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

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Answer 1 · 2015-12-11T17:28:44

The vertex is the point $(\frac{8}{3}, - \frac{106}{3})$ $(8/3, -106/3)$

Expand the expression:

$3 (x + 1) (x - 5) - 4 x + 1 = 3 (x^{2} - 4 x - 5) - 4 x + 1$ $3(x+1)(x-5)-4x+1=3(x^2-4x-5)-4x+1$

$3 x^{2} - 12 x - 15 - 4 x + 1 = 3 x^{2} - 16 x - 14$ $3x^2 -12x-15-4x+1=3x^2-16x-14$

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

$- \frac{b}{2 a} = - \frac{- 16}{2 \cdot 3} = \frac{16}{6} = \frac{8}{3}$ $-b/(2a)=-(-16)/(2*3) = 16/6 = 8/3$

So, the $y$ $y$ coordinate of the vertex is simply $f (\frac{8}{3})$ $f(8/3)$ , which is

$3 \cdot {(\frac{8}{3})}^{2} - 16 \cdot \frac{8}{3} - 14 = - \frac{106}{3}$ $3*(8/3)^2-16*8/3-14=-106/3$

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