Question

What is the vertex form of `y=-32x^2+80x+2`?

Answer 1

Vertex form of equation is `y=-32(x^2-5/4)^2+52`

Explanation:

Vertex form of equation is `y=a(x-h)^2+k`

As we have `y=-32x^2+80x+2`

or `y=-32(x^2-80/32x)+2`

or `y=-32(x^2-5/2x)+2`

or `y=-32(x^2-2xx5/4x+(5/4)^2)+2-(-32)xx(5/4)^2`

or `y=-32(x^2-5/4)^2+2+32xx25/16`

or `y=-32(x^2-5/4)^2+2+50`

or `y=-32(x^2-5/4)^2+52`, where vertex is `(-5/4,-48)`
graph{-32x^2+80x+2 [-10, 10, -60, 60]}

Answer 2

y = - 32(x - 5/4)^2 + 52

Explanation:

`y = - 32x^2 + 80x + 2`
x-coordinate of vertex:
`x = -b/(2a) = 80/64 = 5/4`
y-coordinate of vertex:
`y(5/4) = -32(25/16) + 80(5/4) + 2 = -50 + 100 + 2 = 52`
Vertex form of y:
`y = - 32(x - 5/4)^2 + 52`