# What is the vertex form of y=4x^2+x-4+10 ?

Nov 11, 2017

$y = 4 {\left(x + \frac{1}{8}\right)}^{2} + \frac{95}{16}$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to express in this form use "color(blue)"completing the square}$

$y = 4 {x}^{2} + x + 6$

• " coefficient of "x^2" term must be 1"

$\Rightarrow y = 4 \left({x}^{2} + \frac{1}{4} x + \frac{3}{2}\right)$

• " add/subtract "(1/2"coefficient of x-term")^2

$\text{to } {x}^{2} + \frac{1}{4} x$

$\Rightarrow y = 4 \left({x}^{2} + 2 \left(\frac{1}{8}\right) x \textcolor{red}{+ \frac{1}{64}} \textcolor{red}{- \frac{1}{64}} + \frac{3}{2}\right)$

$\textcolor{w h i t e}{\Rightarrow y} = 4 {\left(x + \frac{1}{8}\right)}^{2} + \left(4 \times \frac{95}{64}\right)$

$\textcolor{w h i t e}{\Rightarrow y} = 4 {\left(x + \frac{1}{8}\right)}^{2} + \frac{95}{16}$