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

Apr 21, 2018

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

#### Explanation:

Because the equation is in the form $y = a \left(x - p\right) \left(x - q\right)$, we know the $x$-intercepts are $p$ and $q$
Therefore, the $x$-intercepts are at $- 5$ and $- 3$
Note the negative signs and the fact that in the case $a = 1$ so it is committed.
This makes sense because the $x$-intercepts are whenever $y = 0$. This occurs when either bracket is $0$.

The $x$-value of the vertex is the midvalue of the intercepts.
let ${V}_{x}$ be the $x$-value of the vertex
${V}_{x} = \frac{\left(- 5\right) + \left(- 3\right)}{2}$
${V}_{x} = - 4$
Subbing in $x = - 4$ to solve for the $y$-value of the vertex:
Let this be ${V}_{y}$:
${V}_{y} = \left(\left(- 4\right) + 5\right) \left(\left(- 4\right) + 3\right)$
${V}_{y} = \left(1\right) \left(- 1\right)$
${V}_{y} = - 1$

Therefore, the vertex is $\left(- 4 , - 1\right)$

To find the $y$-intercept we substitute $x = 0$ into the equation.
$y = \left(5\right) \left(3\right)$
$y = 15$

In conclusion, the $x$-intercepts are $- 5$ and $- 3$, the $y$-intercept is $15$ and the vertex is $\left(- 4 , - 1\right)$

Included below are graphs of the equation:
graph{(x+5)(x+3) [-7.757, 3.6, -2.454, 3.224]}
graph{(x+5)(x+3) [-49.16, 45, -23.16, 23.9]}