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Math Problems
Algebra 1
Transformations of quadratic functions
Given the function
y
=
(
3
x
−
3
+
9
)
(
−
6
−
2
x
−
7
x
3
)
y=\left(3 x^{-3}+9\right)\left(-6-2 x-7 x^{3}\right)
y
=
(
3
x
−
3
+
9
)
(
−
6
−
2
x
−
7
x
3
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
y
=
(
−
10
+
7
x
−
2
+
x
)
(
4
x
−
2
+
3
)
y=\left(-10+7 x^{-2}+x\right)\left(4 x^{-2}+3\right)
y
=
(
−
10
+
7
x
−
2
+
x
)
(
4
x
−
2
+
3
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
f
(
x
)
=
(
−
5
x
3
−
8
x
2
−
1
)
(
9
x
2
+
1
)
f(x)=\left(-5 x^{3}-8 x^{2}-1\right)\left(9 x^{2}+1\right)
f
(
x
)
=
(
−
5
x
3
−
8
x
2
−
1
)
(
9
x
2
+
1
)
, find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
in any form.
\newline
Answer:
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
d
y
d
x
=
−
3
y
\frac{d y}{d x}=-3 y
d
x
d
y
=
−
3
y
, and
y
=
2
y=2
y
=
2
when
x
=
0
x=0
x
=
0
.
\newline
Solve the equation.
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
2
+
e
−
3
x
y=2+e^{-3 x}
y
=
2
+
e
−
3
x
\newline
(B)
y
=
1
+
e
−
3
x
y=1+e^{-3 x}
y
=
1
+
e
−
3
x
\newline
(C)
y
=
2
e
−
3
x
y=2 e^{-3 x}
y
=
2
e
−
3
x
\newline
(D)
y
=
e
−
3
x
y=e^{-3 x}
y
=
e
−
3
x
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d
y
d
x
=
−
4
y
\frac{d y}{d x}=-4 y
d
x
d
y
=
−
4
y
, and
y
=
3
y=3
y
=
3
when
x
=
2
x=2
x
=
2
.
\newline
Solve the equation.
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
3
e
8
−
4
x
y=3 e^{8-4 x}
y
=
3
e
8
−
4
x
\newline
(B)
y
=
3
e
−
4
x
y=3 e^{-4 x}
y
=
3
e
−
4
x
\newline
(C)
y
=
3
e
4
−
4
x
y=3 e^{4-4 x}
y
=
3
e
4
−
4
x
\newline
(D)
y
=
6
e
−
4
x
y=6 e^{-4 x}
y
=
6
e
−
4
x
Get tutor help
d
y
d
x
=
−
4
y
\frac{d y}{d x}=-4 y
d
x
d
y
=
−
4
y
, and
y
=
3
y=3
y
=
3
when
x
=
2
x=2
x
=
2
.
\newline
Solve the equation.
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
3
e
−
4
x
y=3 e^{-4 x}
y
=
3
e
−
4
x
\newline
(B)
y
=
6
e
−
4
x
y=6 e^{-4 x}
y
=
6
e
−
4
x
\newline
(C)
y
=
3
e
8
−
4
x
y=3 e^{8-4 x}
y
=
3
e
8
−
4
x
\newline
(D)
y
=
3
e
4
−
4
x
y=3 e^{4-4 x}
y
=
3
e
4
−
4
x
Get tutor help
Consider the curve given by the equation
y
2
−
6
y
−
9
x
2
−
144
x
=
576
.
y^{2}-6 y-9 x^{2}-144 x=576 \text {. }
y
2
−
6
y
−
9
x
2
−
144
x
=
576
.
It can be shown that
d
y
d
x
=
9
(
x
+
8
)
y
−
3
.
\frac{d y}{d x}=\frac{9(x+8)}{y-3} \text {. }
d
x
d
y
=
y
−
3
9
(
x
+
8
)
.
\newline
Find the
x
x
x
-coordinate of the point where the line tangent to the curve is the
x
x
x
-axis.
\newline
x
=
x=
x
=
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Consider the curve given by the equation
3
x
2
+
6
x
−
y
2
+
8
y
=
16.
3 x^{2}+6 x-y^{2}+8 y=16.
3
x
2
+
6
x
−
y
2
+
8
y
=
16.
It can be shown that
d
y
d
x
=
3
(
x
+
1
)
y
−
4
\frac{d y}{d x}=\frac{3(x+1)}{y-4}
d
x
d
y
=
y
−
4
3
(
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
=
Get tutor help
We are given that
\newline
d
y
d
x
=
1
−
y
2
.
\frac{d y}{d x}=\sqrt{1-y^{2}} \text {. }
d
x
d
y
=
1
−
y
2
.
\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
=
\frac{d^{2} y}{d x^{2}}=
d
x
2
d
2
y
=
Get tutor help
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the translation
5
5
5
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
)
=
____
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