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

##### 1 Answer
Dec 29, 2015

$\left(\frac{7}{3} , - \frac{10}{3}\right)$

#### Explanation:

First expand and simplify to get get one term for each power of x.
$y = 4 {x}^{2} - 12 x + 9 - {x}^{2} - 2 x + 4$
$y = 3 {x}^{2} - 14 x + 13$
$y = 3 \left({x}^{2} - \frac{14 x}{3} + \frac{13}{3}\right)$
Use completing the square to put the expression into vertex form
y = 3(x - 7/3)^2 -49/9 + 13/3) = 3((x-7/3)^2 -10/9)
$y = 3 {\left(x - \frac{7}{3}\right)}^{2} - \frac{10}{3}$
Then the vertex occurs where the bracketed term is zero.
Vertex is $\left(\frac{7}{3} , - \frac{10}{3}\right)$