Question

Find the intercepts of the parabola whose function is $f\left(x\right)=x^2-4x-12.$

Short answer

• The y-intercept is the point (0, −12) .
• The x-intercepts are the points (6, 0) and (−2, 0) .

Step-by-step solution

Step 1. Given information

Consider the function

$f\left(x\right)=x^2-4x-12.$

Step 2. To find the y-intercept, let x = 0 and solve for f (x) .

$$\begin{gathered} f\left(x\right)=x^2-4x-12 \\ f\left(0\right)={\left(0\right)}^2-4\left(0\right)-12 \\ f\left(0\right)=-12 \end{gathered}$$

When x = 0 , then f (0) = −12 .

The y-intercept is the point (0, −12) .

Step 3. To find the x-intercept, let f (x) = 0 and solve for x .

$$\begin{gathered} f\left(x\right)=x^2-4x-12 \\ 0=x^2-4x-12 \end{gathered}$$

Solve by factoring.

$$\begin{gathered} 0=(x-6)(x+2) \\ 0=x-6;0=x+2 \\ 6=x;-2=x \end{gathered}$$

When f (x) = 0 , then x = 6 or x = −2 .

The x-intercepts are the points (6, 0) and (−2, 0) .