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]}