# What is the the vertex of y = 3x^2 + 9x+12 ?

Dec 16, 2015

vertex$= \left(- \frac{3}{2} , \frac{21}{4}\right)$

#### Explanation:

$y = 3 {x}^{2} + 9 x + 12$

Factor out the $3$ from the first two terms.
$y = 3 \left({x}^{2} + 3 x\right) + 12$

To make the bracketed part a trinomial, substitute $c = {\left(\frac{b}{2}\right)}^{2}$ and subtract $c$.
$y = 3 \left({x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2}\right) + 12$

$y = 3 \left({x}^{2} + 3 x + \frac{9}{4} - \frac{9}{4}\right) + 12$

Bring $- \frac{9}{4}$ out of the brackets by multiplying it by the vertical stretch factor, $3$.
$y = 3 \left({x}^{2} + 3 x + \frac{9}{4}\right) + 12 - \left(\frac{9}{4} \cdot 3\right)$

$y = 3 {\left(x + \frac{3}{2}\right)}^{2} + 12 - \left(\frac{27}{4}\right)$

$y = 3 {\left(x + \frac{3}{2}\right)}^{2} + \frac{21}{4}$

Recall that the general equation of a quadratic equation written in vertex form is:

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

where:
$h =$x-coordinate of the vertex
$k =$y-coordinate of the vertex

So in this case, the vertex is $\left(- \frac{3}{2} , \frac{21}{4}\right)$.