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Math Problems
Algebra 2
Transformations of quadratic functions
A curve is defined by the parametric equations
x
(
t
)
=
−
8
e
−
9
t
x(t)=-8 e^{-9 t}
x
(
t
)
=
−
8
e
−
9
t
and
y
(
t
)
=
sin
(
−
5
t
)
y(t)=\sin (-5 t)
y
(
t
)
=
sin
(
−
5
t
)
. Find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Answer:
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A particle moves along a straight line. Its speed is inversely proportional to the square of the distance,
S
S
S
, it has traveled.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
S
d
t
=
k
t
2
\frac{d S}{d t}=\frac{k}{t^{2}}
d
t
d
S
=
t
2
k
\newline
(B)
S
(
t
)
=
k
t
2
S(t)=\frac{k}{t^{2}}
S
(
t
)
=
t
2
k
\newline
(c)
d
S
d
t
=
k
S
2
\frac{d S}{d t}=\frac{k}{S^{2}}
d
t
d
S
=
S
2
k
\newline
(D)
S
(
t
)
=
k
S
2
S(t)=\frac{k}{S^{2}}
S
(
t
)
=
S
2
k
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Consider the curve given by the equation
y
3
−
8
y
2
+
16
y
−
x
2
+
2
x
=
0
.
y^{3}-8 y^{2}+16 y-x^{2}+2 x=0 \text {. }
y
3
−
8
y
2
+
16
y
−
x
2
+
2
x
=
0
.
It can be shown that
d
y
d
x
=
2
(
x
−
1
)
(
3
y
−
4
)
(
y
−
4
)
.
\frac{d y}{d x}=\frac{2(x-1)}{(3 y-4)(y-4)} \text {. }
d
x
d
y
=
(
3
y
−
4
)
(
y
−
4
)
2
(
x
−
1
)
.
\newline
Find the
y
y
y
-coordinate of the point where the line tangent to the curve is the
y
y
y
-axis.
\newline
y
=
y=
y
=
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Let
h
(
x
)
=
x
10
sin
2
(
x
+
1
)
h(x)=\frac{x}{10 \sin ^{2}(x+1)}
h
(
x
)
=
10
s
i
n
2
(
x
+
1
)
x
.
\newline
Select the correct description of the one-sided limits of
h
h
h
at
x
=
−
1
x=-1
x
=
−
1
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
Get tutor help
Let
g
(
x
)
=
2
x
+
3
g(x)=\frac{2}{x+3}
g
(
x
)
=
x
+
3
2
.
\newline
Select the correct description of the one-sided limits of
g
g
g
at
x
=
−
3
x=-3
x
=
−
3
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
−
3
+
g
(
x
)
=
+
∞
and
lim
x
→
−
3
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=+\infty \end{array}
lim
x
→
−
3
+
g
(
x
)
=
+
∞
and
lim
x
→
−
3
−
g
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
−
3
+
g
(
x
)
=
+
∞
and
lim
x
→
−
3
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=-\infty \end{array}
lim
x
→
−
3
+
g
(
x
)
=
+
∞
and
lim
x
→
−
3
−
g
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
−
3
+
g
(
x
)
=
−
∞
and
lim
x
→
−
3
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=+\infty \end{array}
lim
x
→
−
3
+
g
(
x
)
=
−
∞
and
lim
x
→
−
3
−
g
(
x
)
=
+
∞
\newline
(D
\newline
lim
x
→
−
3
+
g
(
x
)
=
−
∞
and
lim
x
→
−
3
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=-\infty \end{array}
lim
x
→
−
3
+
g
(
x
)
=
−
∞
and
lim
x
→
−
3
−
g
(
x
)
=
−
∞
Get tutor help
Let
h
(
x
)
=
−
2
x
(
x
+
1
)
2
h(x)=-\frac{2 x}{(x+1)^{2}}
h
(
x
)
=
−
(
x
+
1
)
2
2
x
.
\newline
Select the correct description of the one-sided limits of
h
h
h
at
x
=
−
1
x=-1
x
=
−
1
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
Get tutor help
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the translation
8
8
8
units up of
f
(
x
)
=
x
2
f(x) = x^2
f
(
x
)
=
x
2
.
\newline
Write your answer in the form
a
(
x
–
h
)
2
+
k
a(x – h)^2 + k
a
(
x
–
h
)
2
+
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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