# What is the vertex of  y= -(2x-1)^2-x^2-2x+3?

Jun 6, 2017

$\left(\frac{1}{5} , \frac{11}{5}\right)$

#### Explanation:

Let's expand everything we've got and see what we're working with:
$y = - {\left(2 x - 1\right)}^{2} - {x}^{2} - 2 x + 3$

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

$y = - \left(\left(2 x - 1\right) \times \left(2 x - 1\right)\right) - {x}^{2} - 2 x + 3$

$y = - \left(4 {x}^{2} - 2 x - 2 x + 1\right) - {x}^{2} - 2 x + 3$

distribute the negative

$y = - 4 {x}^{2} + 4 x - 1 - {x}^{2} - 2 x + 3$

combine like-terms

$y = - 5 {x}^{2} + 2 x + 2$

Now, let's rewrite the standard form into vertex form. To do that, we need to complete the square

$y = - 5 {x}^{2} + 2 x + 2$

factor out the negative $5$

$y = - 5 \left({x}^{2} - \frac{2}{5} x - \frac{2}{5}\right)$

Now we take the middle term ($\frac{2}{5}$) and divide it by $2$. That gives us $\frac{1}{5}$. Now we square it, which gives us $\frac{1}{25}$. Now we have the value that will give us a perfect square. We add $\frac{1}{25}$ to the equation but we cannot randomly introduce a new value in this equation! What we can do is add $\frac{1}{25}$ and then subtract it $\frac{1}{25}$. That way, we haven't actually changed the value of the equation.

So, we have $y = - 5 \left({x}^{2} - \frac{2}{5} x - \frac{2}{5} + \frac{1}{25} - \frac{1}{25}\right)$

$y = - 5 \left(\textcolor{red}{{x}^{2} - \frac{2}{5} x + \frac{1}{25}} - \frac{2}{5} - \frac{1}{25}\right)$

rewrite as a perfect square

$y = - 5 \left({\left(x - \frac{1}{5}\right)}^{2} - \frac{2}{5} - \frac{1}{25}\right)$

combine constants

$y = - 5 \left({\left(x - \frac{1}{5}\right)}^{2} - \frac{11}{25}\right)$

multiply $- \frac{11}{25}$ by $- 5$ to remove one of the parentheses

$y = - 5 {\left(x - \frac{1}{5}\right)}^{2} + \frac{11}{5}$

Now we've got the equation in vertex form.

From here, we can tell the vertex very easily:

$y = - 5 {\left(x \textcolor{b l u e}{- \frac{1}{5}}\right)}^{2} + \textcolor{g r e e n}{\frac{11}{5}}$

Gives us $\left(- \textcolor{b l u e}{- \frac{1}{5}} , \textcolor{g r e e n}{\frac{11}{5}}\right)$, or $\left(\frac{1}{5} , \frac{11}{5}\right)$