Question

How do you find the vertex of `f(x)=-5/2x^2+10x+1/3`?

Answer 1

The vertex is at `(2,31/3)`

Explanation:

The general vertex form for a quadratic is
`color(white)("XXXX")` `f(x) = m(x-a)^2+b`
`color(white)("XXXX")` `color(white)("XXXX")`with its vertex at `(a,b)`

Given `f(x) = - 5/2x^2+10x+1/3`

Extract the `m=-5/2`
`color(white)("XXXX")` `f(x)=(-5/2)(x^2-4x) +1/3`

Complete the square
`color(white)("XXXX")` `f(x)=(-5/2)(x^2-4x+4)+1/3 - (-5/2)*4`

Write as a squared binomial and simplfy
`color(white)("XXXX")` `f(x) = (-5/2)(x-2)^2+31/3`

Note that this is in explicit vertex form.

The graph looks like: graph{-5/2x^2+10x+1/3 [-12.22, 16.25, -2.78, 11.46]}