The aim of this powerpoint is to enable
you to quickly and easily draw the graphs
of parabolas when given their equations.
It will show you how to
• determine whether or not a parabola
will intersect the x-axis
• determine the turning point of a
parabola
• organise the appearance of the
parabola’s equation to give key
information about its features.
The equation of a parabola
The general equation of a parabola is any
equation of the format
y = ax2 + bx + c
where a, b, c are constants.
b and c can equal anything including zero, in
which case the equation may be missing the
second and/or the third term. But a can NEVER
be zero, otherwise the x2 term disappears and you
would no longer have a parabola, but instead,
y
= bx + c which is a straight line!
Here are some equations of parabolas
a
2
1
3
½
1
–2
–1
b
5
1
0
–8
0
1
0
c
3
–4
7
0
0
9
4
Equation
y = 2x2 + 5x + 3
y = x2 + x – 4
y = 3x2 + 7
y = ½x2 – 8x
y = x2
y = – 2x2 + x + 9
y = 4 – x2
Note – the red one is the basic parabola, y = x2.
Note also sometimes the terms are mixed up –
be on the lookout for this! y = 2x – 7 – x2
Special features – LEARN THESE!
If a > 0 it’s upright ; if a < 0 it’s inverted
If b is zero (i.e. no “x”-term) then it’s
symmetrical about the y-axis. No sideways
movement.
If c is zero (i.e. no constant on the end)
then it passes through the origin. The
origin is both an x and y intercept.
The x-value of the VERTEX (turning point)
is the MEAN of the x-values of the two x
intercepts.
Now check the table on the previous page
and identify their features!
a
2
b
5
c
3
Equation
y = 2x2 + 5x + 3
UPRIGHT
1
3
1
0
–4
7
y = x2 + x – 4 UPRIGHT
y = 3x2 + 7 UPRIGHT,
SYMMETRICAL
½
–8
0
y = ½x2 – 8x UPRIGHT,
GOES THROUGH ORIGIN
1
0
0
–2
1
9
y = x2
UPRIGHT, THROUGH
ORIGIN, SYMMETRICAL
y = – 2x2 + x + 9
INVERTED
–1
0
4
y = 4 – x2 INVERTED,
SYMMETRICAL
Drawing parabolas on the TI
Example To draw the graph of y = x2 – 4x + 3
STEP 1 Select Y= and
enter equation
STEP 2 Select WINDOW &
enter Xmin, Xmax
values only. Ignore
y values.
STEP 3 Select ZOOM 0
(i.e. ZOOMFIT)
Drawing parabolas on the TI – cont’d
5
-5
This is a much better
representation as it more
clearly shows x-intercepts
A, B and the location of
the turning point T.
Note we have set WINDOW Xrange as – 5 to 5. A better graph
can be drawn by narrowing this a
bit. Try adjusting WINDOW by
changing Xmin to 0 and Xmax to 4.
Then ZOOM 0.
0
A
B
4
T
There is a fair bit of trial and error (mostly error!) associated with
setting the WINDOW. Begin with -5 to 5, hit ZOOM 0 and then
experiment by adjusting your values to get a better picture.
Using the TABLE facility on the TI
This is a very handy facility as it lets you see the coordinates
of all points lying on your parabola. It is especially useful if
you have to draw a graph by hand and it’s not feasible just to
copy the graph off the screen.
STEP 1 Select Y= and enter
equation as before
STEP 2 Select TBLSET.
This lets you choose
where you start and
what the x-values go
up by (usually 1’s).
Set TblStart = 0
Set ΔTbl = 1
The TABLE facility on the TI – cont’d
STEP 3 Select TABLE
Note the first x value is 0 (we set this up with TblStart) and
the x’s jump by 1’s (we set this up with ΔTbl).
All these points (0, 3), (1, 0), (2, -1) etc lie on your parabola
Note also that you can use the up and down arrow keys to
SCROLL up and down this table to see other points
Using TABLE to locate key points
(0, 3) is the y-intercept
(1, 0) and (3, 0) are
the x-intercepts
(2, -1) is the vertex
We can also determine the vertex from this table.
Note how the y values are symmetrical about the value –1,
i.e. reading down the y-values notice that they follow a
pattern 3, 0, -1, 0, 3….
If this happens, the CENTRE of the pattern is on the axis of
symmetry going through the middle of the parabola, and so is
the turning point. SO, the Vertex is (2, – 1). See next slide
for the visual
y
y-intercept
(0, 3)
x-intercepts
(1,0) & (3,0)
0
x
Vertex (called a “Minimum
Turning Point ” when it is the
lowest point) (2, – 1)
This vertical line, x = 2, is called
the AXIS OF SYMMETRY. It
splits the parabola in half and
contains the Turning Point
A note about the y-intercept
For the parabola y = ax2 + bx + c, the y-intercept is
obtained by substituting x = 0 into the equation.
So, replacing x with 0, we obtain
y = a (0)2 + b(0) + c
i.e.
c
y=c
So the y-intercept of the parabola y = ax2 + bx + c
is equal to
c.
Compare with straight lines y = mx + c which also
have a y-intercept equal to c !
Now try graphing the parabolas in
the table on Slide #4.
So far we have….
• Looked at parabolas whose equations are of
the form y = ax2 + bx + c. This is called
GENERAL FORM
• Investigated some facilities available on the
TI relating to drawing graphs and obtaining
tables of values.
• Used TABLE to identify intercepts and
turning points.
Now we will consider how to draw parabolas
using algebra, rather than relying on the
graphics.
Factorised form for the equation of a
parabola. Also called x-intercept form.
In previous slides we looked at the equation of the
parabola y = x2 – 4x + 3. (called GENERAL FORM)
If we FACTORISE the right hand side, we obtain
y = (x – 1)(x – 3)
This is known as the FACTORISED FORM for
the equation of a parabola and its major use is for
determining x – intercepts. If you’re not using a
graphics, pinpointing the x – intercepts will
enable you to graph the parabola quickly and
easily!
Example
Convert y = x2 – 4x – 5 to factorised form, and
determine its x and y intercepts, turning point,
equation of axis of symmetry.
STEP 1
Finding the
y-intercept
The y-intercept is equal to c and so is
–5
STEP 2
Factorising the right hand side
Finding the
x-intercepts
x2 – 4x – 5 = (x – 5)(x + 1)
so our parabola’s equation becomes
y = (x – 5)(x + 1)
STEP 2
Finding the
x-intercepts
It is important that you
know: x-intercepts lie on
the x-axis and therefore
have no height, i.e. y = 0
y = (x – 5)(x + 1)
So when we make y = 0, this means we make
(x – 5)(x + 1) = 0 and solve this quadratic. The
solutions will be the x-intercepts.
(x – 5)(x + 1) = 0
So x – 5 = 0 OR x + 1 = 0
Which gives x
= 5 and x = – 1
as our x-intercepts
STEP 3
Plotting the
intercepts
So far we know the y-intercept is
(0, – 5) and x-intercepts are (5, 0)
and (– 1, 0). We plot these below:
y
10
8
6
4
2
-10
-8
-6
-4
-2

2
-2
-4
-6
-8
-10

4
6
x
8
10
STEP 4
y
10
Plotting the
axis of symmetry
8
6
The axis of
symmetry cuts
the parabola in
half. It cuts
through the xaxis at the point
halfway
between the two
x-intercepts.
4
2
-10
-8
-6
-4

-2
2
6
4
-2
-4
-6

-8
-10
This point is just the mean of – 1 and 5, i.e.
(– 1 + 5) ÷ 2 which is 2.
We can now draw the axis of symmetry.
x
8
10
STEP 5
Plotting the
Turning Point
y
10
8
The turning point
lies on the axis of
symmetry and so
has an x-value
equal to
2.
6
4
2
-10
-8
-6
-4
-2

-2
-4
It now remains for
us to find the yvalue for the
turning point.
This is done by
substituting x = 2
into the parabola’s
equation
2
6
4
-6

-8
-10
 
y = x2 – 4x – 5
Substituting x = 2
y = 22 – 4(2) – 5 =
–9
Turning pt is (2, – 9)
x
8
10
STEP 6
Joining the dots!
y
10
8
6
4
2
-10
-8
-6
-4
-2

2
6
4
x
8
-2
-4
-6
-8
-10


It’s now just a question of joining the dots. Remember the
parabola is symmetrical about the vertical dotted line. You
can also plot other points for greater accuracy.
10
Example (shorter than the last one!)
Use algebra to determine y-intercept, x-intercepts,
turning point and axis of symmetry of the parabola
y = 4 – 2x – 2x2
STEP 1
Y-INTER
STEP 2
X-INTERS
Rearrange into normal quadratic form
y = – 2x2 – 2x + 4
Y – INTERCEPT = (0, 4)
Note that the a value is – 2, and as
it’s negative the parabola will be
upside-down (inverted). Take out a
common factor, with the negative.
y = – 2x2 – 2x + 4
y = – 2(x2 + x – 2)
y = – 2(x2 + x – 2)
Factorise the brackets
y = – 2(x + 2)(x – 1)
X – INTERCEPTS = ( – 2, 0) AND (1, 0)
STEP 3
TURNING POINT
AXIS OF SYMM
Mean of the two x-intercepts is – ½ .
This is the x – value of the turning point.
So the turning point is ( – ½ , ? )
To find the y-value (?), substitute x = – ½
into original equation y = 4 – 2x – 2x2
TURNING PT IS (– ½ , 4 ½) AND THE
AXIS OF SYMMETRY IS x = – ½
y
Turn pt
(– ½ , 4 ½ )
5.0
y inter
(1, 0)
4
3
2
x inter
(–2 , 0)
-5.0
-4
-3
x inter
(1, 0)
1
-2
-1
1
-1
-2
-3
-4
-5.0
2
3
4
x
5.0
We have investigated two formats for
equations of parabolas…
General form
Factorised form
y = ax2 + bx + c
y = a(x – α)(x – β)
Also known as
expanded form, this
quickly tells you the
y-intercept is c.
The sign of a tells
you whether it’s
upright or inverted.
This is useful as it
tells you the xintercepts are at
x = α and x = β.
We have also worked out that the TURNING
POINT has as its x-coordinate the MEAN
(average) of the two x-intercepts…..
…..and that the AXIS OF SYMMETRY is the
vertical line that splits the parabola in half. The
axis of symmetry contains the TURNING
POINT and it crosses the x-axis halfway
between the two x-intercepts…..
We look at the third version of the equation of the
parabola – the VERTEX (turning point) format.
Investigation….
(1) Expand and simplify (x – 2)2 – 1.
(2) On your graphics, plot y = (x – 2)2 – 1.
(3) Use your graphics to determine the
vertex of y = (x – 2)2 – 1.
Solution….
(1) Expand and simplify (x – 2)2 – 1.
(x – 2)2 – 1
= x2 – 4x + 4 – 1
= x2 – 4x + 3
(2) Enter Y1 = (X – 2)2 – 1
Adjust window
(3) Use your graphics to determine the
vertex of y = (x – 2)2 – 1.
Select 2nd TRACE 3
Move cursor a little to
the LEFT of the vertex.
Hit ENTER
Move cursor a little to
Hit ENTER
the RIGHT of the vertex.
Hit ENTER
The Vertex is therefore at
(2, – 1)