Part II: The Answers
156
Answers
1–100
75.
f x
x
x
x
x
( ) = −
+
−
−
+
1
2
7
2
13
2
3
2
9
4
3
2
The polynomial is fourth-degree and has the x-intercepts at x = –1, 2, and 3, where 3 is
a repeated root. Therefore, the polynomial has the factors (x + 1), (x – 2), and (x – 3)
2
,
so you can write
f x
a x
x
x
( )
(
)(
)(
)
=
+
−
−
1
2
3
2
Use the point (1, 4) to solve for a:
4
1 1 1 2 1 3
4
2 1 4
1
2
2
=
+
−
−
=
−
− =
a
a
a
(
)(
)(
)
( )( )( )
Now you can enter the value of a in the equation of the polynomial and simplify:
f x
x
x
x
x
x
x
x
x
x
x
( )
(
)(
)(
)
= −
+
−
−
= −
− −
(
) − +
(
)
= −
−
+
1
2
1
2
3
1
2
2
6
9
1
2
6
9
2
2
2
4
3
2 2
3
2
2
4
3
2
4
6
9
2
12 18
1
2
7
13
3 18
1
2
7
2
−
+
−
−
+
−
(
)
= −
−
+
+
−
(
)
= −
+
x
x
x
x
x
x
x
x
x
x
x
3 3
2
13
2
3
2
9
−
−
+
x
x
76.
y
x
x
= −
+
+
1
3
2
3
8
2
The parabola has x-intercepts at x = –4 and x = 6, so you know that (x + 4) and
(x – 6) are factors of the parabola. Therefore, you can write
y a x
x
=
+
−
(
)(
)
4
6
Use the point (0, 8) to solve for a:
8
0 4 0 6
8
24
1
3
=
+
−
= −
− =
a
a
a
(
)(
)
Now you can enter the value of a in the equation of the parabola and simplify:
y
x
x
x
x
x
x
= −
+
−
= −
−
−
(
)
= −
+
+
1
3
4
6
1
3
2
24
1
3
2
3
8
2
2
(
)(
)