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Math Problems
Algebra 2
Find the vertex of the transformed function
The equation
s
=
(
t
+
3
)
2
(
t
+
2
)
(
t
+
1
)
(
t
)
(
t
−
1
)
s=(t+3)^{2}(t+2)(t+1)(t)(t-1)
s
=
(
t
+
3
)
2
(
t
+
2
)
(
t
+
1
)
(
t
)
(
t
−
1
)
is graphed on the
s
t
s t
s
t
-plane. What is the product of the unique
t
t
t
-intercepts of the graph?
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13
13
13
. A curve
C
C
C
has parametric equations
\newline
x
=
t
2
+
5
t
2
+
1
y
=
4
t
t
2
+
1
x=\frac{t^{2}+5}{t^{2}+1} \quad y=\frac{4 t}{t^{2}+1}
x
=
t
2
+
1
t
2
+
5
y
=
t
2
+
1
4
t
\newline
Show that all points on
C
C
C
satisfy
\newline
(
x
−
3
)
2
+
y
2
=
4
(x-3)^{2}+y^{2}=4
(
x
−
3
)
2
+
y
2
=
4
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What is the effect on the graph of the function
f
(
x
)
=
x
2
f(x)=x^{2}
f
(
x
)
=
x
2
when
f
(
x
)
f(x)
f
(
x
)
is changed to
f
(
x
)
+
f(x)+
f
(
x
)
+
9
9
9
?
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We are given that
d
y
d
x
=
x
2
−
2
y
\frac{d y}{d x}=x^{2}-2 y
d
x
d
y
=
x
2
−
2
y
.
\newline
Fasind an expression for
d
2
y
d
x
2
\frac{d^{2} y}{d x^{2}}
d
x
2
d
2
y
in terms of
x
x
x
and
y
y
y
.
\newline
d
2
y
d
x
2
=
\frac{d^{2} y}{d x^{2}}=
d
x
2
d
2
y
=
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We are given that
d
y
d
x
=
e
5
y
\frac{d y}{d x}=e^{5 y}
d
x
d
y
=
e
5
y
.
\newline
Find an expression for
d
2
y
d
x
2
\frac{d^{2} y}{d x^{2}}
d
x
2
d
2
y
in terms of
x
x
x
and
y
y
y
.
\newline
d
2
y
d
x
2
=
□
I
+
‾
\frac{d^{2} y}{d x^{2}}=\square \text { I } \overline{+}
d
x
2
d
2
y
=
□
I
+
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The
x
y
xy
x
y
-plane shows the graph of the quadratic equation
y
=
−
(
1
4
)
(
x
−
3
)
2
+
2
y=-(\frac{1}{4})(x-3)^2+2
y
=
−
(
4
1
)
(
x
−
3
)
2
+
2
. Which of the following represents all solutions
(
x
,
y
)
(x,y)
(
x
,
y
)
to the system of equations created by this quadratic equation and the linear equation
y
=
1
2
(
x
+
3
)
−
7
y=\frac{1}{2}(x+3)-7
y
=
2
1
(
x
+
3
)
−
7
?
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Let
f
f
f
be a twice differentiable function. One of these graphs is the graph of
f
f
f
, one is of
f
′
f^{\prime}
f
′
and one is of
f
′
′
f^{\prime \prime}
f
′′
.
\newline
Choose the option that matches each function with its appropriate graph.
\newline
Choose
1
1
1
answer:
\newline
(A) \begin{tabular}{llll}
\newline
A &
f
f
f
&
f
′
f^{\prime}
f
′
&
f
′
′
f^{\prime \prime}
f
′′
\\
\newline
\hline & I & II & III
\newline
\end{tabular}
\newline
(B) \begin{tabular}{llll}
\newline
\hline
B
\mathrm{B}
B
&
f
f
f
&
f
′
f^{\prime}
f
′
&
f
′
′
f^{\prime \prime}
f
′′
\\
\newline
\hline & I & III & II \\
\newline
\hline
\newline
\end{tabular}
\newline
(C) \begin{tabular}{llll}
\newline
f
f
f
1
1
1
&
f
f
f
&
f
′
f^{\prime}
f
′
&
f
′
′
f^{\prime \prime}
f
′′
\\
\newline
\hline & II & III & I
\newline
\end{tabular}
\newline
(D) \begin{tabular}{|llll}
\newline
\hline
f
f
f
5
5
5
&
f
f
f
&
f
′
f^{\prime}
f
′
&
f
′
′
f^{\prime \prime}
f
′′
\\
\newline
\hline & III & I & II \\
\newline
\hline & & &
\newline
\end{tabular}
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The plot below shows the quadratic
y
=
(
x
−
a
)
2
y=(x-a)^{2}
y
=
(
x
−
a
)
2
. Find the value of
a
a
a
.
\newline
a
=
Number
a=\text { Number }
a
=
Number
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3)
f
(
x
)
=
−
(
x
−
8
)
2
+
5
-
(
\begin{array}{l}\text { 3) } f(x)=-(x-8)^{2}+5 \\ \text { - }(\end{array}
3)
f
(
x
)
=
−
(
x
−
8
)
2
+
5
-
(
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The functions
s
(
x
)
s(x)
s
(
x
)
and
t
(
x
)
t(x)
t
(
x
)
are differentiable. The function
u
(
x
)
u(x)
u
(
x
)
is defined as:
u
(
x
)
=
s
(
x
)
t
(
x
)
u(x)= \frac{s(x)}{t(x)}
u
(
x
)
=
t
(
x
)
s
(
x
)
If
s
(
6
)
=
7
s(6)= 7
s
(
6
)
=
7
,
s
′
(
6
)
=
5
s'(6)= 5
s
′
(
6
)
=
5
,
t
(
6
)
=
9
t(6)= 9
t
(
6
)
=
9
, and
t
′
(
6
)
=
2
t'(6)= 2
t
′
(
6
)
=
2
, what is
u
′
(
6
)
u'(6)
u
′
(
6
)
? Simplify any fractions.
u
′
(
6
)
=
u'(6)=
u
′
(
6
)
=
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Find
d
2
t
d
s
2
\frac{d^2 t}{ds^2}
d
s
2
d
2
t
if
t
=
2
s
(
1
−
4
s
)
2
t = 2s(1 - 4s)^2
t
=
2
s
(
1
−
4
s
)
2
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Find the inverse of
\newline
f
(
x
)
=
x
2
−
5
,
x
≤
0
f(x)=x^{2}-5,\,x \leq 0
f
(
x
)
=
x
2
−
5
,
x
≤
0
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Find
k
′
(
x
)
k'(x)
k
′
(
x
)
if
k
(
x
)
=
e
x
(
−
x
5
+
1
3
x
4
5
)
k(x)=e^{x}(-x^{5}+\frac{1}{3}x^{\frac{4}{5}})
k
(
x
)
=
e
x
(
−
x
5
+
3
1
x
5
4
)
.
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When the function
\newline
p
(
x
)
p(x)
p
(
x
)
is divided by
\newline
x
−
1
x-1
x
−
1
the quotient is
\newline
x
2
+
7
+
5
x
−
1
x^{2}+7+\frac{5}{x-1}
x
2
+
7
+
x
−
1
5
. State
\newline
p
(
x
)
p(x)
p
(
x
)
in standard form.
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The parabola
\newline
y
=
x
2
y=x^{2}
y
=
x
2
is reflected across the
\newline
x
x
x
-axis and then scaled vertically by a factor of
5
5
5
.
\newline
What is the equation of the new parabola?
\newline
y
=
□
y=\square
y
=
□
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Given the function
f
(
x
)
=
x
2
+
3
f(x)=x^{2}+3
f
(
x
)
=
x
2
+
3
, find the value of
f
(
−
2
3
)
f\left(-\frac{2}{3}\right)
f
(
−
3
2
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
7
x
2
−
1
f(x)=-7 x^{2}-1
f
(
x
)
=
−
7
x
2
−
1
, find the value of
f
(
3
7
)
f\left(\frac{3}{7}\right)
f
(
7
3
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
1
4
x
2
+
4
f(x)=-\frac{1}{4} x^{2}+4
f
(
x
)
=
−
4
1
x
2
+
4
, find the value of
f
(
−
1
)
f(-1)
f
(
−
1
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
7
6
x
2
−
4
f(x)=\frac{7}{6} x^{2}-4
f
(
x
)
=
6
7
x
2
−
4
, find the value of
f
(
−
2
)
f(-2)
f
(
−
2
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
x
2
−
1
4
f(x)=-x^{2}-\frac{1}{4}
f
(
x
)
=
−
x
2
−
4
1
, find the value of
f
(
1
)
f(1)
f
(
1
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
1
4
x
2
−
4
f(x)=-\frac{1}{4} x^{2}-4
f
(
x
)
=
−
4
1
x
2
−
4
, find the value of
f
(
2
)
f(2)
f
(
2
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
1
+
7
6
x
2
f(x)=-1+\frac{7}{6} x^{2}
f
(
x
)
=
−
1
+
6
7
x
2
, find the value of
f
(
1
7
)
f\left(\frac{1}{7}\right)
f
(
7
1
)
in simplest form.
\newline
Answer:
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Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
x
−
16
f(x)=x-16
f
(
x
)
=
x
−
16
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
1
4
x
+
15
f(x)=-\frac{1}{4} x+15
f
(
x
)
=
−
4
1
x
+
15
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
4
5
x
−
16
f(x)=\frac{4}{5} x-16
f
(
x
)
=
5
4
x
−
16
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
2
x
+
2
f(x)=2 x+2
f
(
x
)
=
2
x
+
2
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
x
−
4
f(x)=-2 x-4
f
(
x
)
=
−
2
x
−
4
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
3
x
+
12
f(x)=-\frac{2}{3} x+12
f
(
x
)
=
−
3
2
x
+
12
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
5
x
+
20
f(x)=-5 x+20
f
(
x
)
=
−
5
x
+
20
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
3
2
x
+
9
f(x)=-\frac{3}{2} x+9
f
(
x
)
=
−
2
3
x
+
9
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
x
−
14
f(x)=-x-14
f
(
x
)
=
−
x
−
14
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
5
x
−
10
f(x)=5 x-10
f
(
x
)
=
5
x
−
10
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
2
3
x
−
12
f(x)=\frac{2}{3} x-12
f
(
x
)
=
3
2
x
−
12
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
5
2
x
−
15
f(x)=\frac{5}{2} x-15
f
(
x
)
=
2
5
x
−
15
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
3
2
x
+
6
f(x)=\frac{3}{2} x+6
f
(
x
)
=
2
3
x
+
6
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
x
+
4
f(x)=-2 x+4
f
(
x
)
=
−
2
x
+
4
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
x
−
8
f(x)=-x-8
f
(
x
)
=
−
x
−
8
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
5
x
+
8
f(x)=-\frac{2}{5} x+8
f
(
x
)
=
−
5
2
x
+
8
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
3
x
+
15
f(x)=-3 x+15
f
(
x
)
=
−
3
x
+
15
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
5
2
x
−
10
f(x)=-\frac{5}{2} x-10
f
(
x
)
=
−
2
5
x
−
10
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
x
+
6
f(x)=-2 x+6
f
(
x
)
=
−
2
x
+
6
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
5
x
−
16
f(x)=-\frac{2}{5} x-16
f
(
x
)
=
−
5
2
x
−
16
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
2
x
−
8
f(x)=2 x-8
f
(
x
)
=
2
x
−
8
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
1
2
x
−
7
f(x)=-\frac{1}{2} x-7
f
(
x
)
=
−
2
1
x
−
7
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
5
x
−
15
f(x)=5 x-15
f
(
x
)
=
5
x
−
15
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
3
2
x
+
9
f(x)=\frac{3}{2} x+9
f
(
x
)
=
2
3
x
+
9
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
x
−
16
f(x)=-x-16
f
(
x
)
=
−
x
−
16
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function of the function
f
(
x
)
=
1
5
x
+
9
f(x)=\frac{1}{5} x+9
f
(
x
)
=
5
1
x
+
9
.
\newline
f
−
1
(
x
)
=
5
x
−
9
f^{-1}(x)=5 x-9
f
−
1
(
x
)
=
5
x
−
9
\newline
f
−
1
(
x
)
=
5
x
−
45
f^{-1}(x)=5 x-45
f
−
1
(
x
)
=
5
x
−
45
\newline
f
−
1
(
x
)
=
1
5
x
−
9
f^{-1}(x)=\frac{1}{5} x-9
f
−
1
(
x
)
=
5
1
x
−
9
\newline
f
−
1
(
x
)
=
1
5
x
−
45
f^{-1}(x)=\frac{1}{5} x-45
f
−
1
(
x
)
=
5
1
x
−
45
Get tutor help
Given
f
(
x
)
=
−
x
2
−
9
x
f(x)=-x^{2}-9 x
f
(
x
)
=
−
x
2
−
9
x
, find
f
(
4
)
f(4)
f
(
4
)
\newline
Answer:
Get tutor help
Given
x
>
0
x>0
x
>
0
, the expression
x
4
3
\sqrt[3]{x^{4}}
3
x
4
is equivalent to
\newline
x
2
x
2
3
x^{2} \sqrt[3]{x^{2}}
x
2
3
x
2
\newline
x
x
x
\newline
x
x
3
x \sqrt[3]{x}
x
3
x
\newline
x
2
x^{2}
x
2
Get tutor help
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