# Thread: analytic method to find the intercepts and asymptotes of the functions

1. ## analytic method to find the intercepts and asymptotes of the functions

I would be so thankful if someone could help me with these equations:
I am to find the intercepts and asymptotes of the functions if there are any and I have to find them analytically:
f(x) = x+2/(x^2)-9

&

f(x) = 3x^3-4x^2-3x-3/x^2-x-2

2. Originally Posted by racersteve47
I would be so thankful if someone could help me with these equations:
I am to find the intercepts and asymptotes of the functions if there are any and I have to find them analytically:
f(x) = x+2/(x^2)-9

&

f(x) = 3x^3-4x^2-3x-3/x^2-x-2
First off, you need to use parenthesis.

Second, to find vertical asymptotes, set the denominator equal to zero.

For slant asymptotes if the numerator is one degree higher than the denominator, do the long division. The linear and constant terms of the result is the slant asymptote.

For horizontal asymptotes the degree of the numerator is the same or less than the degree of the denominator. What happens when x is very large and negative? when x is very small and negative?

There are no other kinds of asymptotes, so this covers everything.

-Dan

3. Originally Posted by racersteve47
I would be so thankful if someone could help me with these equations:
I am to find the intercepts and asymptotes of the functions if there are any and I have to find them analytically:
f(x) = x+2/(x^2)-9
We have a rational function in the form of $R(x)=\frac{P(x)}{Q(x)}$.

Asymptotes

Vertical asymptotes occur when $Q(x)=0$. Thus, in our case, there would be a vertical asymptote when $x^2-9=0\implies \color{red}\boxed{x=\pm 3}$

Horizontal asymptotes occur when:

1) the degree of $P(x)$ is less than $Q(x)$

2) the degree of $P(x)$ and $Q(x)$ are the same.

In this case, we have $R(x)=\frac{x+2}{x^2-9}$. The degree of $P(x)$ is less than the degree of $Q(x)$. Thus, the horizontal asymptote is the line $\color{red}\boxed{y=0}$. If you've taken calculus, you can see this is the case by evaluating $\lim_{x\to{\pm\infty}}\frac{x+2}{x^2-9}$.

Intercepts

You can find the x intercepts by setting $R(x)=0$.

This is the case when $\frac{x+2}{x^2-9}=0\implies x+2=0\implies \color{red}\boxed{x=-2}$. The x intercept is $\color{red}\boxed{(-2,0)}$

You can find the y intercepts by letting x=0. Thus, $R(0)=-\frac{2}{9}$. Thus, the y intercept is $\color{red}\boxed{\bigg(0,-\frac{2}{9}\bigg)}$

The graph verifies what we have done:

Can you try the other one yourself??

f(x) = 3x^3-4x^2-3x-3/x^2-x-2
Hope that this helps.

--Chris

4. Originally Posted by topsquark
First off, you need to use parenthesis.

Second, to find vertical asymptotes, set the denominator equal to zero.

For slant asymptotes if the numerator is one degree higher than the denominator, do the long division. The linear and constant terms of the result is the slant asymptote.

For horizontal asymptotes the degree of the numerator is the same or less than the degree of the denominator. What happens when x is very large and negative? when x is very small and negative?

There are no other kinds of asymptotes, so this covers everything.

-Dan
I was thinking about this again. I need to add a restriction to my general comment on vertical asymptotes. If both the numerator and denominator are zero then factor both the numerator and denominator and do any cancellations. If the resulting rational function has a zero denominator, then you have a vertical asymptote. If the resulting rational function does not have a zero denominator, then the original function has a "hole" at that point.

-Dan