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

User Kwichz
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1 Answer

5 votes

Answer:

vertex (0, 0)

x-intercept: 0

y-intercept: 0

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

User Josemigallas
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