# What is the the vertex of y = -x^2 - 3x+12?

Aug 25, 2016

$y = - {\left(x + \frac{3}{2}\right)}^{2} + \frac{57}{4} \text{ is in the form } y = a {\left(x + b\right)}^{2} + c$

The vertex is at $\left(- \frac{3}{2} , + \frac{57}{4}\right)$

#### Explanation:

This is graph of an inverted parabola. To find the vertex (the maximum turning point) write the equation in the form

$y = a \textcolor{b l u e}{{\left(x + b\right)}^{2}} \textcolor{m a \ge n t a}{+ c}$ which gives us the vertex as $\left(- b , c\right)$

We use the process of "completing the square"

Note:${\left(x - 5\right)}^{2} = {x}^{2} - 10 x + 25 \text{ in particular} {\left(\frac{10}{2}\right)}^{2} = 25$

In the square of a binomial , this relationship is always true.

$\textcolor{red}{{\left(\frac{b}{2}\right)}^{2}} = c$

STEP 1 make the coefficient of ${x}^{2}$ term = +1.

$y = - \left(1 {x}^{2} + 3 x - 12\right) \textcolor{w h i t e}{\times} \left(- 12 \text{ is not the correct value for c}\right)$ .

STEP 2 add and subtract $\textcolor{red}{{\left(\frac{b}{2}\right)}^{2}} \text{ which is } {\left(\frac{3}{2}\right)}^{2}$

$y = - \left[\textcolor{b l u e}{{x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2}} \textcolor{m a \ge n t a}{- {\left(\frac{3}{2}\right)}^{2} - 12}\right)$

STEP 3 Write the first 3 terms as a perfect square

$y = - \left[\textcolor{b l u e}{{\left(x + \frac{3}{2}\right)}^{2}} + \textcolor{m a \ge n t a}{\left(- \left(\frac{9}{4}\right) - \frac{48}{4}\right)}\right]$

STEP 4 Simplify the constant

$y = - \left[\textcolor{b l u e}{{\left(x + \frac{3}{2}\right)}^{2}} \textcolor{m a \ge n t a}{- \frac{57}{4}}\right]$

STEP 5 Remove the outer bracket to get vertex form.

$y = - \textcolor{b l u e}{{\left(x + \frac{3}{2}\right)}^{2}} \textcolor{m a \ge n t a}{+ \frac{57}{4}}$

$y = a \textcolor{b l u e}{{\left(x + b\right)}^{2}} \textcolor{m a \ge n t a}{+ c}$

The vertex is at $\left(- \frac{3}{2} , + \frac{57}{4}\right)$

graph{-x^2 -3x+12 [-3.834, 1.166, 12.52, 15.02]}