How do you find the vertex and the intercepts for $y = (x - 3)^2 - 2$ ?

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Answer 1 · 2017-02-12T12:59:23

Vertex (3, -2)

x-intercepts: $3+sqrt2 and 3-sqrt2$
y-intercept 7

The equation is in vertex form. Its vertex is already evident. It is (3, -2).

For x- intercept put y=0 and solve for x. X-intercepts would be $3+sqrt2$ and $3-sqrt2$

For y- intercept put x=0. It would be y= 7

How do you find the vertex and the intercepts for y = (x - 3)^2 - 2y = (x - 3)^2 - 2?