Mmm, asymptotes: (3x^2 + 5x + 1) / (x^2 + 7x + 10)

[DemiGod]

First post, so.. hi?

So I have the following problem:

3x^2 + 5x + 1
----------------
x^2 + 7x + 10

I can figure out some asymptotes from the graph on my calc usually, but it's is usually i can get the x and not the y, or vice versa. What is the correct, mathematical way to find these asymptotes?

Appreciate the help.

- DemiGod

 

[soroban]

Re: Mmm, asymptotes.

Hello, DemiGod!

Welcome aboard!

\(\displaystyle y \;=\;\frac{3x^2 + 5x + 1}{x^2 + 7x + 10}\)

I can figure out some asymptotes from the graph on my calc usually.
Usually, i can get the x and not the y, or vice versa.
What is the correct mathematical way to find these asymptotes?

For vertical asymptotes, find the values of \(\displaystyle x\) that make the denominator equal to zero.

\(\displaystyle \text{We have: }\;x^2 + 7x + 10 \:=\:0\quad\Rightarrow\quad (x+2)(x+5)\:=\:0\quad\Rightarrow\quad x \:=\:-2,-5\)

\(\displaystyle \text{The vertical asymptotes are: }\;\boxed{x \:=\:-2}\:\text{ and }\:\boxed{x \:=\:-5}\)



For the horizontal asymptote, find the limit of \(\displaystyle y\) as \(\displaystyle x\) approaches \(\displaystyle \infty\).

\(\displaystyle \text{We have: }\;\lim_{x\to\infty}\frac{3x^2 + 5x + 1}{x^2+7x+10}\)


Divide top and bottom by the highest power of \(\displaystyle x\) in the denominator, \(\displaystyle x^2\)

. . \(\displaystyle \lim_{x\to\infty}\frac{\dfrac{3x^2}{x^2} + \dfrac{5x}{x^2} + \dfrac{1}{x^2}}{\dfrac{x^2}{x^2} + \dfrac{7x}{x^2} + \dfrac{10}{x^2}} \;=\; \lim_{x\to\infty}\frac{3 + \frac{5}{x} + \frac{1}{x^2}}{1 + \frac{7}{x} + \frac{10}{x}} \;=\;\frac{3 + 0 + 0}{1 + 0 + 0} \;=\;3\)

\(\displaystyle \text{Therefore, the horizontal asymptote is: }\;\boxed{y \:=\:3}\)