Question

Find the intercepts of the parabola whose function is $f\left(x\right)=x^2-4x-5.$

Short answer

• The y-intercept at the point (0, -5)
• The x-intercepts at the points (5, 0) and (−1, 0) .

Step-by-step solution

Step 1. To find the y-intercept, let x = 0 andsolve for f (x) .

$$\begin{gathered} f\left(x\right)=x^2-4x-5 \\ f\left(0\right)={\left(0\right)}^2-4\left(0\right)-5 \\ f\left(x\right)=-5 \end{gathered}$$

When x = 0 , then f (0) = -5 .

The y-intercept at the point (0, -5) .

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

$$\begin{gathered} f\left(x\right)=x^2-4x-5 \\ 0=x^2-4x-5 \end{gathered}$$

Solve by factoring.

localid="1646062204697" $$\begin{gathered} 0=(x-5)(x+1) \\ 0=x-5;0=x+1 \\ 5=x;-1=x \end{gathered}$$

When f (x) = 0 , then x = 5 or x = −4 .

The x-intercepts are the points (5, 0) and (−1, 0) .