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

Jun 3, 2016

The vertex is at $\left(\frac{1}{8} , \frac{63}{16}\right)$

#### Explanation:

$y = a {\left(x - h\right)}^{2} + k$

The vertex is at the point $\left(h , k\right)$

Rearrange your equation to obtain a form similar to that of the quadratic equation.

$y = 4 {x}^{2} - x + 4$

$y = 4 {x}^{2} - x + \textcolor{red}{\frac{4}{64}} - \textcolor{red}{\frac{4}{64}} + 4$

$y = \left(4 {x}^{2} - x + \textcolor{red}{\frac{4}{64}}\right) - \textcolor{red}{\frac{4}{64}} + 4$

Take $\textcolor{red}{4}$ as a common factor.

$y = 4 \left({x}^{2} - \frac{1}{4} x + \textcolor{red}{\frac{1}{64}}\right) - \textcolor{red}{\frac{4}{64}} + 4$

$y = 4 {\left(x - \frac{1}{8}\right)}^{2} + \frac{4 \times 64 - 4}{64}$

$y = 4 {\left(x - \frac{1}{8}\right)}^{2} + \frac{252}{64}$

$y = 4 {\left(x - \frac{1}{8}\right)}^{2} + \frac{63}{16}$

The vertex is at $\left(\frac{1}{8} , \frac{63}{16}\right)$

graph{4*x^2-x+4 [-7.8, 8.074, -1.044, 6.896]}