# What is the the vertex of y=x^2+15x-30?

Jun 3, 2016

I found: $\left(- 7.5 , - 86.25\right)$

#### Explanation:

There are two ways to find the coordinates of the vertex:

1) knowing that the $x$ coordinate is given as:
${x}_{v} = - \frac{b}{2 a}$ and considering your function in the general form:
$y = a {x}^{2} + b x + c$;

$a = 1$
$b = 15$
$c = - 30$
so:
${x}_{v} = - \frac{15}{2} = - 7.5$
by substituting this value into your original equation you get the corresponding ${y}_{v}$ value:
${y}_{v} = {\left(- \frac{15}{2}\right)}^{2} + 15 \left(- \frac{15}{2}\right) - 30 = \frac{225 - 450 - 120}{4} = - \frac{345}{4} = - 86.25$

2) usig the derivative (but I am not sure you know this procedure):
$y ' = 2 x + 15$
$y ' = 0$
$2 x + 15 = 0$
$x = - \frac{15}{2}$ as before!