What is the vertex of # y = x^2+10x+21 #?

1 Answer
Aug 27, 2016

In the standard form #y=ax^2+bx+c# the #x#-coordinate of the vertex is #-b/(2a)#

In this situation #a=1#, #b=10# and #c=21#, so the #x#-coordinate of the vertex is:
#-b/(2a)=-10/(2xx1) = -5#

Then we simply substitute #x=-5# into the original equation to find the #y#-coordinate of the vertex.

#y=(-5)^2+10(-5)+21=-4#

So the coordinates of the vertex are:
#(-5, -4)#