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Math Problems
Algebra 1
Transformations of absolute value functions: translations and reflections
k
x
2
+
5
x
=
−
7
k x^{2}+5 x=-7
k
x
2
+
5
x
=
−
7
\newline
In the given equation,
k
k
k
is a constant. Which of the following represents all values of
k
k
k
for which the equation has two distinct real solutions?
\newline
Choose
1
1
1
answer:
\newline
(A)
k
<
−
25
28
k<-\frac{25}{28}
k
<
−
28
25
\newline
(B)
k
>
−
25
28
k>-\frac{25}{28}
k
>
−
28
25
and
k
≠
0
k \neq 0
k
=
0
\newline
(C)
k
<
25
28
k<\frac{25}{28}
k
<
28
25
and
k
≠
0
k \neq 0
k
=
0
\newline
(D)
k
>
25
28
k>\frac{25}{28}
k
>
28
25
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g
(
w
)
=
(
w
+
13
)
3
(
w
+
19
)
2
g(w)=(w+13)^{3}(w+19)^{2}
g
(
w
)
=
(
w
+
13
)
3
(
w
+
19
)
2
\newline
The polynomial function
g
g
g
is defined. When
g
(
w
)
g(w)
g
(
w
)
is divided by
(
w
+
16
)
(w+16)
(
w
+
16
)
, the remainder is
r
r
r
. What is the value of
∣
r
∣
|r|
∣
r
∣
?
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(
3
y
−
2
)
(
y
+
a
)
=
3
y
2
+
b
z
(3 y-2)(y+a)=3 y^{2}+b z
(
3
y
−
2
)
(
y
+
a
)
=
3
y
2
+
b
z
\newline
If the given equation is true for all values of
y
y
y
, where
a
a
a
and
b
b
b
are constants, which of the following is the value of
b
b
b
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
38
-38
−
38
\newline
(B)
12
12
12
\newline
(C)
34
34
34
\newline
(D)
36
36
36
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Find the zeros of the function. Enter the solutions from least to greatest.
\newline
f
(
x
)
=
(
x
−
10
)
2
−
49
f(x)=(x-10)^{2}-49
f
(
x
)
=
(
x
−
10
)
2
−
49
\newline
lesser
x
=
x=
x
=
\newline
greater
x
=
x=
x
=
Get tutor help
Find the zeros of the function. Enter the solutions from least to greatest.
\newline
f
(
x
)
=
(
x
−
7
)
2
−
64
f(x)=(x-7)^{2}-64
f
(
x
)
=
(
x
−
7
)
2
−
64
\newline
lesser
x
=
x=
x
=
\newline
greater
x
=
x=
x
=
Get tutor help
Find the zeros of the function. Enter the solutions from least to greatest.
\newline
f
(
x
)
=
(
x
−
4
)
2
−
25
f(x)=(x-4)^{2}-25
f
(
x
)
=
(
x
−
4
)
2
−
25
\newline
lesser
x
=
x=
x
=
\newline
greater
x
=
x=
x
=
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x
2
+
k
x
−
14
=
0
x^{2}+k x-14=0
x
2
+
k
x
−
14
=
0
\newline
In the given equation,
k
k
k
is a constant. The equation has solutions at
7
7
7
and
−
2
-2
−
2
. What is the value of
k
k
k
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
9
-9
−
9
\newline
(B)
−
5
-5
−
5
\newline
(C)
5
5
5
\newline
(D)
9
9
9
Get tutor help
(
t
+
1
)
2
+
c
=
0
(t+1)^{2}+c=0
(
t
+
1
)
2
+
c
=
0
\newline
In the given equation,
c
c
c
is a constant.
\newline
The equation has solutions at
t
=
3
2
t=\frac{3}{2}
t
=
2
3
and
t
=
−
7
2
t=-\frac{7}{2}
t
=
−
2
7
. What is the value of
c
c
c
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
729
4
-\frac{729}{4}
−
4
729
\newline
(B)
−
121
4
-\frac{121}{4}
−
4
121
\newline
(C)
−
25
4
-\frac{25}{4}
−
4
25
\newline
(D)
−
1
-1
−
1
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A polynomial function has zeros at
3
4
\frac{3}{4}
4
3
and
−
1
-1
−
1
. Which of the following must be a factor of the polynomial?
\newline
Choose
1
1
1
answer:
\newline
(A)
x
−
1
x-1
x
−
1
\newline
(B)
x
+
1
x+1
x
+
1
\newline
(C)
3
x
−
4
3 x-4
3
x
−
4
\newline
(D)
3
x
+
4
3 x+4
3
x
+
4
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A third degree polynomial function
y
=
g
(
x
)
y=g(x)
y
=
g
(
x
)
is defined so that
g
(
2
3
)
=
0
g\left(\frac{2}{3}\right)=0
g
(
3
2
)
=
0
and
g
(
0
)
=
5
g(0)=5
g
(
0
)
=
5
. Which of the following must be a factor of
g
g
g
?
\newline
Choose
1
1
1
answer:
\newline
(A)
x
−
5
x-5
x
−
5
\newline
(B)
x
+
5
x+5
x
+
5
\newline
(C)
3
x
−
2
3 x-2
3
x
−
2
\newline
(D)
3
x
+
2
3 x+2
3
x
+
2
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x
x
3
\frac{x}{\sqrt[3]{x}}
3
x
x
\newline
Which of the following is equivalent to the given expression for all
x
≠
0
x \neq 0
x
=
0
?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
1
1
\newline
(B)
x
3
\sqrt[3]{x}
3
x
\newline
(C)
x
2
3
\sqrt[3]{x^{2}}
3
x
2
\newline
(D)
x
3
x^{3}
x
3
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(
x
−
4
)
(
x
−
5
)
=
0
(x-4)(x-5)=0
(
x
−
4
)
(
x
−
5
)
=
0
\newline
If
x
=
s
x=s
x
=
s
and
x
=
t
x=t
x
=
t
are the solutions to the given equation, which of the following is equal to the value of
∣
s
−
t
∣
|s-t|
∣
s
−
t
∣
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
9
-9
−
9
\newline
(B)
−
1
-1
−
1
\newline
(C)
1
1
1
\newline
(D)
9
9
9
Get tutor help
The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is reflected across the
x
x
x
-axis and then scaled vertically by a factor of
1
4
\frac{1}{4}
4
1
.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
−
4
∣
x
∣
y=-4|x|
y
=
−
4∣
x
∣
\newline
(B)
y
=
∣
x
∣
−
4
y=|x|-4
y
=
∣
x
∣
−
4
\newline
(C)
y
=
−
1
4
∣
x
∣
y=-\frac{1}{4}|x|
y
=
−
4
1
∣
x
∣
\newline
(D)
y
=
∣
x
−
4
∣
y=|x-4|
y
=
∣
x
−
4∣
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The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is shifted to the left by
6
6
6
units.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
∣
x
+
6
∣
−
6
y=|x+6|-6
y
=
∣
x
+
6∣
−
6
\newline
(B)
y
=
∣
x
−
6
∣
−
6
y=|x-6|-6
y
=
∣
x
−
6∣
−
6
\newline
(C)
y
=
∣
x
∣
−
6
y=|x|-6
y
=
∣
x
∣
−
6
\newline
(D)
y
=
∣
x
+
6
∣
y=|x+6|
y
=
∣
x
+
6∣
Get tutor help
The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is scaled vertically by a factor of
1
3
\frac{1}{3}
3
1
.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
∣
x
∣
−
3
y=|x|-3
y
=
∣
x
∣
−
3
\newline
(B)
y
=
−
3
∣
x
∣
y=-3|x|
y
=
−
3∣
x
∣
\newline
(C)
y
=
1
3
∣
x
∣
y=\frac{1}{3}|x|
y
=
3
1
∣
x
∣
\newline
(D)
y
=
∣
x
∣
−
1
3
y=|x|-\frac{1}{3}
y
=
∣
x
∣
−
3
1
Get tutor help
The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is scaled vertically by a factor of
6
6
6
.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
∣
x
−
6
∣
y=|x-6|
y
=
∣
x
−
6∣
\newline
(B)
y
=
6
∣
x
∣
y=6|x|
y
=
6∣
x
∣
\newline
(C)
y
=
1
6
∣
x
∣
y=\frac{1}{6}|x|
y
=
6
1
∣
x
∣
\newline
(D)
y
=
∣
x
−
1
6
∣
y=\left|x-\frac{1}{6}\right|
y
=
∣
∣
x
−
6
1
∣
∣
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The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is reflected across the
x
x
x
-axis and then scaled vertically by a factor of
7
7
7
.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
7
∣
x
∣
y=7|x|
y
=
7∣
x
∣
\newline
(B)
y
=
−
7
∣
x
∣
y=-7|x|
y
=
−
7∣
x
∣
\newline
C)
y
=
1
7
∣
x
∣
y=\frac{1}{7}|x|
y
=
7
1
∣
x
∣
\newline
(D)
y
=
−
1
7
∣
x
∣
y=-\frac{1}{7}|x|
y
=
−
7
1
∣
x
∣
Get tutor help
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the reflection across the
y
y
y
-axis of
f
(
x
)
=
∣
x
∣
f(x) = |x|
f
(
x
)
=
∣
x
∣
.
\newline
Write your answer in the form
a
∣
x
−
h
∣
+
k
a|x - h| + k
a
∣
x
−
h
∣
+
k
, where
a
a
a
,
h
h
h
, and
k
k
k
are integers.
\newline
g
(
x
)
=
g(x) =
g
(
x
)
=
______
\newline
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