Question

vertex & x and y intercepts of f(x)=x^2

Answer 1

Answer:

vertex (0, 0)

x-intercept: 0

y-intercept: 0

Step-by-step explanation:

The vertex form of a quadratic function:

$f(x)=a(x-h)^2+k$

(h, k) - vertex

We have

$f(x)=x^2=1(x-0)^2+0$

Therefore the vertex is in (0, 0).

The x-intercept for y = 0. Substitute:

$x^2=0\to x=0$

The y-intercept for x = 0. Substitute:

$y=0^2\to y=0$