# What is the vertex of  y= -3x^2-x-(x-3)^2?

Jul 18, 2017

The vertex of the equation $- 3 {x}^{2} - x - {\left(x - 3\right)}^{2}$ would be at point
$\left(\frac{5}{8} , - \frac{119}{16}\right)$

#### Explanation:

First expand out the ${\left(x - 3\right)}^{2}$ part of the equation into $- 3 {x}^{2} - x - \left({x}^{2} - 6 x + 9\right)$

Then get rid of the parenthesis, $- 3 {x}^{2} - x - {x}^{2} + 6 x - 9$ and combine like terms

$\implies - 4 {x}^{2} + 5 x - 9$
The equation for finding the domain of the vertex is $- \frac{b}{2 a}$

Therefore the domain of the vertex is $- \frac{5}{2 \cdot - 4} = \frac{5}{8}$

Input the domain into the function to get the range

$\implies - 4 {\left(\frac{5}{8}\right)}^{2} + 5 \left(\frac{5}{8}\right) - 9 = - \frac{119}{16}$

Therefore the vertex of the equation is $\left(\frac{5}{8} , - \frac{119}{16}\right)$