# How do you write the quadratic in vertex form given y=-4x^2+12x+5?

May 25, 2018

$y = - 4 {\left(x - \frac{3}{2}\right)}^{2} + 14$

#### Explanation:

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

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

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

$\text{to obtain this form use the method of "color(blue)"completing the square}$

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

$\text{factor out } - 4$

$y = - 4 \left({x}^{2} - 3 x - \frac{5}{4}\right)$

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - 3 x$

$y = - 4 \left({x}^{2} + 2 \left(- \frac{3}{2}\right) x \textcolor{red}{+ \frac{9}{4}} \textcolor{red}{- \frac{9}{4}} - \frac{5}{4}\right)$

$\textcolor{w h i t e}{y} = - 4 {\left(x - \frac{3}{2}\right)}^{2} - 4 \left(- \frac{9}{4} - \frac{5}{4}\right)$

$\textcolor{w h i t e}{y} = - 4 {\left(x - \frac{3}{2}\right)}^{2} + 14 \leftarrow \textcolor{red}{\text{in vertex form}}$