f(x)=(x-1)^(2)-36 At what values of x does the graph of the function intersect the x-axis? Choose 1 answer: (A) x=-7,x=-5 (B) x=7,x=5 (C) x=7,x=-5 (D) f(x) does not intersect the x-axis.

Set f(x) to 0: To find the x-intercepts of the graph of the function, we need to set f(x) to 0 and solve for x. f(x) = (x-1)² - 36 0 = (x-1)² - 36Add 36 to both sides: Now we will add 36 to both sides of the equation to isolate the squared term. 0 + 36 = (x-1)² - 36 + 36 36 = (x-1)²Take square root: Next, we take the square root of both sides of the equation to solve for x-1. Remember that taking the square root of a number yields two solutions, one positive and one negative. √36 = √(x-1)² ±6 = x - 1Solve for x: We will now solve for x by adding 1 to both sides of each equation. 6 + 1 = x - 1 + 1 and -6 + 1 = x - 1 + 1 7 = x and -5 = xIdentify x-intercepts: We have found the two values of x where the graph of the function intersects the x-axis. These values are x = 7 and x = -5.

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f(x)=(x-1)^(2)-36
At what values of
x does the graph of the function intersect the
x-axis?
Choose 1 answer:
(A)
x=-7,x=-5
(B)
x=7,x=5
(C)
x=7,x=-5
(D)
f(x) does not intersect the
x-axis.

$f(x)=(x-1)^{2}-36$ $\newline$ At what values of $x$ does the graph of the function intersect the $x$ -axis? $\newline$ Choose $1$ answer: $\newline$ (A) $x=-7, x=-5$ $\newline$ (B) $x=7, x=5$ $\newline$ (C) $x=7, x=-5$ $\newline$ (D) $f(x)$ does not intersect the $x$ -axis.

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Algebra 1

Domain and range of quadratic functions: equations

Full solution

f(x)=(x-1)^{2}-36

\newline

At what values of

x

does the graph of the function intersect the

x

-axis?

\newline

Choose

1

answer:

\newline

(A)

x=-7, x=-5

\newline

(B)

x=7, x=5

\newline

(C)

x=7, x=-5

\newline

(D)

f(x)

does not intersect the

x

-axis.

Set $f(x)$ to $0$ : To find the x-intercepts of the graph of the function, we need to set $f(x)$ to $0$ and solve for $x$ .
$f(x) = (x-1)^2 - 36$
$0 = (x-1)^2 - 36$
Add $36$ to both sides: Now we will add $36$ to both sides of the equation to isolate the squared term. $\newline$ $0 + 36 = (x-1)^2 - 36 + 36$ $\newline$ $36 = (x-1)^2$
Take square root: Next, we take the square root of both sides of the equation to solve for $x-1$ . Remember that taking the square root of a number yields two solutions, one positive and one negative. $\newline$ $\sqrt{36} = \sqrt{(x-1)^2}$ $\newline$ $\pm 6 = x - 1$
Solve for x: We will now solve for x by adding $1$ to both sides of each equation. $\newline$ $6 + 1 = x - 1 + 1$ and $-6 + 1 = x - 1 + 1$ $\newline$ $7 = x$ and $-5 = x$
Identify x-intercepts: We have found the two values of $x$ where the graph of the function intersects the x-axis. These values are $x = 7$ and $x = -5$ .