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

Dec 11, 2015

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

Explanation:

Expand the expression:

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

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

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

$- \frac{b}{2 a} = - \frac{- 16}{2 \cdot 3} = \frac{16}{6} = \frac{8}{3}$

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

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