# What is the the vertex of y = (x-3)(x+14) +42-10x ?

Nov 27, 2017

The vertex is $\left(- \frac{1}{2} , - \frac{1}{4}\right)$.

#### Explanation:

First expand and collect like terms:
$y = \left({x}^{2} + 14 x - 3 x - 42\right) + 42 - 10 x$
$y = {x}^{2} + 11 x - 42 + 42 - 10 x$
$y = {x}^{2} + x$

Now I'll factor $y$:
$y = x \left(x + 1\right)$

The x-intercepts are $x = 0$ and $x = - 1$. The x-coordinate of the vertex is the average of the x-intercepts, so $x = - \frac{1}{2}$. The y-coordinate is what you get when you substitute this value into the equation: $y = \left(- \frac{1}{2}\right) \left(- \frac{1}{2} + 1\right) = \left(- \frac{1}{2}\right) \left(\frac{1}{2}\right) = - \frac{1}{4}$.

So the vertex is $\left(- \frac{1}{2} , - \frac{1}{4}\right)$.