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Problem The graph of $y=|x|$ is reflected across the $x$ -axis and then scaled vertically by a factor of $\frac{3}{8}$ .

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Reflections of functions

Full solution

Q. Problem The graph of

y=|x|

is reflected across the

x

-axis and then scaled vertically by a factor of

\frac{3}{8}

.

Reflect across x-axis: Reflect $y=|x|$ across the x-axis. $\newline$ To reflect a graph across the x-axis, we change the sign of the $y$ -values. The reflection of $y=|x|$ across the x-axis is $y=-|x|$ .
Scale vertically by $\frac{3}{8}$ : Scale the reflected graph vertically by a factor of $\frac{3}{8}$ . To scale a graph vertically, we multiply the y-values by the scaling factor. The vertical scaling of $y=-|x|$ by a factor of $\frac{3}{8}$ is $y=\left(\frac{3}{8}\right)(-|x|)$ .
Simplify equation: Simplify the equation. $\newline$ The simplified equation of the transformed graph is $y=-\frac{3}{8}|x|$ .
Check final equation: Check the final equation. $\newline$ The original function $y=|x|$ is always non-negative. After reflecting across the $x$ -axis, it becomes non-positive, which is represented by $y=-|x|$ . Then, scaling by $\frac{3}{8}$ should multiply each $y$ -value by $\frac{3}{8}$ , resulting in $y=-\frac{3}{8}|x|$ . This matches our transformation steps, so there is no math error.

More problems from Reflections of functions

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What does the transformation

f(x) \mapsto f(x) - 6

do to the graph of

f(x)

\newline

\newline

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

\newline

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

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

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

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Question

g(x)

g(x)

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f(x) = -4x + 1

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\newline

\text{[g(x) = 4x - 1]}

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\text{[g(x) = -4x + 1]}

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\text{[g(x) = 4x + 1]}

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\text{[g(x) = -4x - 1]}

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Question

What kind of transformation converts the graph of

f(x) = -8(x - 4)^2 + 6

into the graph of

g(x) = -8(x + 6)^2 + 6

\newline

\newline

\text{[A]translation 10 units left}

\newline

\text{[B]translation 10 units right}

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\text{[C]translation 10 units up}

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\text{[D]translation 10 units down}

Posted 2 months ago

Question

g(x)

g(x)

is the reflection across the x-axis of

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

\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

Posted 2 months ago

Question

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

\newline

1

\newline

\frac{1}{2}

\newline

1

\newline

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

\newline

(D) The limit doesn't exist.

Posted 3 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}

\newline

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?

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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.

Posted 1 month ago

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

\quad 30 y=3 x^{2}+C

\newline

3

\quad y=\frac{x^{2}}{10}+C

\newline

Is Aditya's work correct? If not, what is his mistake?

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1

\newline

(A) Aditya's work is correct.

\newline

1

is incorrect. The separation of variables wasn't done correctly.

\newline

2

is incorrect. Aditya didn't integrate

30

\newline

3

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

\frac{30}{3 x^{2}+C}

Posted 2 months ago

Question

Jonael dropped a sandbag from a hot air balloon at a target that is

250

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75

meters below the balloon.

\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 2 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

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-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

Posted 2 months ago

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

x=-3

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