What is the the vertex of #y = x^2+4x+20 #?

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Answer 1 · 2017-02-26T13:09:23

Vertex#->(x,y)=(-2,16)#

The format of the question is already as:

#y=ax^2+bx+c" "->" "y=a(x^2+b/ax)+c#

as #a=1#

#x_("vertex")=(-1/2)xxb/a" "->" "=(-1/2)xx4 = -2#

So by substitution

#y_("vertex")=(-2)^2+4(-2)+20 = 16#

Vertex#->(x,y)=(-2,16)#

Tony B