Question

1. if $f(x) = -x^2 - 2x + 8$, what are the value(s) of $x$ when $f(x) = 5$?

a. $x = -1$
b. $x = -4$, $x = 2$
c. $x = -3$, $x = 1$
d. $x = 0$, $x = 2$

Explanation:

Step1: Set up the equation

Since \( f(x) = -x^2 - 2x + 8 \) and \( f(x) = 5 \), we set them equal: \( -x^2 - 2x + 8 = 5 \)

Step2: Simplify the equation

Subtract 5 from both sides: \( -x^2 - 2x + 3 = 0 \)
Multiply both sides by -1: \( x^2 + 2x - 3 = 0 \)

Step3: Factor the quadratic

Factor \( x^2 + 2x - 3 \): \( (x + 3)(x - 1) = 0 \)

Step4: Solve for x

Set each factor to zero: \( x + 3 = 0 \) or \( x - 1 = 0 \)
So \( x = -3 \) or \( x = 1 \)

Answer:

C. \( x = -3 \), \( x = 1 \)