What is the vertex of y=x^2+12x+18?

Dec 6, 2015

Complete the square to reformulate in vertex form to find that the vertex is at $\left(- 6 , - 18\right)$

Explanation:

Complete the square to reformulate in vertex form:

$y = {x}^{2} + 12 x + 18 = {x}^{2} + 12 x + 36 - 18$

$= {\left(x + 6\right)}^{2} - 18$

So in vertex form we have:

$y = {\left(x + 6\right)}^{2} - 18$

or more fussily:

$y = 1 {\left(x - \left(- 6\right)\right)}^{2} + \left(- 18\right)$

which is in exactly the form:

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

with $a = 1$, $h = - 6$ and $k = - 18$

the equation of a parabola with vertex $\left(- 6 , - 18\right)$ and multiplier $1$

graph{ x^2+12x+18 [-44.92, 35.08, -22.28, 17.72]}