Question

How do you find the y intercept, axis of symmetry and the vertex to graph the function `f(x)=x^2-9`?

Answer 1

Vertex is at `(0,-9)` , y-intercept: `y= -9 or (0, -9)` ,
axis of symmetry is `x=0` or y-axis.

Explanation:

`f(x) = x^2 -9 = a(x-0)^2 -9`. Comparing with standard vertex form

of equation `f(x)=a(x-h)^2+k ; (h,k)` being vertex,

we find here `h=0,k= -9`. So vertex is at `(h,k) or (0,-9)`.

putting `x=0` in the equation we find y-intercept as

`y= 0^2-9 = -9 or (0, -9)`

Axis of symmetry is `x=0` or y-axis.

graph{x^2-9 [-40, 40, -20, 20]} [Ans]