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Section 3.1 Quadratic Functions
Quadratic (second degree) function:
f
(
x
)=
ax
2
+
bx
+
c
Example.
f
(
x
)=3
x
2

2
x

2
–2
–1
0
1
2
–2
–1
1
2
Example.
g
(
x

2
x
2
+3
x
–2
–1
0
1
2
–2
–1
1
2
The graph of a quadratic function is a
parabola
.
Note: The graph of a linear function,
f
(
x
ax
+
b
,isa
straight line
.
By completing the square, the quadratic
y
=
ax
2
+
bx
+
c
can be put in the form
y
=
a
(
x
+
b
2
a
)
2

b
2

4
ac
4
a
. For example, by completing the square, 2
x
2

3
x
+4=2(
x

3
4
)
2
+
23
8
.
When a parabola is the plot of the quadratic
y
=
a
(
x
+
r
)
2

s
, the vertical line given by
x
=

r
is the
axis of symmetry
of the parabola. For example, the axis of symmetry of the plot of
y
=3(
x

4)
2

2
is the line
x
=4.
Graphing a quadratic function without a graphing calculator.
•
Complete the square
•
Locate the vertex
•
Check whether the graph opens upward or downward
•
Determine the
x
and
y
intercepts
•
Draw the graph
Example.
y
=2
x
2
+4
x

3
y
=2[
x
2
+2
x
]

3
y
=2[(
x
+1)
2

1]

3
y
=2(
x
2

5
vertex at (
x, y
)=(

1
,

5)
1
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–4
–2
0
2
–4
–2
2
Example.
y
=

2
x
2
+4
x

3
y
=

2[
x
2

2
x
]

3
y
=

2[(
x

1)
2

1]

3
y
=

2(
x

1)
2

1
vertex at (
x, y
)=(1
,

1)
–2
–1
0
1
2
12
Locating the vertex of a parabola without completing the square:
The vertex of the graph of
y
=
ax
2
+
bx
+
c
is at the point (
x, y
)=(

b
2
a
,f
(

b
2
a
)).
Example.
y
=4
x
2
+2
x

1
a
=4,
b
= 2 implies
x
=

2
8
=

1
4
y
=
f
(

1
4
)=4(

1
4
)
2
+2(

1
4
)

1=

5
4
Locating the
y
intercept of the graph of a function: Set
x
= 0 in the formula of the function and
solve for
y
.
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 Fall '11
 Kutter
 Algebra

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