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Math Problems
Algebra 2
Reflections of functions
An architect is drawing a scale model of a bridge support and begins by graphing
f
(
x
)
=
x
2
f(x)=x^{2}
f
(
x
)
=
x
2
. Through a series of transformations, she finds that
\newline
g
(
x
)
=
−
2
(
x
−
4
)
2
+
3
g(x)=-2(x-4)^{2}+3
g
(
x
)
=
−
2
(
x
−
4
)
2
+
3
best models the bridge support. Which value results in a vertical stretch and reflection in the x-axis of the graph of
\newline
f
(
x
)
=
x
2
f(x)=x^{2}
f
(
x
)
=
x
2
to give the bridge the correct orientation and shape in the drawing?
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Find the surface area of the part of the hyperbolic paraboloid
z
=
x
2
−
y
2
z=x^{2}-y^{2}
z
=
x
2
−
y
2
that lies within the cylinder
x
2
+
y
2
=
1
x^{2}+y^{2}=1
x
2
+
y
2
=
1
in the first octant
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The function is given by:
f
(
x
)
=
(
x
−
2
)
2
−
9
f(x)=(x-2)^{2}-9
f
(
x
)
=
(
x
−
2
)
2
−
9
\newline
At what values of
x
x
x
does the graph of the function intersect the
x
x
x
-axis?
\newline
Choose
1
1
1
answer:
\newline
(A)
x
=
−
5
,
x
=
−
1
x=-5, x=-1
x
=
−
5
,
x
=
−
1
\newline
(B)
x
=
5
,
x
=
1
x=5, x=1
x
=
5
,
x
=
1
\newline
(C)
x
=
−
5
,
x
=
1
x=-5, x=1
x
=
−
5
,
x
=
1
\newline
(D)
x
=
5
,
x
=
−
1
x=5, x=-1
x
=
5
,
x
=
−
1
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What is the image of a circle with equation
(
x
+
2
)
2
+
(
y
−
2
)
2
=
1
(x + 2)^2 + (y - 2)^2=1
(
x
+
2
)
2
+
(
y
−
2
)
2
=
1
when it is rotated through
π
6
\frac{\pi}{6}
6
π
about the origin?
\newline
(a)
(
x
+
(
3
+
1
)
)
2
+
(
y
−
(
3
−
1
)
)
2
=
1
(x + (\sqrt{3} + 1))^2 + (y - ( \sqrt{3} - 1))^2 = 1
(
x
+
(
3
+
1
)
)
2
+
(
y
−
(
3
−
1
)
)
2
=
1
\newline
(b)
(
x
−
(
3
+
1
)
)
2
+
(
y
+
(
3
−
1
)
)
2
=
1
(x - ( \sqrt{3} + 1))^2 + (y + (\sqrt{3} - 1))^2 = 1
(
x
−
(
3
+
1
)
)
2
+
(
y
+
(
3
−
1
)
)
2
=
1
\newline
(c)
(
x
+
(
3
−
1
)
)
2
+
(
y
−
(
3
+
1
)
)
2
=
1
(x + (\sqrt{3} - 1))^2 + (y - ( \sqrt{3} + 1))^2 = 1
(
x
+
(
3
−
1
)
)
2
+
(
y
−
(
3
+
1
)
)
2
=
1
\newline
(d)
(
x
−
(
3
−
1
)
)
2
+
(
y
+
(
3
+
1
)
)
2
=
1
(x - ( \sqrt{3} - 1))^2 + (y + (\sqrt{3} + 1))^2 = 1
(
x
−
(
3
−
1
)
)
2
+
(
y
+
(
3
+
1
)
)
2
=
1
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What is the maximum vertical distance between the line
y
=
x
+
2
y=x+2
y
=
x
+
2
and the parabola
y
=
x
2
y=x^{2}
y
=
x
2
for
−
1
≤
x
≤
2
-1 \leq x \leq 2
−
1
≤
x
≤
2
?
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The
z
z
z
-transform of a sequence
x
(
n
)
x(n)
x
(
n
)
is
\newline
X
(
z
)
=
1
−
4
z
−
1
+
2
z
−
2
1
−
3
z
−
1
+
0.5
z
−
2
X(z)=\frac{1-4 z^{-1}+2 z^{-2}}{1-3 z^{-1}+0.5 z^{-2}}
X
(
z
)
=
1
−
3
z
−
1
+
0.5
z
−
2
1
−
4
z
−
1
+
2
z
−
2
\newline
If the region of convergence includes the unit circle, find the DTFT of
x
(
n
)
x(n)
x
(
n
)
at
ω
=
π
/
2
\omega=\pi / 2
ω
=
π
/2
.
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Progress:
\newline
Match the polynomial on the left with the simplified polynomial on the right.
\newline
Quotation ID:
1
1
1
\newline
{
(
4
x
2
y
−
5
x
y
2
+
2
x
2
y
2
+
2
x
y
)
+
(
2
x
2
y
+
4
x
y
+
2
x
y
2
−
2
x
2
y
2
)
−
6
x
2
y
−
3
x
2
y
2
+
2
x
y
2
−
2
y
−
6
x
2
y
2
+
4
x
2
y
−
2
x
y
2
+
2
x
−
2
y
18
x
3
y
2
+
9
x
2
y
3
−
12
x
3
y
3
+
6
x
y
2
3
x
y
{
6
x
2
y
2
+
3
x
2
y
−
7
x
y
−
2
x
y
2
+
2
y
6
x
2
y
−
3
x
y
2
+
6
x
y
\left\{\begin{array}{l} \left(4x^{2}y-5xy^{2}+2x^{2}y^{2}+2xy\right)+\left(2x^{2}y+4xy+2xy^{2}-2x^{2}y^{2}\right) - 6x^{2}y-3x^{2}y^{2}+2xy^{2}-2y \ -6x^{2}y^{2}+4x^{2}y-2xy^{2}+2x-2y \ \frac{18x^{3}y^{2}+9x^{2}y^{3}-12x^{3}y^{3}+6xy^{2}}{3xy} \ \left\{\begin{array}{l} 6x^{2}y^{2}+3x^{2}y-7xy-2xy^{2}+2y \ 6x^{2}y-3xy^{2}+6xy \end{array}\right. \end{array}\right.
{
(
4
x
2
y
−
5
x
y
2
+
2
x
2
y
2
+
2
x
y
)
+
(
2
x
2
y
+
4
x
y
+
2
x
y
2
−
2
x
2
y
2
)
−
6
x
2
y
−
3
x
2
y
2
+
2
x
y
2
−
2
y
−
6
x
2
y
2
+
4
x
2
y
−
2
x
y
2
+
2
x
−
2
y
3
x
y
18
x
3
y
2
+
9
x
2
y
3
−
12
x
3
y
3
+
6
x
y
2
{
6
x
2
y
2
+
3
x
2
y
−
7
x
y
−
2
x
y
2
+
2
y
6
x
2
y
−
3
x
y
2
+
6
x
y
\newline
Clear
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Find the
y
y
y
-coordinate of the
y
y
y
-intercept of the polynomial function defined below.
\newline
f
(
x
)
=
−
(
x
+
1
)
(
x
−
4
)
2
f(x)=-(x+1)(x-4)^{2}
f
(
x
)
=
−
(
x
+
1
)
(
x
−
4
)
2
\newline
Answer:
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Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the reflection across the x-axis of
f
(
x
)
=
9
∣
x
+
2
∣
−
4
f(x)=9|x+2|-4
f
(
x
)
=
9∣
x
+
2∣
−
4
.
\newline
Choices:
\newline
(
A
)
g
(
x
)
=
9
∣
x
+
2
∣
−
4
]
\text{}(A)g(x) = 9|x+2| - 4\text{]}
(
A
)
g
(
x
)
=
9∣
x
+
2∣
−
4
]
\newline
(
B
)
g
(
x
)
=
9
∣
x
−
2
∣
−
4
]
\text{}(B)g(x) = 9|x-2| - 4\text{]}
(
B
)
g
(
x
)
=
9∣
x
−
2∣
−
4
]
\newline
(
C
)
g
(
x
)
=
−
9
∣
x
−
2
∣
+
4
]
\text{}(C)g(x)=-9|x-2| +4\text{]}
(
C
)
g
(
x
)
=
−
9∣
x
−
2∣
+
4
]
\newline
(D)
g
(
x
)
=
−
9
∣
x
+
2
∣
+
4
]
\text{(D)}g(x) = -9|x + 2| + 4\text{]}
(D)
g
(
x
)
=
−
9∣
x
+
2∣
+
4
]
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