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

Feb 15, 2016

vertex: $\left(x , y\right) = \left(3 , - 9\right)$

#### Explanation:

First simplify the given equation:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{\mathmr{and} a n \ge}{- 3 {x}^{2} - 2 x - 1} + \textcolor{b r o w n}{{\left(2 x - 1\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{\mathmr{and} a n \ge}{- 3 {x}^{2} - 2 x - 1} + \textcolor{b r o w n}{4 {x}^{2} - 4 x + 1}$

$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 6 x$

One of the easiest ways of finding the vertex is to convert the equation into "vertex form":
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$
by "completing the square"

(Note that in this case we can ignore $\textcolor{g r e e n}{m}$ or write it with its implied value of $\textcolor{g r e e n}{1}$).

$\textcolor{w h i t e}{\text{XXXXXX}}$Remember ${\left(x + k\right)}^{2} = {x}^{2} + 2 k x + {k}^{2}$
$\textcolor{w h i t e}{\text{XXXXXX}}$So in this case $k = - 3$
$\textcolor{w h i t e}{\text{XXXXXX}}$ and we will need to add ${\left(- 3\right)}^{2}$ to complete the square

$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 6 x \textcolor{p u r p \le}{+ 9 - 9}$

$\textcolor{w h i t e}{\text{XXX")y=(x-color(red)(3))^2+color(blue)("("-9")}}$

which is in vertex form with the vertex at $\left(\textcolor{red}{3} , \textcolor{b l u e}{\text{("-9")}}\right)$

Here is a graph of the original equation to help verify our result:
graph{-3x^2-2x-1+(2x-1)^2 [-7.46, 12.54, -10.88, -0.88]}