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What kind of transformation converts the graph of $f(x) = -3(x + 5)^2 + 7$ into the graph of $g(x) = -3x^2 + 7$ ? $\newline$ Choices: $\newline$ (A) translation $5$ units down $\newline$ (B) translation $5$ units up $\newline$ (C) translation $5$ units left $\newline$ (D) translation $5$ units right

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Describe function transformations

Full solution

Q. What kind of transformation converts the graph of

f(x) = -3(x + 5)^2 + 7

into the graph of

g(x) = -3x^2 + 7

?

\newline

Choices:

\newline

(A) translation

5

units down

\newline

(B) translation

5

units up

\newline

(C) translation

5

units left

\newline

(D) translation

5

units right

Find Vertex of $f(x)$ : Find the vertex of $f(x) = -3(x + 5)^2 + 7$ . $\newline$ Vertex form is $-3(x - h)^2 + k$ , where $(h, k)$ is the vertex. $\newline$ So, vertex of $f(x)$ is $(-5, 7)$ .
Find Vertex of $g(x)$ : Find the vertex of $g(x) = -3x^2 + 7$ . This is already in vertex form $-3(x - 0)^2 + 7$ . So, vertex of $g(x)$ is $(0, 7)$ .
Compare Vertices: Compare the vertices of $f(x)$ and $g(x)$ .\ Vertex of $f(x) = (-5, 7)$ , Vertex of $g(x) = (0, 7)$ .\ The $y$ -coordinates are the same, so it's a horizontal shift.
Determine Shift Direction: Determine the direction of the shift. $\newline$ The $x$ -coordinate of the vertex of $f(x)$ is $-5$ and for $g(x)$ is $0$ . $\newline$ Moving from $-5$ to $0$ is a shift to the right.
Calculate Shift Distance: Calculate the distance of the shift. $\newline$ |-5 - 0| = |-5| = 5|.\(\newlineThe graph of \$f(x)\) shifts \(5\) units to the right.

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What does the transformation

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

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What kind of transformation converts the graph of

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\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

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

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

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Question

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

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1

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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

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2

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

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3

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

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7

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

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

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(A) Tess's work is correct.

\newline

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

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4

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

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\newline

7

is incorrect. Tess didn't account for adding

3

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Aditya tried to solve the differential equation

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\newline

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

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2

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3

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1

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2

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

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

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2

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Question

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

\newline

The function represents a parabola in the

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appears as a constant or coefficient?

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1

\newline

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

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f(x)

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f(x)

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