f(x)=(x+2)^(2)-64 At what values of x does the graph of the function intersect the x-axis? Choose 1 answer: (A) x=6,x=-10 (B) x=6,x=10 (c) x=-6,x=-10 (D) f(x) does not intersect the x-axis.

Problem Understanding: Understand the problem. We need to find the values of x for which f(x) = 0, because the graph of the function intersects the x-axis where the function value is zero.Equation Setup: Set the function equal to zero and solve for x. f(x) = (x+2)^2 - 64 = 0Quadratic Equation Solution: Solve the quadratic equation. (x+2)^2 = 64 Take the square root of both sides. x+2 = ±√64 x+2 = ±8Solving for x: Solve for x. x = -2 + 8 or x = -2 - 8 x = 6 or x = -10Solution Verification: Check the solutions. Plug x = 6 and x = -10 back into the function to ensure they result in f(x) = 0. f(6) = (6+2)^2 - 64 = 64 - 64 = 0 f(-10) = (-10+2)^2 - 64 = 144 - 64 = 80 There is a math error in the calculation for f(-10). The correct calculation should be: f(-10) = (-10+2)^2 - 64 = (-8)^2 - 64 = 64 - 64 = 0

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f(x)=(x+2)^(2)-64
At what values of
x does the graph of the function intersect the
x-axis?
Choose 1 answer:
(A)
x=6,x=-10
(B)
x=6,x=10
(c)
x=-6,x=-10
(D)
f(x) does not intersect the
x-axis.

$f(x)=(x+2)^{2}-64$ $\newline$ At what values of $x$ does the graph of the function intersect the $x$ -axis? $\newline$ Choose $1$ answer: $\newline$ (A) $x=6, x=-10$ $\newline$ (B) $x=6, x=10$ $\newline$ (C) $x=-6, x=-10$ $\newline$ (D) $f(x)$ does not intersect the $x$ -axis.

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Algebra 1

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Full solution

f(x)=(x+2)^{2}-64

\newline

At what values of

x

does the graph of the function intersect the

x

-axis?

\newline

Choose

1

answer:

\newline

(A)

x=6, x=-10

\newline

(B)

x=6, x=10

\newline

(C)

x=-6, x=-10

\newline

(D)

f(x)

does not intersect the

x

-axis.

Problem Understanding: Understand the problem. $\newline$ We need to find the values of $x$ for which $f(x) = 0$ , because the graph of the function intersects the $x$ -axis where the function value is zero.
Equation Setup: Set the function equal to zero and solve for x. $\newline$ f(x) = (x+ $2$ )^ $2$ - $64$ = $0$
Quadratic Equation Solution: Solve the quadratic equation. $\newline$ $(x+2)^2 = 64$ $\newline$ Take the square root of both sides. $\newline$ $x+2 = \pm\sqrt{64}$ $\newline$ $x+2 = \pm8$
Solving for x: Solve for x. $\newline$ $x = -2 + 8$ or $x = -2 - 8$ $\newline$ $x = 6$ or $x = -10$
Solution Verification: Check the solutions. $\newline$ Plug $x = 6$ and $x = -10$ back into the function to ensure they result in $f(x) = 0$ . $\newline$ $f(6) = (6+2)^2 - 64 = 64 - 64 = 0$ $\newline$ $f(-10) = (-10+2)^2 - 64 = 144 - 64 = 80$ $\newline$ There is a math error in the calculation for $f(-10)$ . The correct calculation should be: $\newline$ $f(-10) = (-10+2)^2 - 64 = (-8)^2 - 64 = 64 - 64 = 0$