f(x)=(x-2)^(2)-9 At what values of x does the graph of the function intersect the x-axis? Choose 1 answer: (A) x=-5,x=-1 (B) x=5,x=1 (c) x=-5,x=1 (D) x=5,x=-1

Set equation to zero: The question prompt is: ＂At what values of x does the graph of the function f(x) = (x-2)^(2) - 9 intersect the x-axis?＂ To find the x-intercepts of the function, we need to set f(x) to zero and solve for x. So, we set the equation (x-2)^(2) - 9 = 0.Solve the equation: Now we solve the equation (x-2)^(2) - 9 = 0. We can rewrite this as (x-2)^(2) = 9. Taking the square root of both sides, we get x - 2 = ±3.Positive root: We solve for x by adding 2 to both sides of the equations. For the positive root: x - 2 = 3 leads to x = 3 + 2, which gives us x = 5. For the negative root: x - 2 = -3 leads to x = -3 + 2, which gives us x = -1.Negative root: We have found two values of x where the function intersects the x-axis: x = 5 and x = -1. These are the x-intercepts of the function.

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f(x)=(x-2)^(2)-9
At what values of
x does the graph of the function intersect the
x-axis?
Choose 1 answer:
(A)
x=-5,x=-1
(B)
x=5,x=1
(c)
x=-5,x=1
(D)
x=5,x=-1

$f(x)=(x-2)^{2}-9$ $\newline$ At what values of $x$ does the graph of the function intersect the $x$ -axis? $\newline$ Choose $1$ answer: $\newline$ (A) $x=-5, x=-1$ $\newline$ (B) $x=5, x=1$ $\newline$ (C) $x=-5, x=1$ $\newline$ (D) $x=5, x=-1$

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

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

\newline

At what values of

x

does the graph of the function intersect the

x

-axis?

\newline

Choose

1

answer:

\newline

(A)

x=-5, x=-1

\newline

(B)

x=5, x=1

\newline

(C)

x=-5, x=1

\newline

(D)

x=5, x=-1

Set equation to zero: The question prompt is: ＂At what values of $x$ does the graph of the function $f(x) = (x-2)^{2} - 9$ intersect the x-axis?＂ $\newline$ To find the x-intercepts of the function, we need to set $f(x)$ to zero and solve for $x$ . $\newline$ So, we set the equation $(x-2)^{2} - 9 = 0$ .
Solve the equation: Now we solve the equation $(x-2)^{2} - 9 = 0$ . $\newline$ We can rewrite this as $(x-2)^{2} = 9$ . $\newline$ Taking the square root of both sides, we get $x - 2 = \pm3$ .
Positive root: We solve for x by adding $2$ to both sides of the equations. $\newline$ For the positive root: $x - 2 = 3$ leads to $x = 3 + 2$ , which gives us $x = 5$ . $\newline$ For the negative root: $x - 2 = -3$ leads to $x = -3 + 2$ , which gives us $x = -1$ .
Negative root: We have found two values of $x$ where the function intersects the $x$ -axis: $x = 5$ and $x = -1$ . These are the $x$ -intercepts of the function.