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The graph of
y=|x| is reflected across the
x-axis and then scaled vertically by a factor of
(3)/(8).
What is the equation of the new graph?
Choose 1 answer:
(A)
y=(8)/(3)|x|
(B)
y=-(8)/(3)|x|
(C)
y=(3)/(8)|x|
(D)
y=-(3)/(8)|x|

The graph of $y=|x|$ is reflected across the $x$ -axis and then scaled vertically by a factor of $\left(\frac{3}{8}\right)$ . $\newline$ What is the equation of the new graph? $\newline$ Choose $1$ answer: $\newline$ (A) $y=\left(\frac{8}{3}\right)|x|$ $\newline$ (B) $y=-\left(\frac{8}{3}\right)|x|$ $\newline$ (C) $y=\left(\frac{3}{8}\right)|x|$ $\newline$ (D) $y=-\left(\frac{3}{8}\right)|x|$

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Transformations of functions

Full solution

Q. The graph of

y=|x|

is reflected across the

x

-axis and then scaled vertically by a factor of

\left(\frac{3}{8}\right)

.

\newline

What is the equation of the new graph?

\newline

Choose

1

answer:

\newline

(A)

y=\left(\frac{8}{3}\right)|x|

\newline

(B)

y=-\left(\frac{8}{3}\right)|x|

\newline

(C)

y=\left(\frac{3}{8}\right)|x|

\newline

(D)

y=-\left(\frac{3}{8}\right)|x|

Reflect across x-axis: Reflect $y=|x|$ across the x-axis. $\newline$ To reflect a graph across the x-axis, we multiply the function by $-1$ . $\newline$ Reflected function: $y = -|x|$
Scale vertically by $\frac{3}{8}$ : Scale the reflected function vertically by a factor of $\frac{3}{8}$ . To scale a function vertically, we multiply the function by the scaling factor. Scaled function: $y = \left(\frac{3}{8}\right)(-|x|)$
Simplify function: Simplify the scaled function. $\newline$ Simplified function: $y = -\left(\frac{3}{8}\right)|x|$

More problems from Transformations of functions

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Question

What kind of transformation converts the graph of

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\newline

\newline

\text{[A]translation 10 units left}

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Question

g(x)

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\newline

\newline

g(x) = 6(x - 7)^2 - 2

\newline

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\newline

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\newline

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Question

\lim _{x \rightarrow \frac{\pi}{4}} \cos (x)=?

\newline

1

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\frac{1}{2}

\newline

1

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\frac{\sqrt{2}}{2}

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Question

Tess tried to solve the differential equation

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\newline

\frac{d y}{d x}=2 x y-6 x

\newline

1

\quad \frac{d y}{d x}=(y-3) 2 x

\newline

2

\quad \int \frac{1}{y-3} d y=\int 2 x d x

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3

\quad \ln |y-3|=x^{2}+C_{1}

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4

\quad e^{\ln |y-3|}=e^{x^{2}}+C_{1}

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5

\quad|y-3|=e^{x^{2}}+C_{1}

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6

\quad y-3= \pm e^{x^{2}}+C_{2}

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7

\quad y= \pm e^{x^{2}}+C

\newline

Is Tess's work correct? If not, what is her mistake?

\newline

1

\newline

(A) Tess's work is correct.

\newline

1

is incorrect. We're not allowed to factor an expression when solving differential equations.

\newline

4

is incorrect. The right-hand side of the equation should be

e^{x^{2}+C_{1}}

\newline

7

is incorrect. Tess didn't account for adding

3

to the right side.

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Question

Aditya tried to solve the differential equation

\frac{d y}{d x}=6 x-30

. This is his work:

\newline

\frac{d y}{d x}=6 x-30

\newline

1

\quad \int 30 d y=\int 6 x d x

\newline

2

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3

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\newline

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1

\newline

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\newline

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2

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\newline

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Question

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\newline

2

of the following expressions represents the distance between the sandbag and the target?

\newline

2

\newline

|-250+75|

\newline

|250+75|

\newline

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Question

f(x) = -(x+1)(x+7)

\newline

The function represents a parabola in the

xy

-plane. Which of the following is an equivalent form of

f

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f

appears as a constant or coefficient?

\newline

1

\newline

f(x) = -(x+4)^{2}+9

\newline

f(x) = -(x+4)^{2}-9

\newline

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\newline

f(x) = -x^{2}-8x-7

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f(x)

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\newline

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\newline

f(x)

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\newline

f(x)

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