# What is the vertex of  y= x^2-x-16+(x-1)^2?

Dec 13, 2017

First, expand the expression and combine like terms:

${x}^{2} - x - 16 + {\left(x - 1\right)}^{2}$

$\setminus \implies {x}^{2} - x - 16 + \left({x}^{2} - 2 x + 1\right)$

$\setminus \implies {x}^{2} + {x}^{2} - x - 2 x - 16 + 1$

$\setminus \implies 2 {x}^{2} - 3 x - 15$

Now that's in the form $a {x}^{2} + b x + c$, the vertex's $x$-coordinate is $\setminus \frac{- b}{2 a}$.

$\setminus \implies \setminus \frac{3}{4}$

Plug that into the original equation to find the $y$-coordinate:

$2 {x}^{2} - 3 x - 15$

$\setminus \implies 2 {\left(\frac{3}{4}\right)}^{2} - 3 \left(\frac{3}{4}\right) - 15$

$\setminus \implies \frac{9}{8} - \frac{9}{4} - \frac{15}{1}$

$\implies - 16.125$

I'm in class rn and will finish this later. Sorry. :/