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What kind of transformation converts the graph of $f(x) = 6(x + 1)^2 + 10$ into the graph of $g(x) = 6(x + 9)^2 + 10$ ? $\newline$ Choices: $\newline$ (A) translation $8$ units down $\newline$ (B) translation $8$ units right $\newline$ (C) translation $8$ units left $\newline$ (D) translation $8$ units up

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Describe function transformations

Full solution

Q. What kind of transformation converts the graph of

f(x) = 6(x + 1)^2 + 10

into the graph of

g(x) = 6(x + 9)^2 + 10

?

\newline

Choices:

\newline

(A) translation

8

units down

\newline

(B) translation

8

units right

\newline

(C) translation

8

units left

\newline

(D) translation

8

units up

Identify Vertex: Identify the vertex of the function $f(x)$ . Compare $f(x) = 6(x + 1)^2 + 10$ with the vertex form of a parabola. Vertex of $f(x)$ : $(-1, 10)$
Compare Functions: Identify the vertex of the function $g(x)$ . Compare $g(x) = 6(x + 9)^2 + 10$ with the vertex form of a parabola. Vertex of $g(x)$ : $(-9, 10)$
Type of Transformation: Determine the type of transformation. $\newline$ Since the $y$ -values of the vertices are the same and only the $x$ -values have changed, the transformation is horizontal.
Direction of Transformation: Determine the direction of the transformation. $\newline$ The $x$ -coordinate of the vertex of $f(x)$ is $-1$ and the $x$ -coordinate of the vertex of $g(x)$ is $-9$ . $\newline$ Since $-9$ is to the left of $-1$ on the number line, the graph has shifted to the left.
Calculate Magnitude: Calculate the magnitude of the transformation. $\newline$ The difference in the x-coordinates of the vertices is |-1 - (-9)| = |8| = 8＂).\(\newlineThe graph of \$f(x)\) shifts \(8\) units to the left to become \(g(x)\).

More problems from Describe function transformations

Question

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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g(x)

g(x)

is the reflection across the

y

f(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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\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}

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\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

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\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}

\newline

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?

\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.

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?

\newline

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

meters below the balloon. At this moment, the sandbag is

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|

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|-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

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

x=-3

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