How do you find the vertex of # f(x)= -3x^2-6x-7#?

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Answer 1 · 2015-06-14T14:56:19

Complete the square to find:

#f(x) = -3x^2-6x-7 = -3(x+1)^2-4#

has vertex #(-1, -4)#

#f(x) = -3x^2-6x-7#

#= -3x^2-6x-3-4#

#= -3(x^2+2x+1)-4#

#= -3(x+1)^2 - 4#

Now #(x+1)^2 >= 0# has it's minimum value #0# when #x = -1#

If #x=-1# then

#f(x) = (x+1)^2-4 = 0^2-4 = 0-4 = -4#

So the vertex is at #(-1, -4)#

In general you can complete the square of any quadratic in #x# as follows:

#ax^2+bx+c = a(x+b/(2a))^2 + (c-b^2/(4a))#