What is the vertex form of  y= (9x-6)(3x+12)-7x^2+5x?

1 Answer
Oct 9, 2017

$y = 20 {\left(x - \left(- \frac{19}{8}\right)\right)}^{2} - \frac{2957}{16}$

Explanation:

Given: $y = \left(9 x - 6\right) \left(3 x + 12\right) - 7 {x}^{2} + 5 x$

Perform the multiplication:

$y = 27 {x}^{2} + 90 x - 72 - 7 {x}^{2} + 5 x$

Combine like terms:

$y = 20 {x}^{2} + 95 x - 72$

This is in the standard Cartesian form:

$y = a {x}^{2} + b x + c$

where $a = 20 , b = 95 , \mathmr{and} c = - 72$

The general vertex form for a parabola of this type is:

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

We know that $a = 20$:

$y = 20 {\left(x - h\right)}^{2} + k$

We know that $h = - \frac{b}{2 a}$

$h = - \frac{95}{2 \left(20\right)}$

$h = - \frac{19}{8}$

$y = 20 {\left(x - \left(- \frac{19}{8}\right)\right)}^{2} + k$

We know that:

$k = 20 {\left(- \frac{19}{8}\right)}^{2} + 95 \left(- \frac{19}{8}\right) - 72$

$k = - \frac{2957}{16}$

$y = 20 {\left(x - \left(- \frac{19}{8}\right)\right)}^{2} - \frac{2957}{16}$