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

1 Answer
Aug 7, 2015

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]}