Question

How do you find the vertex and intercepts for `y=2(x+2)^2-1`?

Answer 1

Vertex (-2, -1)

Explanation:

Coordinate of vertex:
x + 2 = 0 --> x = - 2
y-coordinate of vertex: -1
Vertex (-2, -1)
To find the 2 x-intercepts, make y= 0, and solve the quadratic equation:
`2(x + 2)^2 - 1 = 0`
`(x + 2)^2 = 1/2`
`(x + 2) = +- 1/sqrt2 = +- sqrt2/2`
`x = - 2 +- sqrt2/2 = (- 4 +- sqrt2)/2`