E-Math notes · Chapter 6

Functions and Graphs: O-Level E-Math Notes

Graphs turn equations into pictures you can read. In this chapter you learn the shape of every graph in the syllabus: straight lines, parabolas, power curves and exponential curves. You will sketch them from their equations, read turning points and intercepts, and finish by drawing a graph and then using it, as a Paper 2 question asks.

Your outcome tracker0 / 9

Let's go! Tick an outcome once you can pass its exam check. Saved on this device.

Chapter
6

Figure 1 ·

a > 0

: a bowl with a minimum point

10 outcomes · 37 worked examples · every step shown ✦

Coordinates and ordered pairs

An ordered pair is an address: across first, then up or down.

Linear graphs and gradient

Gradient = vertical change ÷ horizontal change.

Quadratic graphs

The sign of $a$ decides whether the curve is a bowl or an arch.

By the end of this chapter

Plot and read points on the Cartesian plane, and show a relationship between two variables as a set of ordered pairs.
Draw straight-line graphs $y = ax + b$ and find a gradient as vertical change over horizontal change, positive or negative.
Describe quadratic graphs: the effect of the sign of $a$ , maximum and minimum points, and symmetry.
Complete the square, then sketch $y = \pm(x - p)^2 + q$ and $y = \pm(x - a)(x - b)$ .
Sketch $y = ax^n$ for $n = -2, -1, 0, 1, 2, 3$ and sums of up to three such terms, and exponential graphs $y = ka^x$ .
Estimate the gradient of a curve by drawing a tangent, and answer a Paper 2 style 'draw the graph, then use it' question.

Start here

Maths In Real Life

Four real situations, one graph each. The pictures are drawn first, then the graph is placed on top by script, so the graph matches the equation exactly.

$A fountain jet leaving a nozzle and curving down to the ground, with a parabola, axes and labelled turning point drawn over it.$

Maths In Real Life 1

The water jet

A jet of water leaves a nozzle and curves back down under gravity, so its height above the ground is well modelled by a quadratic graph.

A fictional playground fountain shoots a jet of water. At a horizontal distance of $x$ metres from the nozzle the jet is $h$ metres high, where $h = 2.4x - x^2$ .

(a)How far from the nozzle does the jet land?
(b)Find the greatest height of the jet and where it occurs.
(c)Find the height of the jet at $x = 0.6$ m.

The jet is an arch: maximum at the axis of symmetry.

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

\displaystyle 2.4x - x^2 = 0 \Rightarrow x(2.4 - x) = 0

Ground level means

h = 0

x = 0

(the nozzle) or

x = 2.4

Reject the starting point.

(b)

The axis of symmetry is half-way between the two landing points:

x = 1.2

Mirror line of the arch.

\displaystyle -(1.2)^2 + 2.4(1.2) = 1.44

Substitute to find the height.

(c)

\displaystyle -(0.6)^2 + 2.4(0.6) = 1.08

Substitute

x = 0.6

Final answer

(a)

2.4

(b)At

x = 1.2

m from the nozzle the greatest height is

1.44

(c)

1.08

$A cyclist pulling away from a station, with a distance-time curve and a tangent drawn over the picture.$

Maths In Real Life 2

How fast at this instant?

The speed at an instant is the gradient of the distance-time graph at that instant, which is found by drawing a tangent.

A fictional cyclist pulls away from an MRT station. Her distance $s$ metres from the station after $t$ seconds is $s = \dfrac{t^2}{2} + 2t$ . A tangent drawn at $t = 4$ passes through $(2, 4)$ and $(6, 28)$ .

Estimate her speed at $t = 4$ s, in metres per second.

The gradient of the tangent is the speed at

t = 4

Show full solutionHide solution

\displaystyle \text{speed} = \dfrac{28 - 4}{6 - 2} = 6

Gradient of the tangent: metres per second.

Final answer

6

m/s

$Groups of students in a courtyard with an exponential curve and axes drawn over the picture.$

Maths In Real Life 3

A rumour spreads

When something doubles at a steady rate, its graph is exponential: slow at first, then very steep.

A fictional rumour starts in a school. The number of students who know it after $t$ hours is $N = 20 \times 2^t$ .

(a)How many students know it at the start?
(b)How many know it after $5$ hours?
(c)After how many whole hours do more than $1000$ students know it?

Each hour the number doubles.

Show full solutionHide solution

(a)

\displaystyle 20 \times 2^0 = 20

t = 0

(b)

\displaystyle 20 \times 2^5 = 640

t = 5

(c)

\displaystyle t = 5: 640\qquad t = 6: 1280

The first value above

1000

is at

t = 6

Final answer

(a)

20

students

(b)

640

students

(c)

6

hours

$A parabolic walkway arch with axes, the curve and labelled points drawn over the picture.$

Maths In Real Life 4

The covered walkway arch

Arches in buildings and bridges are often parabolas, which spread the load evenly.

A fictional covered walkway has a parabolic arch. Taking the ground as the $x$ -axis and the middle of the arch as $x = 0$ , its height in metres at distance $x$ metres from the middle is $y = 4 - \dfrac{x^2}{4}$ .

(a)Find the width of the arch at ground level.
(b)Find the greatest height of the arch.
(c)A delivery cart is $3$ m wide, so its sides are at $x = \pm 1.5$ . What is the greatest height the cart can have to fit under the arch?

A cart's top corners must stay below the curve at

x = \pm 1.5

Show full solutionHide solution

(a)

\displaystyle 4 - \dfrac{x^2}{4} = 0 \Rightarrow x^2 = 16

Ground level means

y = 0

x = -4

x = 4

: the width is

8

Distance between the two ends.

(b)

The turning point is on the axis

x = 0

, where

y = 4

Maximum point at the middle.

(c)

\displaystyle y = 4 - \dfrac{(1.5)^2}{4} = 3.4375

The arch is lowest at the cart's sides, so use

x = 1.5

Final answer

(a)

8

(b)

4

(c)

3.4375

Supports N6.1 · N6.2

Coordinates and ordered pairs

The Cartesian plane is two number lines at right angles: the horizontal $x$ -axis and the vertical $y$ -axis. They cross at the origin $O$ and split the plane into four quadrants, numbered anticlockwise from the top right. Every point has an address called an ordered pair $(x, y)$ .

The Core Idea

An ordered pair is an address: across first, then up or down.

In $(3, 2)$ the first number is the move along the $x$ -axis and the second is the move up or down. The order matters.

The four quadrants are numbered I to IV, anticlockwise from the top right.

SGSchoolKaki Tip

Check the quadrant from the signs. Both positive is quadrant I. Negative $x$ with positive $y$ is quadrant II.

Diagram for the example below.

Example 6.1

Reading coordinates and quadrants

The diagram shows four points $A$ , $B$ , $C$ and $D$ on a Cartesian plane.

(i)Write down the coordinates of each point.
(ii)State the quadrant in which each point lies.

Show solutionHide solution

(i)

A

3

units right of the

y

-axis and

2

units above the

x

-axis, so

A(3, 2)

Read across to the

x

-value first, then up or down.

B

4

units left of the

y

-axis and

3

units above the

x

-axis, so

B(-4, 3)

Read across to the

x

-value first, then up or down.

C

2

units left of the

y

-axis and

3

units below the

x

-axis, so

C(-2, -3)

Read across to the

x

-value first, then up or down.

D

5

units right of the

y

-axis and

1

unit below the

x

-axis, so

D(5, -1)

Read across to the

x

-value first, then up or down.

(ii)

A

has x positive and y positive;

B

has x negative and y positive;

C

has x negative and y negative;

D

has x positive and y negative.

The two signs together decide the quadrant.

Final answer

(i)

A(3, 2)

B(-4, 3)

C(-2, -3)

D(5, -1)

(ii)

A

is in quadrant I;

B

is in quadrant II;

C

is in quadrant III;

D

is in quadrant IV

Your Turn 6.1

Four points are described below. $P$ is $5$ units left of the $y$ -axis and $2$ units above the $x$ -axis. $Q$ is $1$ unit right of the $y$ -axis and $4$ units below the $x$ -axis. $R$ is $4$ units right of the $y$ -axis and $4$ units above the $x$ -axis. $S$ is $3$ units left of the $y$ -axis and $1$ unit below the $x$ -axis.

(i)Write down the coordinates of each point.
(ii)State the quadrant in which each point lies.

Show worked solutionHide worked solution

(i)

P

: left of the

y

-axis gives

x = -5

, and above the

x

-axis gives

y = 2

, so

P(-5, 2)

Left and down are negative; right and up are positive.

Q

: right of the

y

-axis gives

x = 1

, and below the

x

-axis gives

y = -4

, so

Q(1, -4)

Left and down are negative; right and up are positive.

R

: right of the

y

-axis gives

x = 4

, and above the

x

-axis gives

y = 4

, so

R(4, 4)

Left and down are negative; right and up are positive.

S

: left of the

y

-axis gives

x = -3

, and below the

x

-axis gives

y = -1

, so

S(-3, -1)

Left and down are negative; right and up are positive.

(ii)

P

has x negative and y positive;

Q

has x positive and y negative;

R

has x positive and y positive;

S

has x negative and y negative.

The two signs together decide the quadrant.

Final answer

(i)

P(-5, 2)

Q(1, -4)

R(4, 4)

S(-3, -1)

(ii)

P

is in quadrant II;

Q

is in quadrant IV;

R

is in quadrant I;

S

is in quadrant III

Think About It

A point lies on the

y

-axis. What must its

x

-coordinate be? What about a point on the

x

-axis?

Show the worked answerHide the worked answer

A point on the

y

-axis has not moved left or right from the origin.

Think about the journey from

O

So its

x

-coordinate is

0

, for example

(0, 4)

Across by zero.

In the same way a point on the

x

-axis has not moved up or down, so

y = 0

, for example

(4, 0)

Up or down by zero.

Final answerOn the

y

-axis

x = 0

; on the

x

-axis

y = 0

Now the key idea of this chapter. A relationship between two variables can be shown as a set of ordered pairs. Plot every pair as a point, and the shape of the points tells you what kind of relationship it is.

Example 6.2

A table of values as ordered pairs

Mei Ling saves money every week. After $w$ weeks she has $S$ dollars saved, as shown. $\displaystyle \begin{array}{c|cc} w & 0 & 1 \\ \hline S & 12 & 20 \end{array}$ $\displaystyle \begin{array}{c|cc} w & 2 & 3 \\ \hline S & 28 & 36 \end{array}$ $\displaystyle \begin{array}{c|cc} w & 4 & 5 \\ \hline S & 44 & 52 \end{array}$

(i)Write the data as ordered pairs $(w, S)$ .
(ii)Plot the points with $w$ across and $S$ up.
(iii)Describe the pattern in the points and say what kind of relationship it shows.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

(i)

Each column of the table gives one ordered pair: the top number first, the bottom number second.

Pair the numbers in the same column.

(ii)

The largest value of

S

52

, so the vertical axis must reach at least

52

; the horizontal axis only needs to reach

5

Choose a scale that fits the biggest value.

Mark each pair as a dot: the first is

(0, 12)

and the last is

(5, 52)

Label both axes.

(iii)

From one week to the next,

S

changes by

20 - 12 = 8

28 - 20 = 8

and so on: always

8

Compare the change for equal steps.

An equal change for every equal step means the points lie on a straight line.

Constant change gives a straight line.

The dashed line shows the points are in a straight line.

Final answer

(i)

(0, 12)

(1, 20)

(2, 28)

(3, 36)

(4, 44)

(5, 52)

(ii)Six dots, from

(0, 12)

(5, 52)

, rises steadily.

(iii)

S

changes by

8

every week, so the points lie on a straight line: a linear relationship.

Your Turn 6.2

A prepaid card has a balance of $B$ dollars after $t$ weeks, as shown. $\displaystyle \begin{array}{c|cc} t & 0 & 1 \\ \hline B & 100 & 88 \end{array}$ $\displaystyle \begin{array}{c|cc} t & 2 & 3 \\ \hline B & 76 & 64 \end{array}$ $\displaystyle \begin{array}{c|cc} t & 4 & 5 \\ \hline B & 52 & 40 \end{array}$

(i)Write the data as ordered pairs $(t, B)$ .
(ii)Plot the points with $t$ across and $B$ up.
(iii)Describe the pattern in the points and say what kind of relationship it shows.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

(i)

Each column of the table gives one ordered pair: the top number first, the bottom number second.

Pair the numbers in the same column.

(ii)

The largest value of

B

100

, so the vertical axis must reach at least

100

; the horizontal axis only needs to reach

5

Choose a scale that fits the biggest value.

Mark each pair as a dot: the first is

(0, 100)

and the last is

(5, 40)

Label both axes.

(iii)

From one week to the next,

B

changes by

88 - 100 = -12

76 - 88 = -12

and so on: always

-12

Compare the change for equal steps.

An equal change for every equal step means the points lie on a straight line.

Constant change gives a straight line.

Final answer

(i)

(0, 100)

(1, 88)

(2, 76)

(3, 64)

(4, 52)

(5, 40)

(ii)Six dots, from

(0, 100)

(5, 40)

, falls steadily.

(iii)

B

changes by

-12

every week, so the points lie on a straight line: a linear relationship.

Example 6.3

When the points do not make a line

The rule $y = x^2 - 3$ gives the table of values below. $\displaystyle \begin{array}{c|ccc} x & -3 & -2 & -1 \\ \hline y & \;?\; & 1 & -2 \end{array}$ $\displaystyle \begin{array}{c|ccc} x & 0 & 1 & 2 \\ \hline y & -3 & \;?\; & 1 \end{array}$ $\displaystyle \begin{array}{c|c} x & 3 \\ \hline y & 6 \end{array}$

(i)Complete the table.
(ii)Plot the points and join them with a smooth curve.
(iii)Do the points lie on a straight line? Give a reason.

Show solutionHide solution

Colour keyexpand, rewrite, substitute or solve

(i)

\displaystyle (-3)^2 - 3 = 6

Substitute

x = -3

: square first, then apply the rest.

\displaystyle 1^2 - 3 = -2

Substitute

x = 1

: square first, then apply the rest.

(ii)

Mark every pair as a dot, starting with

(-3, 6)

. Join the dots with one smooth curve: never use a ruler.

A curve has no corners.

(iii)

The

y

-values go

6, 1, -2, -3

and then rise again. They fall and then rise, so the change is not constant.

A straight line changes at a steady rate.

The same points, joined by one smooth curve.

Final answer

(i)The missing

y

-values, from left to right, are

6

and

-2

(ii)A U-shaped curve through the seven points, with its turning point at

(0, -3)

(iii)No: the

y

-values fall and then rise, so the points make a U-shaped curve, not a straight line.

Your Turn 6.3

The rule $y = 5 - x^2$ gives the table of values below. $\displaystyle \begin{array}{c|ccc} x & -3 & -2 & -1 \\ \hline y & -4 & \;?\; & 4 \end{array}$ $\displaystyle \begin{array}{c|ccc} x & 0 & 1 & 2 \\ \hline y & 5 & 4 & \;?\; \end{array}$ $\displaystyle \begin{array}{c|c} x & 3 \\ \hline y & -4 \end{array}$

(i)Complete the table.
(ii)Plot the points and join them with a smooth curve.
(iii)Do the points lie on a straight line? Give a reason.

Show worked solutionHide worked solution

Colour keyexpand, rewrite, substitute or solve

(i)

\displaystyle -(-2)^2 + 5 = 1

Substitute

x = -2

: square first, then apply the rest.

\displaystyle -(2)^2 + 5 = 1

Substitute

x = 2

: square first, then apply the rest.

(ii)

Mark every pair as a dot, starting with

(-3, -4)

. Join the dots with one smooth curve: never use a ruler.

A curve has no corners.

(iii)

The

y

-values go

-4, 1, 4, 5

and then fall again. They rise and then fall, so the change is not constant.

A straight line changes at a steady rate.

Final answer

(i)The missing

y

-values, from left to right, are

1

and

1

(ii)An upside-down U-shaped curve through the seven points, with its turning point at

(0, 5)

(iii)No: the

y

-values rise and then fall, so the points make an upside-down U-shaped curve, not a straight line.

Supports N6.5 · N6.4 · N6.3

Linear functions, their graphs and gradient

A linear function has the form $y = ax + b$ and its graph is a straight line. Two numbers describe the line. The number $b$ is the $y$ -intercept, where the line meets the $y$ -axis. The number $a$ is the gradient, which tells you how steep the line is.

The Core Idea

Gradient = vertical change ÷ horizontal change.

Up as you move right is a positive gradient. Down as you move right is a negative gradient.

The gradient is the same between any two points on the same line, so one rise-over-run measurement describes the whole line.

Case table 1: the sign of the gradient $a$ decides the direction
Gradient	What the line does	Example
$a > 0$ (positive)	Rises as you move to the right	$y = 2x - 1$
$a < 0$ (negative)	Falls as you move to the right	$y = -x + 2$
$a = 0$ (zero)	Flat: it never rises or falls	$y = 2$

SGSchoolKaki Tip

In $y = ax + b$ , $a$ is the gradient and $b$ is the $y$ -intercept. In $y = 4 - 3x$ the gradient is still the number in front of $x$ : $-3$ .

Example 6.4

Drawing a line from a table of values

Draw the graph of $y = 3x - 2$ for $-1 \le x \le 2$ .

(i)Complete a table of values.
(ii)Draw the line and find its gradient with a gradient triangle.
(iii)Write down the $y$ -intercept.

Show solutionHide solution

Colour keyexpand, rewrite, substitute or solve

(i)

\displaystyle x = -1:\quad y = 3(-1) - 2 = -5

Substitute

x = -1

\displaystyle x = 0:\quad y = 3(0) - 2 = -2

Substitute

x = 0

\displaystyle x = 1:\quad y = 3(1) - 2 = 1

Substitute

x = 1

\displaystyle x = 2:\quad y = 3(2) - 2 = 4

Substitute

x = 2

(ii)

Plot the four points and join them with a ruler. They lie on one straight line, so extend it either way.

A straight line: use a ruler.

\displaystyle \text{gradient} = \dfrac{\text{vertical change}}{\text{horizontal change}} = \dfrac{6}{2} = 3

Move

2

across and the line rises

6

(iii)

The line meets the

y

-axis where

x = 0

, at

(0, -2)

Put

x = 0

Across 2, up 6: gradient 3.

Final answer

(i)

\displaystyle \begin{array}{c|cc} x & -1 & 0 \\ \hline y & -5 & -2 \end{array}

\displaystyle \begin{array}{c|cc} x & 1 & 2 \\ \hline y & 1 & 4 \end{array}

(ii)Gradient

= 3

(positive: the line rises).

(iii)

y

-intercept

= -2

Your Turn 6.4

Draw the graph of $y = -2x + 5$ for $-1 \le x \le 2$ .

(i)Complete a table of values.
(ii)Draw the line and find its gradient with a gradient triangle.
(iii)Write down the $y$ -intercept.

Show worked solutionHide worked solution

Colour keyexpand, rewrite, substitute or solve

(i)

\displaystyle x = -1:\quad y = -2(-1) + 5 = 7

Substitute

x = -1

\displaystyle x = 0:\quad y = -2(0) + 5 = 5

Substitute

x = 0

\displaystyle x = 1:\quad y = -2(1) + 5 = 3

Substitute

x = 1

\displaystyle x = 2:\quad y = -2(2) + 5 = 1

Substitute

x = 2

(ii)

Plot the four points and join them with a ruler. They lie on one straight line, so extend it either way.

A straight line: use a ruler.

\displaystyle \text{gradient} = \dfrac{\text{vertical change}}{\text{horizontal change}} = \dfrac{-4}{2} = -2

Move

2

across and the line falls

4

(iii)

The line meets the

y

-axis where

x = 0

, at

(0, 5)

Put

x = 0

Final answer

(i)

\displaystyle \begin{array}{c|cc} x & -1 & 0 \\ \hline y & 7 & 5 \end{array}

\displaystyle \begin{array}{c|cc} x & 1 & 2 \\ \hline y & 3 & 1 \end{array}

(ii)Gradient

= -2

(negative: the line falls).

(iii)

y

-intercept

= 5

Diagram for the example below.

Example 6.5

A negative gradient from two points

The graph shows a straight line through the points $P$ and $Q$ .

(i)Read off the coordinates of $P$ and $Q$ .
(ii)Find the gradient of the line.

Show solutionHide solution

(i)

P

2

left of the

y

-axis and

5

up, so

P(-2, 5)

Q

2

right and

1

down, so

Q(2, -1)

Across first, then up or down.

(ii)

\displaystyle \text{horizontal change} = 2 - (-2) = 4

Going from

P

Q

: right is positive.

\displaystyle \text{vertical change} = -1 - (5) = -6

The line falls, so the change is negative.

\displaystyle \text{gradient} = \dfrac{-6}{4} = -1.5

Vertical change over horizontal change.

Falling as you move right: a negative gradient.

Final answer

(i)

P(-2, 5)

and

Q(2, -1)

(ii)Gradient

= -1.5

(negative: the line falls).

Think About It

A line has gradient

0

. Draw it in your head. What does

y = ax + b

become?

Show the worked answerHide the worked answer

A gradient of zero means no rise at all, so the line is horizontal.

Rise = 0.

\displaystyle y = 0x + b = b

The

ax

term disappears.

So the equation is

y = b

: a flat line through the

y

-intercept.

A horizontal line.

Final answerA flat line with equation

y = b

Diagram for the example below.

Example 6.6

Writing the equation of a line

The line in the diagram passes through $(0, 4)$ and $(3, -2)$ . Write down its equation in the form $y = ax + b$ .

Show solutionHide solution

The line meets the

y

-axis at

(0, 4)

, so

b = 4

b

is the

y

-intercept.

\displaystyle a = \dfrac{-2 - (4)}{3 - 0} = \dfrac{-6}{3} = -2

Rise over run between the two points.

\displaystyle y = -2x + 4

Put

a

and

b

into

y = ax + b

\displaystyle \text{Check at } x = 3: \quad -2(3) + 4 = -2

The second point must fit.

Check: the line passes through both given points.

Final answer

y = -2x + 4

Example 6.7

A line as a model

A fictional ride app charges a fixed amount plus a rate per kilometre. The cost $C$ dollars for $d$ kilometres is a straight line through $(0, 3.5)$ and $(8, 9.5)$ .

(i)Find the gradient and say what it means.
(ii)Write down the equation connecting $C$ and $d$ .
(iii)Find the cost for $12$ kilometres, to the nearest cent.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solveround, estimate or compare

(i)

\displaystyle \text{gradient} = \dfrac{9.5 - 3.5}{8 - 0} = \dfrac{6}{8} = 0.75

Dollars gained for each extra kilometre.

(ii)

The line meets the

C

-axis at

3.5

, so the fixed charge is 3.5 dollars.

The

y

-intercept is the fixed amount.

\displaystyle C = 0.75d + 3.5

Gradient times the variable, plus the intercept.

(iii)

\displaystyle C = 0.75(12) + 3.5 = 12.5

Substitute

d = 12

To the nearest cent that is 12.50 dollars.

Money answers have two decimal places.

Gradient = cost per kilometre;

y

-intercept = fixed charge.

Final answer

(i)Gradient

= 0.75

: the cost rises by 0.75 dollars for every extra kilometre.

(ii)

C = 0.75d + 3.5

(iii)12.50 dollars

Your Turn 6.7

A fictional gym charges a joining fee plus a rate per visit. The cost $C$ dollars for $v$ visits is a straight line through $(0, 4)$ and $(10, 10)$ .

(i)Find the gradient and say what it means.
(ii)Write down the equation connecting $C$ and $v$ .
(iii)Find the cost for $15$ visits, to the nearest cent.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solveround, estimate or compare

(i)

\displaystyle \text{gradient} = \dfrac{10 - 4}{10 - 0} = \dfrac{6}{10} = 0.6

Dollars gained for each extra visit.

(ii)

The line meets the

C

-axis at

4

, so the fixed charge is 4 dollars.

The

y

-intercept is the fixed amount.

\displaystyle C = 0.6v + 4

Gradient times the variable, plus the intercept.

(iii)

\displaystyle C = 0.6(15) + 4 = 13

Substitute

v = 15

To the nearest cent that is 13.00 dollars.

Money answers have two decimal places.

Final answer

(i)Gradient

= 0.6

: the cost rises by 0.6 dollars for every extra visit.

(ii)

C = 0.6v + 4

(iii)13.00 dollars

Supports N6.6.a · N6.6.b · N6.6.c · N6.2 · N6.3

Quadratic graphs: shape, turning point and symmetry

A quadratic function has the form $y = ax^2 + bx + c$ with $a \ne 0$ . Its graph is a smooth, symmetrical curve called a parabola. The number $a$ decides the shape, $c$ is the $y$ -intercept, and $a$ and $b$ together place the turning point.

The Core Idea

The sign of $a$ decides whether the curve is a bowl or an arch.

If $a > 0$ the turning point is a minimum. If $a < 0$ it is a maximum.

Every parabola is symmetrical about the vertical line through its turning point, called the axis of symmetry. Fold the page along that line and the two halves match.

Case table 2: the sign of $a$ in $y = ax^2 + bx + c$
Sign of $a$	Picture	The turning point is a…	Value of $y$ at the turning point
$a > 0$		minimum point (lowest point)	The least value of $y$
$a < 0$		maximum point (highest point)	The greatest value of $y$

a > 0

: a bowl with a minimum point

a < 0

: an arch with a maximum point, the mirror image in the

x

-axis

Two points at the same height are mirror images of each other, so the axis of symmetry is exactly half-way between them. You will use this idea again and again.

SGSchoolKaki Tip

Read $a$ from the coefficient of $x^2$ , wherever the term is written. In $y = 3 - 2x - x^2$ the value of $a$ is $-1$ , not $3$ , so the graph is an arch.

Example 6.8

Table of values, graph and features

Draw the graph of $y = x^2 - 6x + 5$ for $0 \le x \le 6$ . Part of the table is shown. $\displaystyle \begin{array}{c|ccc} x & 0 & 1 & 2 \\ \hline y & 5 & 0 & \;?\; \end{array}$ $\displaystyle \begin{array}{c|ccc} x & 3 & 4 & 5 \\ \hline y & -4 & \;?\; & 0 \end{array}$ $\displaystyle \begin{array}{c|c} x & 6 \\ \hline y & 5 \end{array}$

(i)Complete the table.
(ii)Draw the graph.
(iii)Write down the turning point and the equation of the axis of symmetry.
(iv)Write down the $x$ -intercepts and the $y$ -intercept.

Show solutionHide solution

Colour keyexpand, rewrite, substitute or solve

(i)

\displaystyle 2^2 - 6(2) + 5 = -3

Substitute

x = 2

\displaystyle 4^2 - 6(4) + 5 = -3

Substitute

x = 4

(ii)

Plot every pair, then join the points with one smooth curve. Do not use a ruler.

A parabola is smooth with no corners.

(iii)

The lowest point on the curve is

(3, -4)

. The axis of symmetry is the vertical line through it.

Read it from the table and the graph.

(iv)

The curve crosses the

x

-axis where

y = 0

: the table shows this at

x = 1

and

x = 5

Intercepts are where the curve meets an axis.

It crosses the

y

-axis where

x = 0

: the table gives

y = 5

Put

x = 0

Turning point, intercepts, axis of symmetry and the equation: all read from the same curve.

Final answer

(i)The missing

y

-values, from left to right, are

-3

and

-3

. The values are symmetrical about the middle of the table.

(ii)A smooth curve through all the points with its minimum point at

(3, -4)

(iii)Turning point

(3, -4)

(a minimum); axis of symmetry

x = 3

(iv)

x

-intercepts

1

and

5

;

y

-intercept

5

Your Turn 6.8

Draw the graph of $y = x^2 - 2x - 3$ for $-2 \le x \le 4$ . Part of the table is shown. $\displaystyle \begin{array}{c|ccc} x & -2 & -1 & 0 \\ \hline y & 5 & \;?\; & -3 \end{array}$ $\displaystyle \begin{array}{c|ccc} x & 1 & 2 & 3 \\ \hline y & -4 & \;?\; & 0 \end{array}$ $\displaystyle \begin{array}{c|c} x & 4 \\ \hline y & 5 \end{array}$

(i)Complete the table.
(ii)Draw the graph.
(iii)Write down the turning point and the equation of the axis of symmetry.
(iv)Write down the $x$ -intercepts and the $y$ -intercept.

Show worked solutionHide worked solution

Colour keyexpand, rewrite, substitute or solve

(i)

\displaystyle (-1)^2 - 2(-1) - 3 = 0

Substitute

x = -1

\displaystyle 2^2 - 2(2) - 3 = -3

Substitute

x = 2

(ii)

Plot every pair, then join the points with one smooth curve. Do not use a ruler.

A parabola is smooth with no corners.

(iii)

The lowest point on the curve is

(1, -4)

. The axis of symmetry is the vertical line through it.

Read it from the table and the graph.

(iv)

The curve crosses the

x

-axis where

y = 0

: the table shows this at

x = -1

and

x = 3

Intercepts are where the curve meets an axis.

It crosses the

y

-axis where

x = 0

: the table gives

y = -3

Put

x = 0

Final answer

(i)The missing

y

-values, from left to right, are

0

and

-3

. The values are symmetrical about the middle of the table.

(ii)A smooth curve through all the points with its minimum point at

(1, -4)

(iii)Turning point

(1, -4)

(a minimum); axis of symmetry

x = 1

(iv)

x

-intercepts

-1

and

3

;

y

-intercept

-3

Example 6.9

A parabola with the terms out of order

The graph of $y = 3 - 2x - x^2$ is drawn.

(i)Without drawing, say whether the turning point is a maximum or a minimum.
(ii)Find the turning point, the axis of symmetry and the greatest value of $y$ .
(iii)Write down the $y$ -intercept.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

Rewrite in order:

y = -x^2 - 2x + 3

. The coefficient of

x^2

a = -1

, which is negative.

Find the

x^2

term wherever it sits.

A negative

a

means the graph is an arch.

Sign of

a

decides the shape.

(ii)

The graph crosses the

x

-axis at

x = -3

and

x = 1

. The axis is half-way between them.

Axis of symmetry = midpoint of the

x

-intercepts.

\displaystyle x = \dfrac{-3 + (1)}{2} = -1

Halfway between the two intercepts.

\displaystyle -(-1)^2 - 2(-1) + 3 = 4

Substitute

x = -1

to find the height of the turning point.

(iii)

\displaystyle -(0)^2 - 2(0) + 3 = 3

Substitute

x = 0

: only the constant term is left, and it is the

y

-intercept.

An arch: the greatest value of

y

is the height of the turning point.

Final answer

(i)A maximum point (the graph is an arch).

(ii)Turning point

(-1, 4)

(maximum); axis

x = -1

; the greatest value of

y

4

(iii)

y

-intercept

= 3

Your Turn 6.9

The graph of $y = 6 - 8x + 2x^2$ is drawn.

(i)Without drawing, say whether the turning point is a maximum or a minimum.
(ii)Find the turning point, the axis of symmetry and the least value of $y$ .
(iii)Write down the $y$ -intercept.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

Rewrite in order:

y = 2x^2 - 8x + 6

. The coefficient of

x^2

a = 2

, which is positive.

Find the

x^2

term wherever it sits.

A positive

a

means the graph is a bowl.

Sign of

a

decides the shape.

(ii)

The graph crosses the

x

-axis at

x = 1

and

x = 3

. The axis is half-way between them.

Axis of symmetry = midpoint of the

x

-intercepts.

\displaystyle x = \dfrac{1 + (3)}{2} = 2

Halfway between the two intercepts.

\displaystyle 2(2)^2 - 8(2) + 6 = -2

Substitute

x = 2

to find the height of the turning point.

(iii)

\displaystyle 2(0)^2 - 8(0) + 6 = 6

Substitute

x = 0

: only the constant term is left, and it is the

y

-intercept.

Final answer

(i)A minimum point (the graph is a bowl).

(ii)Turning point

(2, -2)

(minimum); axis

x = 2

; the least value of

y

-2

(iii)

y

-intercept

= 6

Example 6.10

Using the mirror line

A parabola passes through $P(-2, 7)$ , $Q(6, 7)$ and $A(4, -5)$ .

(i)Find the equation of the axis of symmetry.
(ii)Find the coordinates of the mirror image of $A$ in the axis of symmetry.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

P

and

Q

are at the same height, so they are mirror images of each other in the axis of symmetry.

Equal heights on a parabola come in mirror pairs.

\displaystyle x = \dfrac{-2 + 6}{2} = 2

The axis is half-way between

P

and

Q

(ii)

A

2

units to the right of the axis, so its mirror image is

2

units to the left, at the same height.

Mirror images are the same distance from the axis.

\displaystyle x = 2(2) - (4) = 0

New

x

= twice the axis value minus the old

x

Mirror images are the same height and the same distance from the axis.

Final answer

(i)

x = 2

(ii)

(0, -5)

Your Turn 6.10

A parabola passes through $P(-4, 5)$ , $Q(2, 5)$ and $A(1, 0)$ .

(i)Find the equation of the axis of symmetry.
(ii)Find the coordinates of the mirror image of $A$ in the axis of symmetry.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

P

and

Q

are at the same height, so they are mirror images of each other in the axis of symmetry.

Equal heights on a parabola come in mirror pairs.

\displaystyle x = \dfrac{-4 + 2}{2} = -1

The axis is half-way between

P

and

Q

(ii)

A

2

units to the right of the axis, so its mirror image is

2

units to the left, at the same height.

Mirror images are the same distance from the axis.

\displaystyle x = 2(-1) - (1) = -3

New

x

= twice the axis value minus the old

x

Final answer

(i)

x = -1

(ii)

(-3, 0)

Example 6.11

Shape and intercept without drawing

For each graph, say whether the turning point is a maximum or a minimum, and write down the $y$ -intercept.

(i) $y = 2x^2 - 5x + 1$
(ii) $y = -x^2 + 5x$

Show solutionHide solution

(i)

a = 2

is positive, so the graph is a bowl.

Sign of

a

Put

x = 0

: the

y

-intercept is the constant term,

1

Constant term =

y

-intercept.

(ii)

a = -1

is negative, so the graph is an arch.

Sign of

a

There is no constant term, so the graph passes through the origin and the

y

-intercept is

0

Constant term =

y

-intercept.

Final answer

(i)Minimum point;

y

-intercept

= 1

(ii)Maximum point;

y

-intercept

= 0

Think About It

Why can a parabola that opens upward never have a highest point?

Show the worked answerHide the worked answer

An upward parabola has

a > 0

, so both arms point up.

Sign of

a

The arms never stop rising, so no point is the highest.

Going up for ever.

The only extreme point is the lowest one: the minimum at the turning point.

One extreme point only.

Final answerBoth arms keep rising for ever.

Links To

For a shorter, free version of this topic, see the free E-Math revision page on functions and graphs

Locked · Chapter 6

You've read the free 26%. Unlock the rest of Chapter 6.

Still to read in this chapter:

Completing the square and (x - p)² + q
Factorised forms
Power functions
Exponential graphs
Gradient of a curve
Draw the graph, then use it
Formula panel
Exercise

$9 one-time · all 19 chapters · no subscription

Worked examples, formulas and exam checks.
The whole E-Math syllabus.

ONE PRICE19chapters

Premium members get all 19 chapters included.

Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 6 of 19