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

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Answer 1 · 2015-12-06T11:01:05

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

Complete the square to reformulate in vertex form:

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

$= (x+6)^2-18$

So in vertex form we have:

$y = (x+6)^2-18$

or more fussily:

$y = 1(x-(-6))^2+(-18)$

which is in exactly the form:

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

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

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

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

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