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

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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)#.