y
4
2
O
x
2
4
2
2
6
4
f
(
x
)
g
(
x
)
10.
vertex (

3, 2), axis of symmetry
x
=

3, maximum
y
=
2, domain
(

Ŭ¶ Ŭ·¶ TCPIG µ

Ŭ¶ ¸·
12.
f
(
x
)
=
3
__
4
(
x

2)
2
+
1
14.
g
(
x
)
=
(
x

2)
2

4
16.
vertex (3, 6),
y
intercept (0, 15)
y
16
12
8
4
O
x
4
8
8
4
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Algebra 2

11

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Topic 2
PearsonRealize.com
18.
y
=
x
2

3
x
+
7
20.
Answers may
vary. Sample: Calculate the vertex to
find the maximum height of the ball.
You can use the vertex to find the
horizontal distance from the starting
point and multiply by 2 to find the
total distance.
22.
x
=

3 or
x
=
9
24.
x
=

1.5 or
x
=
0.5
26.
x
<

5 or
x
>
6
28.
x

6
30.
13

i
32.
3
__
2

1
__
2
i
34.
They did not change
i
2
VQ Ũ±¹
µ¸ Ũ ²
i
)(4
+
i
)
=
±± Ũ ±º
i
36.
(
x

8)
2
=
28
38.
x
=
12
±
ū
____
226
40.
x
=
2
±
ū
___
41
____
2
42.
5.04 seconds; The other root is
negative, which is not an appropriate
value for this situation.
44.
x
=
8 ± 2
ū
___
10
46.
x
=
9 ±
ū
___
71
_______
2
48.
2 real solutions
50.
k
=
± 8
52.
below 25 kilometers per hour
and above 79 kilometers per hour
54.
0
y
8
6
4
2
O
x
1
2
1
2
56.
(

1, 6), (0, 7)
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Topic 3
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Lesson 31
2.
Allie did not write the polynomial
function in standard form; instead,
she said the exponent of the first term
was the degree of the polynomial.
The degree is 6.
4.
For a polynomial
function with an even degree, as
x

and
x
+
,
y
values
either both approach
+
or both
approach

. If the leading
coefficient is positive, both ends
approach
+
. If negative, both
ends approach

. A polynomial
function with an odd degree has end
behavior for positive and negative
x
in opposite directions. If the leading
coefficient is positive, as
x

,
y
–
, and as
x
+
,
y
+
. If the
leading coefficient is negative, the end
behavior is in the opposite direction.
6.
4
8.
As
x

,
y
+
. As
x
+
,
y
+
.
10.
x
=

4,
x
=

2,
x
=
1,
x
=
3
12.
approximately (

0.5, 28)
14.
f
(
x
)
=

2
x
5
+
2
x
4

2
x
3
+
x
2

x
+
1;
The polynomial function must have a
negative leading coefficient and an odd
degree because of the end behavior.
The function must have a degree of at
least 5 because there are 6 terms. The
last term of the function is 1 because
the
y
intercept is (0, 1).
y
O
x
4
2
4
2
4
2
4
16.
Check students’ graphs. Sample:
y
O
x
4
2
4
2
4
2
4
2
y
O
x
4
2
4
2
1
2
1
y
O
x
4
2
4
2
4
2
4
2
18.
f
(
x
)
=
2
x
5
+
x
4
+
5
x
3

3
x
2
+
x

6;
5; 6; 2
20.
f
(
x
)
=
–
x
4

x
3
+
5
x
2
+
9
x
+
12; 4; 5; –1
22.
The leading
coefficient, 7, is positive, so the graph
opens upward. The degree is 4, which
is even, so the end behaviors are the
same. As
x
becomes infinitely positive
or negative, the
y
values approach
+
.
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Topic 3
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24.
zeros:
x
=

3,
x
=

1,
x
=
2; turning
points between

3 and

1, between

1
and 2
y
O
x
4
4
2
8
8
4
26.
y
O
x
4
4
20
10
10
28. a.
f
(
x
)
=
4
x
3

38
x
2
+
88
x
b.
reasonable domain: 0
<
x
<
4
y
O
x
4
6
2
2
20
20
40
60
c.
x
intercepts: 0, 4, and 5.5; 0 and 4
represent the side lengths of the cut
squares that will result in a box with
0 volume. The intercept at 5.5 is not
meaningful, because it is not possible
to cut two 5.5inch corners from an
8inch side.
d.
About 1.5 in. will create
a box with a volume of about 60 in.