# What is the vertex form of y=(x+5)(x+3)?

Nov 27, 2015

$y = {\left(x + 4\right)}^{2} - 1$

#### Explanation:

Step 1: Foil (multiply) the right hand side of the equation

$y = \left(x + 5\right) \left(x + 3\right)$
$\Rightarrow y = {x}^{2} + 5 x + 3 x + 15$
$\implies \textcolor{red}{y = {x}^{2} + 8 x + 15}$

Step 2: We can write the vertex form by several methods
Reminder: vertex form is $\textcolor{b l u e}{y = a {\left(x - h\right)}^{2} + k}$

$\implies$ Method 1: By completing square
$\implies \textcolor{red}{y = {x}^{2} + 8 x + 15}$ $\implies$ re-write

We make a perfect trinomial in the form of
$\implies {a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$
$\implies {a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$

$y = \left({x}^{2} + 8 x + \textcolor{g r e e n}{16}\right) \textcolor{g r e e n}{- 16} + 15$
$16 = {\left[\frac{1}{2} \left(8\right)\right]}^{2}$

$y = {\left(x + 4\right)}^{2} - 1$ Vertex form completed

$\implies$ Method 2: Using formula
$h = {x}_{v e r t e x} = - \frac{b}{2 a}$
$k = {y}_{v e r t e x} = y \left(- \frac{b}{a b}\right)$

From this$\implies \textcolor{red}{y = {x}^{2} + 8 x + 15}$
We have $a = 1$ ; $b = 8$ , $c = 15$

$h = {x}_{v e r t e x} = - \frac{8}{2 \cdot 2} = \textcolor{red}{-} 4$

$k = {y}_{v e r t e x} = y \left(- 4\right) = {\left(- 4\right)}^{2} + 8 \left(- 4\right) + 15$
$y \left(- 4\right) = 16 - 32 + 15 = \textcolor{red}{- 1}$
vertex form is $\textcolor{b l u e}{y = 1 {\left(x - \left(- 4\right)\right)}^{2} + \left(- 1\right)}$
simplify ${\textcolor{red}{y = 1 \left(x + 4\right)}}^{\textcolor{red}{2}} - 1$