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

Dec 4, 2017

the vertex is at $\left(- 0.875 , 9.0625\right)$

Explanation:

y = −3x^2 −x −3 −(x −3)^2

Simplify the RHS
$y = - 3 {x}^{2} - x - 3 - {x}^{2} - 6 x + 9$
$y = - 4 {x}^{2} - 7 x + 6$

The general quadratic form is $y = a x 2 + b x + c$
The vertex can be found at $\left(h , k\right)$
where $h = - \frac{b}{2} a$

Substitute in what we know
$h = - \frac{- 7}{2 \cdot - 4} = - \frac{7}{8} = - 0.875$

Substititue the value of $h$ for $x$ in the original equation
$y = - 4 {\left(- \frac{7}{8}\right)}^{2} - 7 \left(- \frac{7}{8}\right) + 6 = 9.0625$

the vertex is at $\left(- 0.875 , 9.0625\right)$