What is the the vertex of #y = 3(x+1)(x-5) -4x+1#?

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Answer 1 · 2015-12-11T17:28:44

The vertex is the point #(8/3, -106/3)#

Expand the expression:

#3(x+1)(x-5)-4x+1=3(x^2-4x-5)-4x+1#

#3x^2 -12x-15-4x+1=3x^2-16x-14#

Once your parabola is in the form #ax^2+bx+c#, the vertex has #x# coordinate #-b/(2a)#, so we have

#-b/(2a)=-(-16)/(2*3) = 16/6 = 8/3#

So, the #y# coordinate of the vertex is simply #f(8/3)#, which is

#3*(8/3)^2-16*8/3-14=-106/3#