When you complete a table of values for y = 2x – 3 ,

You get a STRAIGHT LINE

When you complete a table of values for y = x2 + 5x – 6 ,

You get a PARABOLA, which is a CURVE. It’s the x2 that makes the graph curve. If the x2

wasn’t there, you would just have a line!

So let’s try plotting y = x2 – 5x + 6 using a table of values.

First choose a few x-values and make sure they’re in order and spaced 1 apart.

X Y

-2

-1

0

1

2

3

4

5

6

7

8

Now one by one, put these into the equation y = x2 - 5x + 6, being sure to bracket negatives when squaring them (they become positives!!)

Begin with x = -2

(–2)2 – 5 (– 2) + 6

= 4 + 10 + 6

20 =

So our first point is (2, 20)

X Y

-2 20

-1

0

1

2

3

4

5

6

7

8

Now continue the same process for the rest of the numbers.

12

6

2

0

0

2

6

12

20

30

And plot them on a grid and join with a smooth curve

x

y

-10

-6

2

6

10

14

18

22

26

30

There is an alternative to the sometimes laborious “box” method of plotting parabolas. This alternative uses the y-intercept, the x-intercepts and the turning point to get a graph. Once you become practised at this you will find it quicker than boxes and substitution.

When you solve a quadratic such as

0 = x2 + x - 6 You are actually finding the two x-intercepts of the parabola

y = x2 + x - 6 because these are the points where y = 0

x2 + x - 6 = 0( x + 3 )(x – 2) = 0

x = -3 or x = 2

This means that the parabola

y = x2 + x - 6 will have x-

intercepts equal to -3 and 2

So here we go….

y = 0 at these 2 pts

x

y

-15

-13

-11

-9

-7

-5

-3

1

3

5

x-intercepts are -3 and 2

x

y

-15

-13

-11

-9

-7

-5

-3

1

3

5

y = x2 + x - 6

Plot the graph of y = x2 – 5x + 6, by first finding the x and y intercepts, and its turning point.

STEP 1

Start with the y-intercept as it’s the easiest. It’s the y-value you get when you make x = 0. It’s also very easy to recognise it as the number on its own, i.e. +6.

x y

0 6

So we now have our first point for our parabola…..

x

y

-5

-4

-3

-2

-1

1

2

3

4

5

(0, 6)y-intercept

STEP 2Now we’ll do the x-intercepts – they require a bit more work. See “IMPORTANT INFO” on Slide #2. They’re the x-values you get when you make y = 0. So this means we have to solve the quadratic equation 0 = x2 – 5x + 6 .

The process is a familiar one!!! Factorise then solve

x2 – 5x + 6 = 0

First we factorise the quadratic

Δ x ◊ = 6

Δ + ◊ = -5gives -2 and -3

(x – 2)(x – 3) = 0so either

x – 2 = 0

or x – 3 = 0which gives

x = 2

x = 3

These are the x-intercepts

x y

0 6

2 0

3 0

So we now have 2 more points for our parabola…..

x

y

-5

-4

-3

-2

-1

1

2

3

4

5

(0, 6)y-intercept

STEP 3

Now for the TURNING POINT

If your parabola is written as y = ax2 + b x + c, then the turning point’s coordinates can be found from the formula

abac

ab

44

,2

2

So in y = x2 – 5x + 6, first write down what a, b and c are:

b = -5

1 – 5

a = 1 c = 6

+ 6

abac

ab

44

,2

2

a = 1, b = -5, c = 6so we substitute these values into our formula and use the calculator to work it out…. and this will

give us our turning point

14)5(614

,12

5 2

25.0,5.2 For the bright sparks! You may have noticed that the x-value of the turning point is always midway between the 2 x-intercepts. This is another reliable way of finding the turning point and avoids using the formula above! Here, x-intercepts were 2 and 3, so the turning point’s x-value is 2.5 ! Find y by substituting 2.5 into the formula y = x2 – 5x + 6 and this will give y = -0.25

x y

0 6

2 0

3 0

2.5 -0.25

So we now have another point for our parabola…..

x

y

-5

-4

-3

-2

-1

1

2

3

4

5

y-intercept

(2.5, –0.25)

(0, 6)

x

y

-5

-4

-3

-2

-1

1

2

3

4

5

Now you can plot the graph! You may wish to plot a few extra points to help with accuracy.

axis of symmetry is the vertical line x = 2.5

Plot the graph of y = x2 + 2x - 8 using intercepts and turning point. State equation of axis of symmetry.

STEP 1 – y intercept.This is the number on its own, -8

x y

0 -8

STEP 2 – x intercepts.

Found by making y = 0 and solving x2 + 2x – 8 = 0

x2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x + 4 = 0 OR x – 2 = 0

x = - 4 OR x = 2

-4 0 2 0

STEP 3 – Turning point.

abac

ab

44

,2

2

a = 1, b = 2, c = -8

14

2814,

122 2

9,1

x y

0 -8

-4 0 2 0-1 -9

y = x2 + 2x - 8

x

y

-15

-13

-11

-9

-7

-5

-3

1

3

5

(0, -8) y-intercept

(2, 0) x-intercept

(-4, 0) x-intercept

(-1, -9) turning ptaxis of symmetry x = -1

Plot the graph of y = 3x - x2 using intercepts and turning point.

STEP 1 – y intercept.This is the “invisible” number on its own, 0

x y

0 0

STEP 2 – x intercepts.

Found by making y = 0 and solving 3x – x2 = 0

3x - x2 = 0Sign change ….x2 – 3x = 0

x = 0 OR x – 3 = 0

x = 0 OR x = 3

0 0 3 0x(x – 3) = 0

Note that here we get a repeat!! (0,0)

STEP 3 – Turning point.

abac

ab

44

,2

2

a = -1, b = 3, c = 0

14

3014,

123 2

25.2,5.1

x y

0 0

0 0 3 0

y = 3x - x2

1.5 2.25

x

y

-15

-13

-11

-9

-7

-5

-3

1

3

5y-intercept AND one of the x-intercepts (0,0)

(3,0) x intercept

(1.5, 2.25) turning

pt

axis of symmetry x = 1

Of the 3 parabolas we graphed, only the last one had a negative in front of the x2.

Its graph was upside down!

The entrance to a railway tunnel forms a parabolic arch. At any distance d across the base of the tunnel, starting from the left, the height h is found using the equation

h = -d 2 + 6d

d

h

We can represent the tunnel as a graph….see over

x

y

0

1

2

3

4

5

6

7

8

9

10

h = -d2 + 6d

becomes

y = -x2 + 6x

for graph drawing purposes

From this diagram you can see that…………

the base of the tunnel is 6m wide

the height of the tunnel is 9m at its highest point

A train is 4m wide and 4m tall. Would it be able to pass through?

YES! But only just!

4m

x

y

0

1

2

3

4

5

6

7

8

9

10

How high is the tunnel 1m along the base from either side of the tunnel?The base points are at 1 and 5. Moving up to meet the graph, we can see that these will be 5 up. The tunnel is therefore 5m high at points 1m in from either side

In all these questions you must aim to eventually end up with a STANDARD QUADRATIC FORMAT, which looks like

ax2 + bx + c = 0

where the a, b and c are just numbers, and most importantly, you have ZERO on one side!

Solve the equation 432 xx

Must get 0 on one side, so subtract the 4:

x2 + 3x – 4 = 0

This is now in regular quadratic format so factorise and follow normal routine….

(x + 4)(x – 1) = 0

x + 4 = 0 or x – 1 = 0

Answers

x = -4

x = 1

Solve the equation xx 62

Must get 0 on one side, so subtract the x and place it in the middle between the x2 and the -6:x2 - x – 6 = 0

This is now in regular quadratic format so factorise and follow normal routine….

(x - 3)(x + 2) = 0

x - 3 = 0 or x + 2 = 0

Answers

x = 3

x = -2

Solve the equation 2422 xx

Must get 0 on one side, so add x2:

0 = x2 + 2x - 24

This is now in regular quadratic format so factorise and follow normal routine….

(x + 6)(x - 4) = 0

x + 6 = 0 or x - 4 = 0

Answers

x = -6

x = 4

Solve the equationx

x6

5

STEP 1 – get rid of any fractions. In this case, multiply all three terms by x

652 xx

STEP 2 – make 0 on right hand side. In this case, take 6 from both sides

0652 xxSTEP 3 – Factorise and solve using familiar routine 0)1)(6( xx

So…. x = 6 or x = -1