# What is the vertex of y= 12x^2 - 18x - 6 ?

Aug 16, 2016

$P = \left(\frac{3}{4} , - \frac{51}{4}\right)$

#### Explanation:

$P = \left(h , k\right) \text{ Vertex coordinates}$

$y = a {x}^{2} + b x + c$
$a = 12 \text{ ; "b=-18" ; } c = - 6$

$y = 12 {x}^{2} - 18 x - 6$
$h = - \frac{b}{2 a}$

$h = \frac{18}{2 \cdot 12} = \frac{18}{24} = \frac{3}{4}$

$k = 12 \cdot {\left(\frac{3}{4}\right)}^{2} - 18 \cdot \frac{3}{4} - 6$

$k = 12 \cdot \frac{9}{16} - \frac{54}{4} - 6$

$k = \frac{27}{4} - \frac{54}{4} - \frac{24}{4} = \frac{27 - 78}{4} = - \frac{51}{4}$

$P = \left(\frac{3}{4} , - \frac{51}{4}\right)$