The graph of a
quadratic
function is
called a
parabola.
The parent
function is
given as
y x
2
This is the
parent graph
of all
quadratic
functions.
(-3,9)
A table of
values can be
constructed
from the graph
as given to the
(-2,4)
right.
(3,9)
(2,4)
(1,1)
(-1,1)
(0,0)
y x
2
x
-3
-2
-1
0
1
2
3
y
9
4
1
0
1
4
9
y x
(-3,9)
2
All other quadratic functions
can be expressed in the form:
(3,9)
y  a (x  h )2  k
This is called the standard form.
(-2,4)
(2,4)
The general form is given as:
y  ax 2  bx  c
(1,1)
(-1,1)
(0,0)
(h,k)
In standard form,
y = a  x  h 2 + k
(h,k) identifies the
vertex of the parabola.
(1,8)
(5,8)
In standard form,
y = a  x  h 2 + k
a affects the direction the
parabola opens and how wide
or narrow it will open.
(2,2)
(4,2)
(3,0)
y = 2  x  3 2 + 0
Since a=2 and it is positive, the
parabola opens up and
the y-values are all 2 times
larger than on the parent
graph.
(3,0)
(2,-2)
In standard form,
(4,-2)
y = a  x  h 2 + k
If a is negative, the
parabola will open down.
Since a=-2 and it is negative,
the-parabola
opens down
2
-2
- 2and
(5,-8)
(1,-8)
- 2 the y-values are all 2 times
2
-2
larger than on the parent
2
y = - 2 -x2 3 2 + 0
graph.
(3,8)
(4,6)
(2,6)
The points where the
parabola intersects the xaxis are called the
Roots or Zeros of the
function.
These roots occur when the
y-value is equal to zero.
0  2(x  3)2  8
(1,0)
1
1
1
Solving for x we get the
(5,0)
5
values:
5 1 51
X=
1 15
5 X=5
5
2
y  2(x  3)  8
Example:
Graph
y  3(x  5)  2
2
(5,
-2)
The vertex is
(5,
-2)
(5,
-2)
(5,
-2)
(5,
-2)
The graph opens (5,
upward
-2)
(5,
-2)
(5,
-2)
because 3 is positive.
(5,
(5,-2)
-2)
(5,
-2)
(5,
(5,-2)
-2)-2)
The y-values are
(5,
(5,(5,
-2)
multiplied by 3.
-2)
(5,
-2)
(5,
-2)-2)
(5,
(5,(5,
-2)
-2)
(5,
(5,-2)
-2)
Over 1
up 3
The zeros of the
function can be found
by setting y=0.
0  3(x  5)  2
2
Now solve for x.
0  3(x  5)2  2
0  2  3(x  5)2  2  2
2  3(x  5)2
1
1
2  3(x  5)2
3
3
2
 (x  5)2
3
2

 (x  5)2
3

2
 x 5
3
The roots or zeros are:
5
2
 x 55
3
(4.33, 0) and (5.66,0)
5
2
 x 55
3
2
x
3
5.66  x
5
4.33  x