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.
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
-2
-1
0
1
2
3
4
5
6
7
8
Y
Now continue the same process for
the rest of the numbers.
And plot them on a grid and join with a
smooth curve
y
30
26
22
18
14
10
6
2



x

-6
-10






X
-2
-1
0
1
2
3
4
5
6
7
8
Y
20
12
6
2
0
0
2
6
12
20
30
There is an alternative to the sometimes
laborious “box” method of plotting parabolas.
This alternative uses the y-intercept, the xintercepts 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 = 0 at these 2
pts
y = x2 + x - 6
because these are the points where y = 0
So here we go….
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 xintercepts equal to -3
and 2
y = x2 + x - 6
y
5
3
1





x





-3
-5
-7
-9
-11
-13
-15
x-intercepts are -3
and 2
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 yintercept as it’s the
easiest. It’s the yvalue you get when
you make x = 0. It’s
also very easy to
recognise it as the
number on its own,
i.e. +6.
So we now have our first point
for our parabola…..
x
y
0
6
y-intercept
 (0, 6)
y
5
4
3
2
1
x






-1
-2
-3
-4
-5




STEP 2
Now 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
so either
which gives
Δx◊=6
Δ + ◊ = -5
(x – 2)(x – 3) = 0
x – 2 = 0 or x – 3 = 0
x=2
x=3
gives -2 and -3
These are the
x-intercepts
So we now have 2 more points
for our parabola…..
x
y
0
2
3
6
0
0
y-intercept
 (0, 6)
y
5
4
3
2
1






-1
-2
-3
-4
-5




x


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
 b 4ac  b 2 
 

,
4a
 2a

So in y = 1 x2 – 5
5x + 6
6, first write down what a, b
and c are:
a = 1 b = -5 c = 6
a = 1, b = -5, c = 6
so we substitute these values into our formula
and use the calculator to work it out….
and this will give
us our turning
point
 b 4ac  b 
 

,
4a
 2a

  5 4  1  6  ( 5)2 

  
,
4 1
 21

2
 2.5,0.25
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
So we now have another point
for our parabola…..
x
y
0
2
3
2.5
6
0
0
-0.25
y-intercept
 (0, 6)
y
5
4
axis of symmetry is the
vertical line x = 2.5
3
2
1











x


-1
-2
(2.5, –0.25)
-3
-4
-5
Now you can plot the graph! You may wish
to plot a few extra
points to help with accuracy.
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,
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
x
0
y
-8
-4
2
0
0
-8
STEP 3 – Turning point.
a = 1, b = 2, c = -8
 b 4ac  b 2 
 

,
4a
 2a

2

2 4  1  8  2 

  
,
4 1
 21

  1,9 
y = x2 + 2x - 8
x
0
y
-8
-4
2
-1
0
0
-9
y
5
3
1





x





-3
-5
-7
-9
axis of
symmetry x = -1
-11
-13
-15
 (0, -8) y-intercept
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
STEP 2 – x intercepts.
Found by making y = 0 and solving
3x – x2 = 0
3x - x2 = 0
Sign change ….x2 – 3x = 0
x(x – 3) = 0
x = 0 OR x – 3 = 0
x = 0 OR x = 3
x
0
y
0
0
3
0
0
Note that here we get a
repeat!! (0,0)
STEP 3 – Turning point.
a = -1, b = 3, c = 0
 b 4ac  b 2 
 

,
4a
 2a

2

3
4  1  0  3 

  
,
4  1
 2  1



 1.5, 2.25
y = 3x - x2
x
0
y
0
0 0
3 0
1.5 2.25
y
y-intercept AND one
of the x-intercepts
(0,0)
5
3
1





x

-3
-5
-7
-9
-11
axis of
symmetry x = 1-13
-15




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
2
h = -d
+ 6d
d
We can represent the tunnel as a graph….see
over
y
10
From this diagram you
can see that…………
9
h = -d2 + 6d
8
7
the base of the tunnel
is 6m wide
becomes
6
the height of the tunnel
is 9m at its highest point
y = -x2 + 6x
5
for graph
drawing purposes
4
3
A train is 4m wide and
4m tall. Would it be able
to pass through?
4m
2
YES! But only just!
1
x
0











y
10
9
How high is the tunnel 1m along the
base from either side of the tunnel?
8
7
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
6
5
4
3
2
1
x
0











In all these questions you must aim
to eventually end up with a
STANDARD QUADRATIC FORMAT,
which looks like
2
ax
+ bx + c = 0
where the a, b and c are just
numbers, and most importantly,
you have ZERO on one side!
Solve the equation x 2  3x  4
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 + 4)(x – 1) = 0
x+4=0
or
Answers
x–1=0
x=1
Solve the equation x 2  6  x
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 - 3)(x + 2) = 0
x-3=0
or
Answers
x+2=0
x = -2
Solve the equation  x 2  2x  24
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 + 6)(x - 4) = 0
x+6 =0
or
Answers
x -4=0
x=4
6
Solve the equation x  5 
x
STEP 1 – get rid of any fractions. In this
case, multiply all three terms by x
x 2  5x  6
STEP 2 – make 0 on right hand side. In this
case, take 6 from both sides
x 2  5x  6  0
STEP 3 – Factorise and solve using familiar routine
(x  6)(x  1)  0
So….
x = 6 or x = -1