What is the vertex of #f(x)=x^2+4x+3#?

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Answer 1 · 2014-09-26T22:13:35

The vertex of #f# is #(-2,-1)#.

Let us look at some details.

The vertex of #f(x)=ax^2+bx+c# can be found by

#(-b/{2a},f(-b/{2a}))#.

For #f(x)=x^2+4x+3#,

#-b/{2a}=-4/{2(1)}=-2#

So, the vertex is

#(-2,f(-2))=(-2, -1)#