# What is the vertex form of y= x^2 + 5x + 6 ?

Jul 24, 2017

Vertex form is ${\left(x + \frac{5}{2}\right)}^{2} - \frac{1}{4}$.

#### Explanation:

Vertex from Standard Form

$y = {x}^{2} + 5 x + 6$ is the standard form for a quadratic equation, $a {x}^{2} + b x + 6$, where $a = 1$, $b = 5$, and $c = 6$.

The vertex form is $a {\left(x - h\right)}^{2} + k$, and the vertex is $\left(h , k\right)$.

In the standard form, $h = \frac{- b}{2 a}$, and $k = f \left(h\right)$.

Solve for $h$ and $k$.

$h = \frac{- 5}{2 \cdot 1}$

$h = - \frac{5}{2}$

Now plug in $- \frac{5}{2}$ for $x$ in the standard form to find $k$.

$f \left(h\right) = k = {\left(- \frac{5}{2}\right)}^{2} + \left(5 \times - \frac{5}{2}\right) + 6$

Solve.

$f \left(h\right) = k = \frac{25}{4} - \frac{25}{2} + 6$

The LCD is 4.

Multiply each fraction by an equivalent fraction to make all of the denominators $4$. Reminder: $6 = \frac{6}{1}$

$f \left(h\right) = k = \frac{25}{4} - \left(\frac{25}{2} \times \frac{2}{2}\right) + \left(\frac{6}{1} \times \frac{4}{4}\right)$

Simplify.

$f \left(h\right) = k = \frac{25}{4} - \frac{50}{4} + \frac{24}{4}$

Simplify.

$f \left(h\right) = k = - \frac{1}{4}$

Vertex $\left(- \frac{5}{2} , - \frac{1}{2}\right)$

Vertex form: $a {\left(x - h\right)}^{2} + k$

$1 {\left(x + \frac{5}{2}\right)}^{2} - \frac{1}{4}$

${\left(x + \frac{5}{2}\right)}^{2} - \frac{1}{4}$