# What is the vertex of y=2(x -1)^2 -4x ?

##### 1 Answer
May 16, 2018

Vertex at $\left(2 , - 6\right)$

#### Explanation:

Method 1: convert the equation into vertex form
Note: vertex form is $y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ for a parabola with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

$y = 2 {\left(x - 1\right)}^{2} - 4 x \textcolor{w h i t e}{\text{xxxxxxxx}}$...as given

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

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

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

completing the square
$y = 2 \left({x}^{2} - 4 x + 4\right) - 6$
we added $3$ to the previous $1$ but this is multiplied by $2$ so we need to subtract $2 \times 3 = 6$ to keep this equivalent.

y=color(green)2(x-color(red)2)^2+color(blue)(""(-6))
which is the vertex form with vertex at $\left(\textcolor{red}{2} , \textcolor{b l u e}{- 6}\right)$

Method 2: Note that the slope (derivative) of the parabola at the vertex is zero

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

expanding:
$y = 2 {x}^{2} - 8 x + 2$

at the vertex
$y ' = 4 x - 8 = 0$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow \textcolor{red}{x = 2}$ at the vertex

Substituting $2$ for $x$ back in the original equation gives
$\textcolor{b l u e}{y} = 2 {\left(2 - 1\right)}^{2} - 4 \cdot 2 = 2 - 8 \textcolor{b l u e}{= - 6}$

Again, giving the vertex at
$\textcolor{w h i t e}{\text{XXX}} \left(\textcolor{red}{2} , \textcolor{b l u e}{- 6}\right)$

Method 3: Use a graphing calculator/software package