# What is the the vertex of y = -12x^2+6x-18 ?

Sep 30, 2017

$\text{vertex } = \left(\frac{1}{4} , - \frac{69}{4}\right)$

#### Explanation:

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

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

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

$\text{to express "y=-12x^2+6x-18" in this form}$

$\text{we can use the method of "color(blue)"completing the square}$

• " ensure the coefficient of "x^2" term is 1"

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

$\Rightarrow y = - 12 {x}^{2} + 6 x - 18$

$\textcolor{w h i t e}{\Rightarrow y} = - 12 \left({x}^{2} - \frac{1}{2} x + \frac{3}{2}\right)$

$\textcolor{w h i t e}{\Rightarrow y} = - 12 \left({x}^{2} + 2 \left(- \frac{1}{4}\right) x \textcolor{red}{+ \frac{1}{16}} \textcolor{red}{- \frac{1}{16}} + \frac{3}{2}\right)$

$\textcolor{w h i t e}{\Rightarrow y} = - 12 {\left(x - \frac{1}{4}\right)}^{2} - \frac{69}{4} \leftarrow \text{ in vertex form}$

$\text{with " h=1/4" and } k = - \frac{69}{4}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{4} , - \frac{69}{4}\right)$