What is the vertex of #f(x)=x^2#?

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Answer 1 · 2014-09-16T00:01:08

The vertex is #(0,0)#

Convert this equation to vertex form:

#f(x)=x^2#

#f(x)=(x+0)^2#

#f(x)=(x+0)^2+0#

#f(x)=(x-h)^2+k -> vertex -> (h,k)#

Or use the expressions: #(-b/(2a),f(-b/(2a)))#

#f(x)=ax^2+bx+c#

#f(x)=x^2#

#f(x)=x^2+0x+0#

#a=1, b=0, c=0#

#x=-b/(2a)=-0/(2(0))=0#

#y=f(-b/(2a))=f(0)=(0)^2=0#

#vertex -> (0,0)#