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

Aug 7, 2015

The vertex is at $\left(2 , \frac{31}{3}\right)$

Explanation:

The general vertex form for a quadratic is
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = m {\left(x - a\right)}^{2} + b$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$with its vertex at $\left(a , b\right)$

Given $f \left(x\right) = - \frac{5}{2} {x}^{2} + 10 x + \frac{1}{3}$

Extract the $m = - \frac{5}{2}$
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- \frac{5}{2}\right) \left({x}^{2} - 4 x\right) + \frac{1}{3}$

Complete the square
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- \frac{5}{2}\right) \left({x}^{2} - 4 x + 4\right) + \frac{1}{3} - \left(- \frac{5}{2}\right) \cdot 4$

Write as a squared binomial and simplfy
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- \frac{5}{2}\right) {\left(x - 2\right)}^{2} + \frac{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]}