# What is the vertex form of y=4x^2-17x+60 ?

Mar 18, 2016

Complete the square:
The vertex is ${V}_{y} \left(\textcolor{red}{\frac{17}{8}} , \textcolor{red}{\frac{671}{16}}\right)$

#### Explanation:

We can convert by completing the square on the first two terms, but first we need to have a "1" in front of the x-squared.

A standard form of parabola is:
$f \left(x\right) = a {x}^{2} + b x + c$
The vertex form for the same equation is:
$f \left(x\right) = a \left(x - \textcolor{red}{h}\right) + \textcolor{red}{k}$
Where the point $V \left(\textcolor{red}{h} , \textcolor{red}{k}\right)$ is the vertex f(x)

$y = 4 \left({x}^{2} - \frac{17}{4} x\right) + 60$

Add (b/2)^2 to complete the square

$y = 4 \left({x}^{2} - \frac{17}{4} x + \frac{289}{64}\right) + 60 - \frac{289}{16}$

The -289/16 is necessary to balance out the 4(289/64) that we added.

Factor the parentheses and find an LCD to add the 60 and -289/16

$y = 4 {\left(x - \frac{17}{8}\right)}^{2} + \frac{960}{16} - \frac{289}{16}$

$y = 4 {\left(x - \frac{17}{8}\right)}^{2} + \frac{671}{16}$