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

Question

3665 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-16T07:01:08

vertex $=(-3/2, 21/4)$

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

Factor out the $3$ from the first two terms.
$y=3(x^2+3x)+12$

To make the bracketed part a trinomial, substitute $c=(b/2)^2$ and subtract $c$ .
$y=3(x^2+3x+(3/2)^2-(3/2)^2)+12$

$y=3(x^2+3x+9/4-9/4)+12$

Bring $-9/4$ out of the brackets by multiplying it by the vertical stretch factor, $3$ .
$y=3(x^2+3x+9/4)+12-(9/4*3)$

$y=3(x+3/2)^2+12-(27/4)$

$y=3(x+3/2)^2+21/4$

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

$y=a(x-h)^2+k$

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

So in this case, the vertex is $(-3/2,21/4)$ .

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