f(x)=(x-4)^(2)-16 At what values of x does the graph of the function intersect the x-axis? Choose 1 answer: (A) x=8,x=0 (B) x=-8,x=0 (c) x=8,x=-8 (D) f(x) does not intersect the x-axis.

Problem Understanding: Understand the problem. We need to find the values of x where the function f(x) intersects the x-axis. The x-axis is where the value of the function f(x) is zero.Setting the Equation: Set the function equal to zero to find the x-intercepts. 0 = (x-4)^(2) - 16Solving the Equation: Solve the equation for x. Add 16 to both sides of the equation to isolate the squared term. (x-4)^(2) = 16Taking the Square Root: Take the square root of both sides. √((x-4)^(2)) = ±√(16) x - 4 = ±4Final Solution: Solve for x. x = 4 + 4 or x = 4 - 4 x = 8 or x = 0

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f(x)=(x-4)^(2)-16
At what values of
x does the graph of the function intersect the
x-axis?
Choose 1 answer:
(A)
x=8,x=0
(B)
x=-8,x=0
(c)
x=8,x=-8
(D)
f(x) does not intersect the
x-axis.

$f(x)=(x-4)^{2}-16$ $\newline$ At what values of $x$ does the graph of the function intersect the $x$ -axis? $\newline$ Choose $1$ answer: $\newline$ (A) $x=8, x=0$ $\newline$ (B) $x=-8, x=0$ $\newline$ (C) $x=8, x=-8$ $\newline$ (D) $f(x)$ does not intersect the $x$ -axis.

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

f(x)=(x-4)^{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=8, x=0

\newline

(B)

x=-8, x=0

\newline

(C)

x=8, x=-8

\newline

(D)

f(x)

does not intersect the

x

-axis.

Problem Understanding: Understand the problem. $\newline$ We need to find the values of $x$ where the function $f(x)$ intersects the $x$ -axis. The $x$ -axis is where the value of the function $f(x)$ is zero.
Setting the Equation: Set the function equal to zero to find the x-intercepts. $\newline$ $0 = (x-4)^{2} - 16$
Solving the Equation: Solve the equation for $x$ . $\newline$ Add $16$ to both sides of the equation to isolate the squared term. $\newline$ $(x-4)^{2} = 16$
Taking the Square Root: Take the square root of both sides. $\newline$ $\sqrt{(x-4)^{2}} = \pm\sqrt{16}$ $\newline$ $x - 4 = \pm4$
Final Solution: Solve for $x$ . $x = 4 + 4 \text{ or } x = 4 - 4$ $x = 8 \text{ or } x = 0$