# What is the vertex form of  y= (3x-8)(-6x-2)+2x^2+3x?

Jun 2, 2018

Vertex $\left(\frac{45}{32} , \frac{3049}{64}\right)$

#### Explanation:

$y = \left(3 x - 8\right) \left(- 6 x - 2\right) + 2 {x}^{2} + 3 x$

$y = - 18 {x}^{2} - 6 x + 48 x + 16 + 2 {x}^{2} + 3 x$

$y = - 16 {x}^{2} + 45 x + 16$

The vertex is found by $x = - \frac{b}{2 a}$

By looking at the general formula of a quadratic function,
$y = a {x}^{2} + b x + c$ , we can see that $b = 45$ and $a = - 16$

$x = - \frac{b}{2 a} = - \frac{45}{2 \times - 16} = \frac{45}{32}$

Sub $x = \frac{45}{32}$ into $y = - 16 {x}^{2} + 45 x + 16$

$y = - 16 {\left(\frac{45}{32}\right)}^{2} + 45 \left(\frac{45}{32}\right) + 16$
$y = \frac{3049}{64}$

Vertex $\left(\frac{45}{32} , \frac{3049}{64}\right)$

You can from the graph below that the vertex is quite high up.

graph{(3x-8)(-6x-2)+2x^2+3x [-10, 10, -5, 5]}