How do you find the vertex of  y=x^2+10x+21?

Mar 26, 2018

$\text{vertex} = \left(- 5 , - 4\right)$

Explanation:

$x = - \frac{b}{2 a}$

$x = - \frac{10}{2 \left(1\right)}$

$x = - 5$

Sub $- 5$ into the equation

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

$y = - 4$

The formula $- \frac{b}{2 a}$ is used to find the axis of symmetry which is
always the $x$ value of the vertex. Once you find the $x$ value of the vertex, you simply substitute that value into the quadratic equation and find the $y$ value, which in this case, is the vertex.

Mar 26, 2018

(-5,-4)

Explanation:

You have to use the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2} a$
which becomes
$x = - \frac{b}{2 a} \pm \left(\frac{\sqrt{{b}^{2} - 4 a c}}{2 a}\right)$
We know that $- \frac{b}{2 a}$ is constant and that the other part is plussing and minusing from it

So it is the vertex and as $a = 1 b = 10 c = 21$ i.e. just the coefficents of all the terms in sequence.

The vertex must be $- \frac{10}{2 \cdot 1}$ so the x co-ordinate of the vertex is $- 5$

Plug in $f \left(- 5\right)$ and you get the y co-ordinate
$f \left(- 5\right) = {\left(- 5\right)}^{2} + 10 \left(- 5\right) + 21$ becomes $f \left(- 5\right) = 25 - 50 + 21$
so $f \left(- 5\right) = - 4$

so the co-ordiantes of the vertex are (-5,-4)