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f(x)=(x+6)^(2)-16
At what values of x does the graph of the function intersect the x-axis?
Choose 1 answer:
(A) x=-2,x=10
(B) x=2,x=-10
(C) x=-2,x=-10
(D) f(x) does not intersect the x-axis.

$f(x)=(x+6)^2-16$ $\newline$ At what values of $x$ does the graph of the function intersect the $x$ -axis? $\newline$ Choose $1$ answer: $\newline$ (A) $x=-2$ , $x=10$ $\newline$ (B) $x=2$ , $x=-10$ $\newline$ (C) $x=-2$ , $x=-10$ $\newline$ (D) $f(x)$ does not intersect the $x$ -axis.

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Reflections of functions

Full solution

Q.

f(x)=(x+6)^2-16

\newline

At what values of

x

does the graph of the function intersect the

x

-axis?

\newline

Choose

1

answer:

\newline

(A)

x=-2

,

x=10

\newline

(B)

x=2

,

x=-10

\newline

(C)

x=-2

,

x=-10

\newline

(D)

f(x)

does not intersect the

x

-axis.

Set Equation and Solve: To find the $x$ -intercepts of the graph of the function, we need to set $f(x)$ to $0$ and solve for $x$ . So, we set the equation $(x+6)^2 - 16 = 0$ .
Solve for x: Now we solve the equation for x. $\newline$ $(x+6)^2 - 16 = 0$ $\newline$ $(x+6)^2 = 16$
Take Square Root: Next, we take the square root of both sides of the equation. $\newline$ $\sqrt{(x+6)^2} = \pm\sqrt{16}$ $\newline$ $x + 6 = \pm4$
Two Equations to Solve: We now have two equations to solve for $x$ : $1.$ $x + 6 = 4$ $2.$ $x + 6 = -4$
Solve for x (Equation $1$ ): Solving the first equation for x: $\newline$ $x + 6 = 4$ $\newline$ $x = 4 - 6$ $\newline$ $x = -2$
Solve for x (Equation $2$ ): Solving the second equation for x: $\newline$ $x + 6 = -4$ $\newline$ $x = -4 - 6$ $\newline$ $x = -10$

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