Introduction of Quadratic Functions and its graph

1
Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 8-1
Quadratic
Functions
Chapter 8

2
§ 8.5
Graphing Quadratic
Functions

3
Quadratic Functions
Quadratic Function
A quadratic function is a function that can be written
in the form
f(x) = ax2 + bx + c
For real numbers a, b, and c, with a ≠ 0.

4
The graph of every quadratic function is a
parabola.
Definitions
The vertex is the lowest point on a parabola
that opens upward, or the highest point on
a parabola that opens downward.

5
Graphs of quadratic equations have symmetry
about a line through the vertex. This line is called
the axis of symmetry.
The sign of a, the numerical coefficient of the
squared term, determines whether the parabola
will open upward or downward.
Definitions

6
Vertex of a Parabola
Vertex of a Parabola
The parabola represented by the function f(x) = ax2 +
bx + c will have vertex
Since we often find the y-coordinate of the vertex by
substituting the x-coordinate of the vertex into f(x), the
vertex may also be designated as





 

a
b
ac
a
b
4
4
,
2
2













a
b
f
a
b
2
,
2

7
Axis of Symmetry of a Parabola
Axis of Symmetry
For a quadratic function of the form f(x) = ax2 + bx + c,
the equation of the axis of symmetry of the parabola is
a
b
x
2



8
x-Intercepts of a Parabola
x-Intercepts of a Parabola
To find the x-intercepts (if there are any) of a quadratic
function, solve the equation ax2 + bx + c = 0 for x.
This equation may be solved by factoring, by using the
quadratic formula, or by completing the square.

9
Graph Quadratic Functions
Example Consider the quadratic function
y = – x2 + 8x – 12.
a. Determine whether the parabola opens upward
or downward.
b. Find the y-intercept.
c. Find the vertex.
d. Find the equation of the axis of symmetry.
e. Find the x-intercepts, if any.
f. Draw the graph.
continued

10
a. Since a is -1, which is less than 0, the parabola
opens downward.
b. To find the y-intercept, set x = 0 and solve for y.
continued
12
12
)
0
(
8
)
0
( 2






y
The y-intercept is (0, 12)

11
c. First, find the x-coordinate, then find the y-
coordinate of the vertex. From the function, a = -1,
b = 8, and c = -12.
continued
4
)
1
(
2
8
2





a
b
x
Since the x-coordinate of the vertex is not a fraction,
we will substitute x = 4 into the original function to
determine the y-coordinate of the vertex.
4
12
32
16
12
)
4
(
8
)
4
(
12
8
2
2













y
x
x
y
The vertex is (4, 4).

12
continued
d. Since the axis of symmetry is a vertical line
through the vertex, the equation is found using the
same formula used to find the x-coordinate of the
vertex (see part c). Thus, the equation of the axis of
symmetry is x = 4.

13
continued
e. To find the x-intercepts, set y = 0.
2
x
6
0
2
or
0
6
0
)
2
)(
6
(
0
12
8
2












x
x
x
x
x
x
x
Thus, the x-intercepts are (2, 0) and (6, 0). These
values could also be found by the quadratic formula
(or by completing the square).

14
f. Draw the graph.

15
Solve Maximum and Minimum Problems
A parabola that opens upward has a minimum value
at its vertex, and a parabola that opens downward has
a maximum value at its vertex.

16
Understand Translations of Parabolas
-4
x
y
4
4
  2
1
2
h x x

  2
f x x

  2
2
g x x

Start with the basic graph of f(x) = ax2 and translate, or shift, the
position of the graph to obtain the graph of the function you are
seeking.
Notice that the value of a in the graph f(x) = ax2 determines the
width of the parabola. As |a| gets larger, the parabola gets
narrower, and as |a| gets smaller, the parabola gets wider.

17
-4
x
y
4
4
  2
( 2)
h x x
 
  2
f x x

  2
( 2)
g x x
 
seeking.
If h is a positive real number, the graph of g(x) = a(x – h)2 will be
shifted h units to the right of the graph g(x) = ax2. If h is a negative
real number, the graph of g(x) = a(x – h)2 will be shifted |h| units to
the left.

18
-4
x
y
4
4
  2
1
h x x
 
  2
f x x

  2
1
g x x
 
seeking.
In general, the graph of g(x) = ax2 + k is the graph of f(x) = ax2
shifted k units up if k is a positive real number and |k| units down if
k is a negative real number.

19
Parabola Shifts
For any function f(x) = ax2, the graph of g(x) = a(x-h)2 + k will
have the same shape as the graph of f(x). The graph of g(x)
will be the graph of f(x) shifted as follows:
• If h is a positive real number, the graph will be shifted h
units to the right.
• If h is a negative real number, the graph will be shifted |h|
units to the left.
• If k is a positive real number, the graph will be shifted k
units up.
• If k is a negative real number, the graph will be shifted |k|
units down.

20
Axis of Symmetry and Vertex of a Parabola
The graph of any function of the form
f(x) = a(x – h)2 + k
will be a parabola with axis of symmetry x = h and vertex at
(h, k).

21
Write Functions in the Form f(x) = a(x – h)2 + k
If we wish to graph parabolas using translations, we
need to change the form of a function from f(x) = ax2
+ bx + c to f(x) = a(x – h)2 + k. To do this we complete
the square as we discussed in Section 8.1.
Example Given f(x) = x2 – 6x + 10,
a) Write f(x) in the form of f(x) = a(x – h)2 + k.
b) Graph f(x).

22
a) We use the x2 and -6x terms to obtain a perfect
square trinomial.
Now we take half the coefficient of the x-term and
square it.
We then add this value, 9, within the parentheses.
10
)
6
(
)
( 2


 x
x
x
f
9
)
6
(
2
1
2






 
10
9
)
9
6
(
)
( 2




 x
x
x
f
continued

23
By doing this we have created a perfect square
trinomial within the parentheses, plus a constant
outside the parentheses. We express the perfect
square trinomial as the square of a binomial.
The function is now in the form we are seeking.
1
)
3
(
)
( 2


 x
x
f
continued

24
b) Graph f(x).

Introduction of Quadratic Functions and its graph

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