Question

what are the x-intercepts of the function $f(x) = -x^2 - x + 2$?
○ $(-2, 0)$ and $(1, 0)$
○ $(2, 0)$ and $(1, 0)$
○ $(-2, 0)$ and $(-1, 0)$
○ $(0, -2)$ and $(0, 1)$

Explanation:

Step1: Set \( f(x) = 0 \)

To find the \( x \)-intercepts, we set \( f(x)=0 \), so we have the equation \( -x^{2}-x + 2=0 \). Multiply both sides by - 1 to simplify: \( x^{2}+x - 2=0 \)

Step2: Factor the quadratic equation

We factor the quadratic \( x^{2}+x - 2 \). We need two numbers that multiply to - 2 and add to 1. The numbers are 2 and - 1. So, \( x^{2}+x - 2=(x + 2)(x - 1)=0 \)

Step3: Solve for \( x \)

Using the zero - product property, if \( (x + 2)(x - 1)=0 \), then either \( x+2 = 0 \) or \( x - 1=0 \).

• If \( x+2=0 \), then \( x=-2 \).
• If \( x - 1=0 \), then \( x = 1 \).

So the \( x \)-intercepts occur when \( x=-2 \) and \( x = 1 \), and the coordinates of the \( x \)-intercepts are \( (-2,0) \) and \( (1,0) \)

Answer:

\( (-2, 0) \) and \( (1, 0) \) (corresponding to the first option: \( \boldsymbol{(-2, 0)} \) and \( \boldsymbol{(1, 0)} \))