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f(x)=(x+4)^(2)-25
At what values of
x does the graph of the function intersect the
x-axis?
Choose 1 answer:
(A)
x=1,x=-9
(B)
x=1,x=9
(c)
x=-1,x=-9
(D)
f(x) does not intersect the
x-axis.

$f(x)=(x+4)^{2}-25$ $\newline$ At what values of $x$ does the graph of the function intersect the $x$ -axis? $\newline$ Choose $1$ answer: $\newline$ (A) $x=1, x=-9$ $\newline$ (B) $x=1, x=9$ $\newline$ (C) $x=-1, x=-9$ $\newline$ (D) $f(x)$ does not intersect the $x$ -axis.

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Domain and range of quadratic functions: equations

Full solution

Q.

f(x)=(x+4)^{2}-25

\newline

At what values of

x

does the graph of the function intersect the

x

-axis?

\newline

Choose

1

answer:

\newline

(A)

x=1, x=-9

\newline

(B)

x=1, x=9

\newline

(C)

x=-1, x=-9

\newline

(D)

f(x)

does not intersect the

x

-axis.

Given function: We are given the function $f(x) = (x + 4)^2 - 25$ . To find the x-intercepts, we need to set $f(x)$ to $0$ and solve for $x$ . $\newline$ $0 = (x + 4)^2 - 25$
Moving $25$ to other side: Now we will solve the equation by moving $25$ to the other side of the equation. $(x + 4)^2 = 25$
Taking square root: Next, we take the square root of both sides of the equation. Remember that taking the square root of a number yields two solutions, one positive and one negative. $\newline$ $x + 4 = \pm\sqrt{25}$
Positive root: Since the square root of $25$ is $5$ , we have: $\newline$ $x + 4 = \pm 5$
Negative root: Now we will solve for $x$ by creating two separate equations, one for the positive root and one for the negative root. $\newline$ For the positive root: $\newline$ $x + 4 = 5$ $\newline$ $x = 5 - 4$ $\newline$ $x = 1$
Solving for positive root: For the negative root: $\newline$ $x + 4 = -5$ $\newline$ $x = -5 - 4$ $\newline$ $x = -9$

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