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The graph of
y=|x| is reflected across the
x-axis and then scaled vertically by a factor of 7 .
What is the equation of the new graph?
Choose 1 answer:
(A)
y=7|x|
(B)
y=-7|x|
(C)
y=(1)/(7)|x|
(D)
y=-(1)/(7)|x|

The graph of $y=|x|$ is reflected across the $x$ -axis and then scaled vertically by a factor of $7$ . $\newline$ What is the equation of the new graph? $\newline$ Choose $1$ answer: $\newline$ (A) $y=7|x|$ $\newline$ (B) $y=-7|x|$ $\newline$ (C) $y=\frac{1}{7}|x|$ $\newline$ (D) $y=-\frac{1}{7}|x|$

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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

7

.

\newline

What is the equation of the new graph?

\newline

Choose

1

answer:

\newline

(A)

y=7|x|

\newline

(B)

y=-7|x|

\newline

(C)

y=\frac{1}{7}|x|

\newline

(D)

y=-\frac{1}{7}|x|

Reflect across x-axis: Reflect $y=|x|$ across the x-axis. $\newline$ Reflection across the x-axis changes $y$ to $-y$ . $\newline$ Transformed function: $y' = -|x|$
Scale vertically by $7$ : Scale the reflected graph vertically by a factor of $7$ . Scaling a function vertically by a factor of $k$ multiplies the output by $k$ . Transformed function: $y'' = 7 \times y$
Substitute and simplify: Substitute $y'$ with $-|x|$ into the scaled function. $y'' = 7 \cdot (-|x|)$ Simplify the expression. $y'' = -7 \cdot |x|$

More problems from Transformations of functions

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\text{translates it } 6 \text{ units down}

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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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\text{[B]translation 10 units right}

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Question

g(x)

g(x)

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\newline

\newline

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

\newline

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

\newline

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

\newline

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

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Question

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

\newline

1

\newline

\frac{1}{2}

\newline

1

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

\newline

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Posted 10 months ago

Question

Tess tried to solve the differential equation

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

. This is her work:

\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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\newline

3

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\newline

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\newline

1

\newline

(A) Aditya's work is correct.

\newline

1

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\newline

2

is incorrect. Aditya didn't integrate

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\newline

3

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

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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

|-75+250|

Posted 9 months ago

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

y

-intercept of the graph of

f

appears as a constant or coefficient?

\newline

1

\newline

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

\newline

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

\newline

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

\newline

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

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Question

Determine whether the function

f(x)

is continuous at

x=-3

\newline

f(x)=\left\{\begin{array}{ll} 18-x^{2}, & x \leq-3 \\ 15+3 x, & x>-3 \end{array}\right.

\newline

f(x)

is discontinuous at

x=-3

\newline

f(x)

is continuous at

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