# Thread: End behaviour of a rational function

1. ## End behaviour of a rational function

Hi,

I hope someone can help. I'm trying to understand why in this example does x approach -3 from the positive direction starting at the interval of -2? Why doesn't it, for example, begin at -1?

Here's also a link to the video: https://www.khanacademy.org/math/alg...onal-functions

Sincerely,
Olivia

2. ## Re: End behaviour of a rational function

The sign changes if and only if a single factor changes sign. In the interval $(-\infty,-3)$, all factors are negative. On the interval $(-3,-2)$ only $x+3$ is positive. So if you want to evaluate the sign of the limit, you want to look on an interval where the sign stays the same.

3. ## Re: End behaviour of a rational function

I am truly not sure I understand the question.

$0 < \alpha .$

$\dfrac{(-\ 3 - \alpha)^2 + 3(-\ 3 - \alpha) + 2}{(-\ 3 - \alpha) + 3} = \dfrac{(-\ 1)^2(3 + \alpha)^2 - 9 - 3\alpha + 2}{-\ \alpha} =$

$\dfrac{9 + 6 \alpha + \alpha^2 - 9 - 3 \alpha + 2}{-\ \alpha} = \dfrac{3 \alpha + \alpha^2 + 2}{-\ \alpha} = -\ 3 - \alpha - \dfrac{2}{\alpha}.$

$\therefore \displaystyle \lim_{\alpha \rightarrow 0} \left ( -\ 3 - \alpha - \dfrac{2}{\alpha} \right ) = -\ 3 - 0 -\ \infty = -\ \infty.$ Take as an example.

$-\ 3 - 0.0001 - \dfrac{2}{.0001} = - \ 3.0001 - 20000 = -\ 20003.0001 \text { and}\\ -\ 3 - 0.00001 - \dfrac{2}{.00001} = -\ 3 - .0.00001 - 200000 = -\ 200003.00001 .$

Growing by leaps and bounds in the negative direction.

$\dfrac{(-\ 3 + \alpha)^2 + 3(-\ 3 + \alpha) + 2}{(-\ 3 + \alpha) + 3} = \dfrac{9 - 6 \alpha + \alpha^2 - 9 + 3 \alpha + 2}{\alpha} =$

$\dfrac{-\ 3 \alpha + \alpha^2 + 2}{\alpha} = -\ 3 + \alpha + \dfrac{2}{\alpha}.$

$\therefore \displaystyle \lim_{\alpha \rightarrow 0} \left ( -\ 3 + \alpha + \dfrac{2}{\alpha} \right ) = -\ 3 + 0 + \infty = \infty.$

Take as an example.

$-\ 3 + 0.0001 + \dfrac{2}{.0001} = - \ 3 + 20000.0001 = 19997.0001 \text { and}\\ -\ 3 + 0.00001 + \dfrac{2}{.00001} = -\ 3 + 200000.00001 = 199997.00001 .$

Growing by leaps and bounds in the positive direction.

4. ## Re: End behaviour of a rational function

If you look and listen closely in the video at the 5:00 mark, he chose to approach the Vertical Asymptote starting from -2 and study the range of -3 ---- -2 to make sure that there's no sign changes in the numerator (changes in graph behavior locally). He could have picked a larger number such as -1 or 100 like you said but he chose x= -2 because (-2 is an x intercept in the numerator) and sometimes a line can bounce or go through an intercept (depending on it's multiplicity) and that can affect the way it looks and behaves around an asymptote.