What is the vertex of  y= -(x-6)^2-4x^2-2x-2?

Jan 31, 2016

$\left(1 , - 33\right)$

Explanation:

We start out with $y = - {\left(x - 6\right)}^{2} - 4 {x}^{2} - 2 x - 2$.

The first thing we want to do is combine like terms, but there aren't any... yet . We need to expand ${\left(x - 6\right)}^{2}$, which we do by rewriting it as $\left(x - 6\right) \cdot \left(x - 6\right)$ and multiply through to create ${x}^{2} - 12 x + 36$.

We plug that into where ${\left(x - 6\right)}^{2}$ used to be, and we see this: $y = - \left({x}^{2} - 12 x + 36\right) - 4 {x}^{2} - 2 x - 2$. Distribute the $-$ into the $\left({x}^{2} - 12 x + 36\right)$, changing it to $- {x}^{2} + 12 x - 36 - 4 {x}^{2} - 2 x - 2$.

NOW we can combine like terms.

$- {x}^{2} - 4 {x}^{2}$ becomes $- 5 {x}^{2}$
$12 x - 2 x$ becomes $10 x$
$- 36 - 2$ becomes $- 38$.
Put it all together and we have $- 5 {x}^{2} + 10 x - 38$. This is not factorable, so we will solve by completing the square. To do that, the coefficient of ${x}^{2}$ must be 1, so we factor out $- 5$. The equation now becomes $- 5 \left({x}^{2} - 2 x + \frac{38}{5}\right)$. To complete the square, we have to find the value that will make ${x}^{2} - 2 x$ factorable. We do that by taking the middle term, $- 2 x$, dividing it by two ($- \frac{2}{2} = - 1$), and squaring the answer you got ($- {1}^{2} = 1$).

We then rewrite the equation as $y = - 5 \left({x}^{2} - 2 x + 1 + \frac{38}{5}\right)$.

But wait!

We can't just stick a random number in the equation! What we do to one side we must do to the other. Now, I don't know about you, but I don't really want to change $y$. I like having it isolated, but we still have to deal with adding a $1$ to only one side of the equation.

But you know, we could just subtract an $- 1$ , which would cancel out the $1$ so it wouldn't effect the equation. Let's do that!

Now the equation reads: $y = - 5 \left({x}^{2} - 2 x \textcolor{red}{+ 1 - 1} + \frac{38}{5}\right)$. We can simplify ${x}^{2} - 2 x + 1$ to ${\left(x - 1\right)}^{2}$ and simplify $- 1 + \frac{35}{5}$ to just $\frac{33}{5}$. We can simplify the equation to $- 5 \left({\left(x - 1\right)}^{2} + \frac{33}{5}\right)$. The last step is to multiply the $- 5 \cdot \frac{33}{5}$, and because the $5$s divide out (like so: $- \cancel{5} \cdot \left(\frac{33}{\cancel{5}}\right)$), all that is left is -33.

Putting it all together, we have $y = - 5 {\left(x - 1\right)}^{2} - 33$.

This is actually in vertex form. All we have to do to find the vertex is take the $y = - 5 {\left(x \textcolor{red}{- 1}\right)}^{2} \textcolor{b l u e}{- 33}$ and put it into coordinate-pair form: $\left(\textcolor{red}{1} , \textcolor{b l u e}{- 33}\right)$.

NOTE the $\textcolor{red}{x}$ value changed signs once I took it out of the equation. Remember this as it happens every time.