(b)

$\begin{aligned}x&=\frac{-1.5+(-1)}{2}\\ &=-1.25\end{aligned}$

The line of symmetry is $x=-1.25$ .

(c)

The minimum point lies on the line of symmetry, so substitute $x=-1.25$ :

$\begin{aligned}y&=2(-1.25)^2+5(-1.25)+3\\ &=3.125-6.25+3\\ &=-0.125\end{aligned}$

I do not agree with Jane. The minimum $y$ -value is $-0.125$ , not $-0.1$ .

Q6Outram Secondary2025EOY P2 Q6●●○ Standard8 marks

Plotting a quadratic graph and a line on the same grid

The variables $x$ and $y$ are connected by the equation $y=x^2-6x+7$ . Some corresponding values of $x$ and $y$ are given in the following table

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$y$	$7$	$2$	$-1$	$-2$	$q$	$2$	$7$

(a)

Find the value of $q$ .

(b)

Using a scale of 2 cm to represent 1 unit on the $x$ -axis and 2 cm to represent 1 unit on the $y$ -axes, draw the graph of $y=x^2-6x+7$ for $0\leq x\leq 6$ .

(c)

Use your graph.

(i)

write down the equation of the line of symmetry.

(ii)

find the value of $y$ when $x=2.5$ .

(d)

On the graph, draw the line $y=x+1$ . Write down the $x$ -coordinates of the points where this line intersects the graph $y=x^2-6x+7$ .

Blank sheet of graph paper with 1 cm squares

Show worked solution▾

(a)

$\begin{aligned}q&=4^2-6(4)+7\\ &=16-24+7\\ &=-1\end{aligned}$

(b)

Draw the $x$ -axis from $0$ to $6$ and the $y$ -axis from $-2$ to $7$ , with 2 cm for each unit on both axes. Plot the seven points from the completed table:

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$y$	$7$	$2$	$-1$	$-2$	$-1$	$2$	$7$

Join them with one smooth U-shaped curve. It is symmetrical about $x=3$ , with its lowest point at $(3,-2)$ , and passes through $(0,7)$ and $(6,7)$ . Label it $y=x^2-6x+7$ .

(c)(i)

Points with equal $y$ -values, such as $(2,-1)$ and $(4,-1)$ , sit either side of the line of symmetry through the minimum point $(3,-2)$ :

$\begin{aligned}x&=3\end{aligned}$

(c)(ii)

Draw a vertical line from $x=2.5$ to the curve and read across. As a check by substitution:

$\begin{aligned}y&=2.5^2-6(2.5)+7\\ &=6.25-15+7\\ &=-1.75\end{aligned}$

(d)

The line $y=x+1$ passes through $(0,1)$ , $(3,4)$ and $(6,7)$ . Draw it with a ruler; it meets the curve at two points. Reading the $x$ -coordinates:

$\begin{aligned}x=1\quad&\text{and}\quad x=6\end{aligned}$

Check: at the intersections $x^2-6x+7=x+1$ , so $x^2-7x+6=0$ , $(x-1)(x-6)=0$ , giving $x=1$ or $x=6$ .

Q7Greenridge Secondary2025EOY P2 Q6●●○ Standard9 marks

Maximum point, line of symmetry and intersection with a line

(a)

The variables $x$ and $y$ are connected by the equation $y=4-3x-x^2$ . Some corresponding values of $x$ and $y$ are given in the table below.

$x$	$-5$	$-4$	$-3$	$-2$	$-1$	$0$	$1$
$y$	$p$	$0$	$4$	$6$	$6$	$4$	$0$

Calculate the value of $p$ .

(b)

On the grid, draw the graph of $y=4-3x-x^2$ for $-5\leq x\leq 1$ .

Blank grid with x from -5 to 1 and y from -7 to 6

(c)

Using your graph, write down the maximum value of $y=4-3x-x^2$ .

(d)

Write down the equation of the line of symmetry of the curve $y=4-3x-x^2$ .

(e)

(i)

On the grid, draw the line $y=x+5$ for $-5\leq x\leq 1$ .

(ii)

Write down the $x$ -coordinates of the points of intersection of the two graphs.

Show worked solution▾

(a)

$\begin{aligned}p&=4-3(-5)-(-5)^2\\ &=4+15-25\\ &=-6\end{aligned}$

(b)

Plot the seven points $(-5,-6)$ , $(-4,0)$ , $(-3,4)$ , $(-2,6)$ , $(-1,6)$ , $(0,4)$ , $(1,0)$ and join them with a smooth n-shaped curve. The curve rises a little above $6$ between $x=-2$ and $x=-1$ , so do not flatten the top.

Plotted curve y = 4 - 3x - x squared from x = -5 to 1 with the line y = x + 5 crossing it twice

(c)

The highest point of the curve is at $x=-1.5$ . Reading across, the maximum value is about $6.3$ . (By substitution, $4-3(-1.5)-(-1.5)^2=6.25$ .)

(d)

$(-2,6)$ and $(-1,6)$ have the same $y$ -value, so the line of symmetry is halfway between them:

$\begin{aligned}x&=\frac{-2+(-1)}{2}=-1.5\end{aligned}$

(e)(i)

The line $y=x+5$ passes through $(-5,0)$ , $(0,5)$ and $(1,6)$ ; it is drawn on the graph above.

(e)(ii)

Reading the $x$ -coordinates where the line crosses the curve:

$\begin{aligned}x\approx-3.7\quad&\text{and}\quad x\approx-0.3\end{aligned}$

Check: at the intersections $4-3x-x^2=x+5$ , which rearranges to $x^2+4x+1=0$ . Substituting $x=-0.3$ gives $0.09-1.2+1=-0.11$ , close to $0$ , as expected for a reading from a graph.

Q8Chung Cheng High (Yishun)2025EOY Q20●●● Challenging9 marks

Quadratic curve and line: intersections and a triangle

The variables $x$ and $y$ are connected by the equation $y=4x^2-8x-21$ .

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$11$	$p$	$-21$	$-25$	$-21$	$-9$	$11$

Find the value of $p$ .

(a)

On the grid, draw the graph of $y=4x^2-8x-21$ for $-2\leq x\leq4$ .

Blank grid with x from -3 to 5 and y from -27.5 to 17.5

(c)

On the same grid, draw the line $4x+y=-13$ for $-2\leq x\leq3$ .

(d)

Using your graph,

(i)

write down the coordinates of the points of intersection of the line and the curve.

(ii)

A triangle is formed by the points of intersection points in (d)(i) and $(3,-9)$ . Calculate the area of this triangle.

Show worked solution▾

$p$

$\begin{aligned}p&=4(-1)^2-8(-1)-21\\ &=4+8-21\\ &=-9\end{aligned}$

(a)

Plot the seven points from the table and join them with a smooth U-shaped curve with its minimum at $(1,-25)$ . Label the curve.

Plotted curve y = 4x squared - 8x - 21 and the line 4x + y = -13 meeting at (-1, -9) and (2, -21), with a dash-dot line from (-1, -9) to (3, -9)

(c)

Rewrite the line as $y=-13-4x$ and work out a few points:

$x$	$-2$	$0$	$3$
$y$	$-5$	$-13$	$-25$

Plot these and join them with a ruler, as shown above.

(d)(i)

The line meets the curve at

$\begin{aligned}(-1,-9)\quad&\text{and}\quad(2,-21)\end{aligned}$

Check: $4x^2-8x-21=-13-4x$ gives $4x^2-4x-8=0$ , so $x^2-x-2=0$ and $(x+1)(x-2)=0$ .

(d)(ii)

The points $(-1,-9)$ and $(3,-9)$ lie on the horizontal line $y=-9$ , so take that side as the base. The height is the vertical distance from $(2,-21)$ to $y=-9$ :

$\begin{aligned}\text{Base}&=3-(-1)=4\\ \text{Height}&=-9-(-21)=12\\ \text{Area}&=\frac12\times4\times12\\ &=24\text{ units}^2\end{aligned}$

Q9Ngee Ann Secondary2025EOY Q15●●○ Standard6 marks

Height of a ball from a quadratic graph

A ball bearing is thrown upwards where $h$ is the height of the ball above the ground in metres and $t$ is the time of the flight in seconds. The motion is given by $h=-t^2+6t+6$ . Some corresponding values of $t$ and $h$ are given in the following table.

$t$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$h$	$6$	$p$	$14$	$15$	$14$	$11$	$6$

(a)

Find the value of $p$ .

(b)

On the grid opposite, draw the graph of $h=-t^2+6t+6$ for $0\leq t\leq6$ .

(c)

Use the graph to find

(i)

the maximum height of the ball bearing,

(ii)

the possible times, in seconds, when the ball bearing is at 10m.

(d)

Is the ball thrown from the ground level? Explain your answer.

Blank grid with t from 0 to 6 seconds and h from 0 to 16 metres

Show worked solution▾

(a)

$\begin{aligned}p&=-(1)^2+6(1)+6\\ &=11\end{aligned}$

(b)

Plot $(0,6)$ , $(1,11)$ , $(2,14)$ , $(3,15)$ , $(4,14)$ , $(5,11)$ , $(6,6)$ and join them with a smooth curve.

Plotted curve h = -t squared + 6t + 6 rising from 6 m at t = 0 to 15 m at t = 3 and back to 6 m at t = 6

(c)(i)

The highest point of the curve is at $t=3$ , so the maximum height is $15$ m.

(c)(ii)

Draw the horizontal line $h=10$ . It meets the curve twice, on the way up and on the way down:

$\begin{aligned}t\approx0.8\text{ s}\quad&\text{and}\quad t\approx5.2\text{ s}\end{aligned}$

(d)

No. At $t=0$ the height is $h=6$ m, so the ball starts $6$ m above the ground, not at ground level.

Q10Woodgrove Secondary2025EOY P2 Q20●●○ Standard7 marks

Flight path of a bird: landing point and maximum height

In a computer game, the flight path of a bird is modelled by the equation $y=-x^2+6x+1$ , where $y$ is the height of the bird above the ground at a horizontal distance, $x$ units, from the starting point. The landing point of the bird is at $y=0$ .

Illustration of a bird launched from the top of a block at the starting point, flying along a dashed curved path to a landing point on the ground

Some corresponding values of x and y are given in the table below.

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$6.5$
$y$	$1$	$6$	$9$	$10$	$9$	$6$	$p$	$q$

(a)

Find the value of $p$ and $q$ .

(b)

On the grid opposite, draw the graph of $y=-x^2+6x+1$ for $0\leq x\leq 6.5$ .

(c)

Use your graph to find

(i)

the coordinates of the landing point,

(ii)

the maximum height of the bird above the ground.

Blank grid with x from 0 to 8 and y from -8 to 14

Show worked solution▾

(a)

$\begin{aligned}p&=-(6)^2+6(6)+1\\ &=-36+36+1\\ &=1\\ q&=-(6.5)^2+6(6.5)+1\\ &=-42.25+39+1\\ &=-2.25\end{aligned}$

(b)

Plot the eight points and join them with a smooth n-shaped curve, symmetrical about $x=3$ .

Plotted curve y = -x squared + 6x + 1 from x = 0 to 6.5, peaking at (3, 10) and crossing the x-axis just after x = 6

(c)(i)

The bird lands where the curve crosses the $x$ -axis ( $y=0$ ), between $x=6$ (where $y=1$ ) and $x=6.5$ (where $y=-2.25$ ). Reading from the graph, the landing point is about

$\begin{aligned}(6.15,\,0)\end{aligned}$

(c)(ii)

The highest point of the curve is $(3,10)$ , so the maximum height is $10$ units.

Q11St Margaret’s Secondary2025EOY Q19●●● Challenging8 marks

Ball thrown from a building: reading times and heights

A ball it thrown upwards from the top of a building. The height $h$ metres, of the ball above the building after $t$ seconds is given by the equation $h=25t-5t^2$ . Some corresponding values of $h$ and t are given in the table below.

$t$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$h$	$0$	$a$	$30$	$30$	$20$	$0$	$b$

(a)

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

(b)

Draw the graph of $h=25t-5t^2$ for $0\leq t\leq 6$ on the grid provided on the next page.

(c)

Using your graph, find the length of the time when the object is 26 m above the height of the building.

(d)

Find the time when the object will reach the greatest height.

(e)

Given that the ball hits the ground 6 seconds after it was thrown, state the height of the building from the graph.

Blank grid with t from -3 to 7 and h from -27 to 80

Show worked solution▾

(a)

$\begin{aligned}a&=25(1)-5(1)^2=20\\ b&=25(6)-5(6)^2=150-180=-30\end{aligned}$

(b)

Plot the seven points and join them with a smooth n-shaped curve. It passes back through $h=0$ at $t=5$ (level with the top of the building) and continues below the $t$ -axis to $(6,-30)$ .

Plotted curve h = 25t - 5t squared from t = 0 to 6 with dashed lines at h = 26 and at t = 2.5

(c)

Draw the horizontal line $h=26$ . It meets the curve at about $t=1.48$ and $t=3.53$ . The ball is at least $26$ m above the building between these times:

$\begin{aligned}\text{Length of time}&\approx3.53-1.48\\ &=2.05\text{ s}\end{aligned}$

(d)

The curve is symmetrical: $h=0$ at $t=0$ and $t=5$ , and $h=30$ at $t=2$ and $t=3$ . The greatest height is halfway:

$\begin{aligned}t&=\frac{0+5}{2}=2.5\text{ s}\end{aligned}$

(e)

At $t=6$ the graph gives $h=-30$ , so the ball is $30$ m below the top of the building when it hits the ground. The building is $30$ m tall.

Q12Regent Secondary2025EOY P2 Q8●●● Challenging6 marks

Path of an arrow: greatest height and distances

An arrow was fired from a raised platform towards a field. The vertical height of the arrow, $y$ metres above the ground, at a horizontal distance $x$ metres from the platform, is given by $y=6+x(3-\dfrac15x)$ , $x\geq 0$ .

The table below shows some corresponding values of $x$ and $y$ .

$x$	$0$	$2$	$4$	$6$	$9$	$12$	$14$	$16$	$18$
$y$	$6$	$11.2$	$14.8$	$16.8$	$16.8$	$13.2$	$8.8$	$p$	$-4.8$

(a)

Find the value of $p$ .

(b)

On the grid opposite, draw the graph of $y=6+x(3-\dfrac15x)$ for $0\leq x\leq 18$ .

Blank grid with x from 0 to 18 and y from -6 to 18

(c)

Use your graph to find

(i)

the greatest height reached by the arrow,

(ii)

the horizontal distance, $x$ travelled by the arrow before it hits the ground,

(iii)

the horizontal distance for which the arrow was more than 14 metres above the ground.

Show worked solution▾

(a)

$\begin{aligned}p&=6+16\left(3-\frac15(16)\right)\\ &=6+16(-0.2)\\ &=2.8\end{aligned}$

(b)

Plot the nine points $(0,6)$ , $(2,11.2)$ , $(4,14.8)$ , $(6,16.8)$ , $(9,16.8)$ , $(12,13.2)$ , $(14,8.8)$ , $(16,2.8)$ and $(18,-4.8)$ , and join them with a smooth n-shaped curve. The curve is symmetrical about $x=7.5$ (halfway between $x=6$ and $x=9$ , which have equal heights), with its highest point a little above $17$ , and it crosses the $x$ -axis between $x=16$ and $x=18$ .

(c)(i)

The highest point is at $x=7.5$ . Reading across, the greatest height is about

$\begin{aligned}17.3\text{ m}\end{aligned}$

(By substitution, $6+7.5(3-1.5)=17.25$ .)

(c)(ii)

The arrow hits the ground where the curve crosses the $x$ -axis:

$\begin{aligned}x&\approx16.8\text{ m}\end{aligned}$

(c)(iii)

Draw the horizontal line $y=14$ . It meets the curve at about $x=3.5$ and $x=11.5$ :

$\begin{aligned}\text{Horizontal distance}&\approx11.5-3.5\\ &=8\text{ m}\end{aligned}$

Q13Queensway Secondary2025EOY P2 Q10●●● Challenging7 marks

Area of a rectangular garden as a quadratic graph

(a)

Sarah wants to build a rectangular garden. The total perimeter of the garden is 36 meters. The breadth of the garden is $x$ m, and the area of the garden is $y$ m $^2$ .

Show that $y=18x-x^2$ .

(b)

The variables $x$ and $y$ are connected by the equation $y=18x-x^2$ . Some corresponding values of $x$ and $y$ are given in the table below.

$x$	$0$	$2$	$4$	$6$	$8$	$10$	$12$
$y$	$0$	$32$	$56$	$72$	$80$	$80$	$p$

(i)

Find the value of $p$ .

(ii)

On the grid next page, draw the graph of $y=18x-x^2$ for $0\leq x\leq 12$ .

(iii)

Use your graph to estimate the breadth(s) of the garden when the area is 75 m $^2$ .

Blank grid with x from 0 to 12 and y from 0 to 90

Show worked solution▾

(a)

Length and breadth together make half the perimeter, $36\div2=18$ m, so the length is $(18-x)$ m:

$\begin{aligned}y&=x(18-x)\\ &=18x-x^2\quad\text{(shown)}\end{aligned}$

(b)(i)

$\begin{aligned}p&=18(12)-12^2\\ &=216-144\\ &=72\end{aligned}$

(b)(ii)

Plot the seven points and join them with a smooth curve rising from $(0,0)$ to a maximum at $(9,81)$ and coming down to $(12,72)$ .

Plotted curve y = 18x - x squared from x = 0 to 12 through (8, 80) and (10, 80)

(b)(iii)

Draw the horizontal line $y=75$ . It meets the curve twice:

$\begin{aligned}x\approx6.5\text{ m}\quad&\text{or}\quad x\approx11.5\text{ m}\end{aligned}$

Note: the school answer is $6.4$ m or $11.4$ m. Substituting shows the curve is below $75$ at $x=6.4$ ( $y=74.24$ ) and above it at $x=6.6$ ( $y=75.24$ ), so the first crossing is about $6.55$ m, and the second, by symmetry about $x=9$ , is about $11.45$ m.

Unsure how to read a solution off a graph, or which line to draw? Graph work is exactly what we practise in our Secondary 2 Maths programme, every method behind these questions taught step by step.

Frequently asked questions

How do I find the line of symmetry of a quadratic graph?▾

Find two points on the curve with the same

y

-value, such as the two

x

-intercepts, and take the

x

-value halfway between them. For

y=(x+5)(3-x)

the intercepts are

-5

and

3

, so the line of symmetry is

x=-1

. See Q4 and Q5.

How do I tell whether a parabola has a maximum or a minimum point?▾

Look at the sign of the

x^2

term. If it is positive, as in

y=x^2-6x+7

, the curve is U-shaped with a minimum point. If it is negative, as in

h=-t^2+6t+6

, the curve is n-shaped with a maximum point.

What do the points where a line meets a curve tell me?▾

Their

x

-coordinates are the solutions of the equation you get by setting the two expressions equal. In Q6, the line

y=x+1

meets

y=x^2-6x+7

x=1

and

x=6

, which solve

x^2-7x+6=0

Sec 2 Maths: Linear & Quadratic Graphs, practice questions & worked solutions

Key methods

Questions & worked solutions

Frequently asked questions

Related Sec 2 topics