# How do you find the vertex and intercepts for y=(x+3)(x+5)?

Jan 26, 2016

y-intercept: $15$
x-intercepts: $\left(- 3\right)$ and $\left(- 5\right)$
vertex: $\left(- 4 , - 1\right)$

#### Explanation:

Starting with the intercepts:

The y-intercept is the value of $y$ when $x = 0$.
$y = \left(0 + 3\right) \left(0 + 5\right) = 15$

The x-intercept(s) is/are the values of $x$ when $y = 0$
$0 = \left(x + 3\right) \left(x + 5\right)$
$\Rightarrow x = - 3$ or $x = - 5$

Determining the vertex.
Method 1:

For a parabola with a vertical axis,
the x coordinate of the vertex will be half way between the two x-intercepts.
i.e. the x-coordinate of the vertex will be $\frac{\left(- 3\right) + \left(- 5\right)}{2} = - 4$
By substituting $\left(- 4\right)$ for $x$ in the given equation
$\textcolor{w h i t e}{\text{XXX}} y = \left(- 4 + 3\right) \left(- 4 + 5\right) = - 1$
So the vertex will be at $\left(- 4 , - 1\right)$

Method 2:

Convert the given equation into vertex form: $y = m {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{g r e e n}{b}$
for a parabola with vertex at $\left(\textcolor{red}{a} , \textcolor{g r e e n}{b}\right)$
$\textcolor{w h i t e}{\text{XXX}} y = \left(x + 3\right) \left(x + 5\right)$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = {x}^{2} + 8 x + 15$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = {x}^{2} + 8 x \textcolor{b l u e}{+ 16} + 15 \textcolor{b l u e}{- 16}$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = {\left(x + 4\right)}^{2} + \left(- 1\right)$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = {\left(x - \left(\textcolor{red}{- 4}\right)\right)}^{2} + \left(\textcolor{g r e e n}{- 1}\right)$