Question

How do you find the vertex and the intercepts for `y = -8x^2 + 32x`?

Answer 1

Vertex is `(2,32)` and intercepts are at `(0,0)` and `(4,0)`.

Explanation:

Vertex form of quadratic function is `y=a(x-h)^2+k`, where `(h,k)` is the vertex

Hence `y=-8x^2+32x`

= `-8(x^2-4x)`

= `-8(x^2-4x+4)+32`

= `-8)x-2)^2+32`

Hence vertex is `(2,32)`

for intercepts on `x`-axis, put `y=0` i.e. `-8x^2+32x=0` or `-8x(x-4)=0` i.e. `x=0` or `x=4`

and for intercepts on `y`-axis, put `x=0` and we have `y=0`

Hence intercepts are at `(0,0)` and `(4,0)`
graph{-8x^2+32x [-10, 10, -50, 50]}