# What is the the vertex of y = (x+6)(x+4) -x+12?

Jun 23, 2016

${y}_{\min} = \frac{63}{4}$ at $x = - \frac{9}{2}$

#### Explanation:

$y = \left(x + 6\right) \left(x + 4\right) - x + 12$
$y = {x}^{2} + 10 x + 24 - x + 12$
$y = {x}^{2} + 9 x + 36$
$y = {\left(x + \frac{9}{2}\right)}^{2} - \frac{81}{4} + 36$
$y = {\left(x + \frac{9}{2}\right)}^{2} + \frac{63}{4}$

${y}_{\min} = \frac{63}{4}$ at $x = - \frac{9}{2}$

Jun 23, 2016

The vertex is (-9/2;63/4)

#### Explanation:

let's rewrite the equation in the equivalent form:

$y = {x}^{2} + 4 x + 6 x + 24 - x + 12$

$y = {x}^{2} + 9 x + 36$

Then let's find the vertex coordinates by the following:

${x}_{V} = - \frac{b}{2 a}$

where a=1; b=9

so

${x}_{V} = - \frac{9}{2}$

and

${y}_{V} = f \left(- \frac{9}{2}\right)$

that's

$y = {\left(- \frac{9}{2}\right)}^{2} + 9 \left(- \frac{9}{2}\right) + 36$

$y = \frac{81}{4} - \frac{81}{2} + 36$

$y = \frac{81 - 162 + 144}{4}$

$y = \frac{63}{4}$