Question

Find the intercepts of the parabola whose function is $f\left(x\right)=x^2+2x-8$.

Short answer

• The y-intercept is the point (0, −8) .
• The x-intercepts are the points$(-4,0)and(2,0)$
  

Step-by-step solution

Step 1. Given Information

The parabola whose function is $f\left(x\right)=x^2+2x-8$

We have to find the intercepts of the parabola.

Step 2. Solve

To find the y-intercept, let x = 0 and solve for f (x) .

$$\begin{gathered} f\left(x\right)=x^2+2x-8 \\ f\left(0\right)={\left(0\right)}^2+2\left(0\right)-8 \\ f\left(0\right)=-8 \end{gathered}$$

When x = 0 , then f (0) = −8 .

The y-intercept is the point (0, −8) .

To find the x-intercept, let f (x) = 0 and solve for x .

localid="1646151346632" $$\begin{gathered} f\left(x\right)=x^2+2x-8 \\ 0=x^2+2x-8 \end{gathered}$$

Solve by factoring:

$$\begin{gathered} 0=x^2+2x-8 \\ 0=(x+4)(x-2) \\ 0=x+4;0=x-2 \\ -4=x;2=x \end{gathered}$$

When $f\left(x\right)=0$, then $x=-4orx=2$.

The x-intercepts are the points$(-4,0)and(2,0)$