Question

How do you find the intercepts, vertex and graph `f(x)=x^2-9`?

Answer 1

vertex (0,-9)

x intercepts (3,0) and (-3,0)

y intercept (0,-9)

Explanation:

`f(x)=y= x^2-9` represents a vertical parabola opening up. Written in standard form as `(y+9)= (x-0)^2`, shows that its vertex is at (0,-9).

For intercepts, let x=0, which gives intercept y=-9. For x intercept put y=0, which gives `x^2 =9`. Thus there would be two x -intercepts (3,0) and (-3,0). The graph of parabola would be like this :