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

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

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