Answer:
Step-by-step explanation:
The vertex form of a quadratic function:
[tex]f(x)=a(x-h)^2+k[/tex]
(h, k) - vertex
We have
[tex]f(x)=x^2=1(x-0)^2+0[/tex]
Therefore the vertex is in (0, 0).
The x-intercept for y = 0. Substitute:
[tex]x^2=0\to x=0[/tex]
The y-intercept for x = 0. Substitute:
[tex]y=0^2\to y=0[/tex]