Question

How do you find the vertex and intercepts for `y=2(x+2)^2-13`?

Answer 1

Find vertex and intercepts for y = 2(x + 2)^2 - 13

Explanation:

y = 2(x + 2)^2 - 13.
This is the intercept form, therefor the coordinates of vertex are: Vertex (-2, -13).
To find y-intercept, make x = 0 --> `y = 2(2)^2 - 13 = 8 - 13 = -5`
To find x-intercepts, make y = 0 --> `0 = 2(x + 2)^2 - 13`
`(x + 2)^2 = 13/2 --> x + 2 = +- sqrt13/sqrt2`
`x = -2 +- sqrt26/2`