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

Question

3428 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-27T07:43:45

In the standard form $y=ax^2+bx+c$ the $x$ -coordinate of the vertex is $-b/(2a)$

In this situation $a=1$ , $b=10$ and $c=21$ , so the $x$ -coordinate of the vertex is:
$-b/(2a)=-10/(2xx1) = -5$

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

$y=(-5)^2+10(-5)+21=-4$

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

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