Question

How do you find the vertex and intercepts for `y=2x^2-16x+27`?

Answer 1

`y = 2(x - 4)^2 - 5" "` Vertex is at `(4, -5)`
`y-`intercept is at 27

Explanation:

`y = 2x^2 - 16x + 27`
`y = 2(x^2 - 8x + 27/2) " "`now complete the square

`y = 2[color(blue)(x^2 - 8x + 16) -16 +27/2] " " rArr` +16-16 = 0

`y = 2[color(blue)((x - 4)^2) -16 +27/2] " "` `-16+27/2 = -5/2`

`y = 2(x-4)^2 - 2xx5/2`

`y = 2(x - 4)^2 - 5`