# What is the vertex of y=x^2+12x+26?

Apr 12, 2017

The vertex is at $\left(- 6 , - 10\right)$

#### Explanation:

You can find the vertex (turning point) by first finding the line that is the axis of symmetry.

$x = \frac{- b}{2 a} = \frac{- 12}{2 \left(1\right)} = - 6 \text{ } \leftarrow$ This is the $x$-value of the vertex.

Now find $y$.

$y = {x}^{2} + 12 x + 26$

$y = {\left(- 6\right)}^{2} + 12 \left(- 6\right) + 26$
$y = 36 - 72 + 26$
$y = - 10 \text{ } \leftarrow$ This is the $y$-value of the vertex.

The vertex is at $\left(- 6 , - 10\right)$

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You can also find the vertex by completing the square to get the equation in vertex form: $y = a {\left(x + b\right)}^{2} + c$

$y = {x}^{2} + 12 x + 26$
$y = {x}^{2} + 12 x \textcolor{red}{+ {6}^{2}} \textcolor{red}{- {6}^{2}} + 26 \text{ } \textcolor{red}{{\left(\frac{b}{2}\right)}^{2} = {\left(\frac{12}{2}\right)}^{2}}$

$y = {\left(x + 6\right)}^{2} - 10$

Vertex is at $\left(- b , c\right) \text{ } \rightarrow \left(- 6 , - 10\right)$