# What is the vertex form of y=8x^2 + 19x + 12 ?

Oct 16, 2016

$y = 8 {\left(x - - \frac{19}{16}\right)}^{2} + \frac{23}{32}$

#### Explanation:

The equation is in the standard form, $y = a {x}^{2} + b x + c$ where $a = 8 , b = 19 , \mathmr{and} c = 12$

The x coordinate, h, of the vertex is:

$h = - \frac{b}{2 a}$

$h = - \frac{19}{2 \left(8\right)} = - \frac{19}{16}$

To find the y coordinate, k, of the vertex, evaluate the function at the value of h:

$k = 8 \left(- \frac{19}{16}\right) \left(- \frac{19}{16}\right) + 19 \left(- \frac{19}{16}\right) + 12$

$k = \left(\frac{1}{2}\right) \left(- 19\right) \left(- \frac{19}{16}\right) + 19 \left(- \frac{19}{16}\right) + 12$

$k = - {19}^{2} / 32 + 12$

$k = - \frac{361}{32} + 12$

$k = - \frac{361}{32} + \frac{384}{32}$

$k = \frac{23}{32}$

The vertex form of the equation of a parabola is:

$y = a {\left(x - h\right)}^{2} + k$

Substitute our values into that form:

$y = 8 {\left(x - - \frac{19}{16}\right)}^{2} + \frac{23}{32}$