What is the vertex of #y=x^2+12x+26#?

1 Answer
Apr 12, 2017

The vertex is at #(-6, -10)#

Explanation:

You can find the vertex (turning point) by first finding the line that is the axis of symmetry.

#x = (-b)/(2a) = (-12)/(2(1)) = -6" "larr# This is the #x#-value of the vertex.

Now find #y#.

#y = x^2 +12x+26#

#y = (-6)^2 +12(-6) +26#
#y= 36-72+26#
#y=-10" "larr# This is the #y#-value of the vertex.

The vertex is at #(-6, -10)#

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You can also find the vertex by completing the square to get the equation in vertex form: #y = a(x+b)^2 +c#

#y = x^2 +12x+26#
#y= x^2 +12x color(red)(+6^2) color(red)(-6^2) +26" " color(red)((b/2)^2 =(12/2)^2)#

#y = (x+6)^2 -10#

Vertex is at #(-b, c)" "rarr (-6, -10)#