# How do you graph y=(x-1)/(x+5) using asymptotes, intercepts, end behavior?

Oct 29, 2016

Find the VA, HA, and intercepts, then graph as shown below.

#### Explanation:

$y = \frac{x - 1}{x + 5}$

The vertical asymptote (VA) is found by setting the denominator equal to zero.

$x + 5 = 0$

$x = - 5$ is the VA

The horizontal asymptote (HA) is found by comparing the degree of the numerator to the degree of the denominator.

If the degree of the denominator is greater than the degree of the numerator, the HA is $y = 0$.

If the degree of the denominator is equal to the degree of the numerator, the HA is the leading coefficient of the numerator divided by the leading coefficient of the denominator.

In this example, the degree of both the numerator and denominator is $\textcolor{red}{1}$ and the leading coefficient of both is also $\textcolor{b l u e}{1}$.

$y = \frac{\textcolor{b l u e}{1} {x}^{\textcolor{red}{1}} - 1}{\textcolor{b l u e}{1} {x}^{\textcolor{red}{1}} + 5}$

The HA is $y = \frac{\textcolor{b l u e}{1}}{\textcolor{b l u e}{1}} = 1$

If the degree of the denominator is one less than the degree of the numerator, there is an oblique asymptote(OA). This function does not have an OA.

To find the $y$ intercept, set $x = 0$.

$y = \frac{0 - 1}{0 + 5} = - \frac{1}{5}$

The $y$ intercept is $\left(0 , - \frac{1}{5}\right)$

To find the $x$ intercept, find the value that makes $y = 0$. This value is found by setting the numerator equal to zero.

$x - 1 = 0$

$x = 1$

The $x$ intercept is $\left(1 , 0\right)$

To summarize:

VA: $x = - 5$
HA: $y = 1$
x-intercept: $\left(1 , 0\right)$
y-intercept: $\left(0 , - \frac{1}{5}\right)$

To find end behavior:
First graph the intercepts. Note that the $y$ intercept is "more negative" and to the left of the $x$ intercept. This indicates that as you approach the VA from the right, $y$ approaches negative infinity.

Because the VA has an "odd degree" of $\textcolor{red}{1}$, i.e. ${\left(x + 5\right)}^{\textcolor{red}{1}}$, the end behavior around the VA is "odd". An "odd" end behavior indicates that the end behavior is "opposite" around the VA. So, as you approach the VA from the left, $y$ approaches positive infinity.