# What is the vertex of  y = x^2+10x+21 ?

Aug 27, 2016

In the standard form $y = a {x}^{2} + b x + c$ the $x$-coordinate of the vertex is $- \frac{b}{2 a}$

In this situation $a = 1$, $b = 10$ and $c = 21$, so the $x$-coordinate of the vertex is:
$- \frac{b}{2 a} = - \frac{10}{2 \times 1} = - 5$

Then we simply substitute $x = - 5$ into the original equation to find the $y$-coordinate of the vertex.

$y = {\left(- 5\right)}^{2} + 10 \left(- 5\right) + 21 = - 4$

So the coordinates of the vertex are:
$\left(- 5 , - 4\right)$