What is the the vertex of #y = x^2 - x - 6 #?

1 Answer
Ken C.
Mar 15, 2016

#(1/2, -13/2)#

Explanation:

Vertex of a parabola in the form #ax^2+bx+c# is given by:
#x=-b/(2a)#
Note this is only gives the #x#-coordinate; we will have to evaluate this value to get the #y#-coordinate.

Our parabola #x^2-x-6# has #a=1#, #b=-1#, and #c=-6#. Using the vertex formula above, we see:
#x=-(-1)/(2(1))=1/2#

Evaluating #y# at this value:
#y=(1/2)^2-(1/2)-6=1/4-1/2-6=-13/2#

Therefore our vertex occurs at the point #(1/2, -13/2)#.

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