# What is the vertex of y=3(x -2)^2 -4x ?

Jan 5, 2016

$\left(\frac{8}{3} , - \frac{148}{9}\right)$

#### Explanation:

You need to expand the expression and simplify it before converting it from standard form to vertex form by completing the square. Once it is in vertex form you can deduce the vertex.
$y = 3 {\left(x - 2\right)}^{2} - 4 x$
$y = 3 \left({x}^{2} - 4 x + 4\right) - 4 x$
$y = 3 {x}^{2} - 12 x + 12 - 4 x$
$y = 3 {x}^{2} - 16 x + 12$
$y = 3 \left({x}^{2} - \frac{16}{3} x\right) + 12$
Now complete the square
$y = 3 {\left(x - \frac{8}{3}\right)}^{2} - \frac{256}{9} + 12$
$y = 3 {\left(x - \frac{8}{3}\right)}^{2} - \frac{256 + 108}{9}$
$y = 3 {\left(x - \frac{8}{3}\right)}^{2} - \frac{148}{9}$
The vertex occurs the bracketed term is zero and is therefore $\left(\frac{8}{3} , - \frac{148}{9}\right)$