# Question about asymptotes of rational function

• Sunwoo Bae
In summary, the graphing calculator may not be accurate and it would be time consuming to graph a more complicated function.
Sunwoo Bae
Homework Statement
sketch the graph of (x^3-3x^2+1)/x^3, making use of any suitable information you can obtain from the function and its first and second deerivative
Relevant Equations
If both polynomials are the same degree, divide the coefficients of the highest degree terms

I tried graphing the function in the calculator, and the graph seems to have a horizontal asymptote at y=0, not at y=1. Why is this so?

Thanks for helping out.

Sunwoo Bae said:
Homework Statement:: sketch the graph of (x^3-3x^2+1)/x^3, making use of any suitable information you can obtain from the function and its first and second derivative
Relevant Equations:: If both polynomials are the same degree, divide the coefficients of the highest degree terms

View attachment 263848
View attachment 263849

I tried graphing the function in the calculator, and the graph seems to have a horizontal asymptote at y=0, not at y=1. Why is this so?

Thanks for helping out.
That graph looks consistent with having a horizontal asymptote of ##y=1## .

You were given the hint to divide the coefficients of the highest degree terms. That gives ##1## , which indicates that there is a horizontal asymptote: ##y=1## .

Better than that is to express the function as follows.

##y=\dfrac{x^3}{x^3} + \dfrac{-3x^2+1}{x^3}##

##y=1 + \dfrac{-3x^2+1}{x^3}##

You may be able to see from this, what the horizontal asymptote is. Also, using this expression, it may be easier to get the derivatives (1st & 2nd) as asked for.

Last edited:
scottdave
You get the horizontal asymptotes by seeing if the function has any limits as ##x\to\pm\infty##.
I think that what you meant is that there's vertical asymptote at ##x=0##.

SammyS
archaic said:
I think that what you meant is that there's vertical asymptote at ##x=0##
It's pretty clear to me that he meant exactly what he said.
I tried graphing the function in the calculator, and the graph seems to have a horizontal asymptote at y=0, not at y=1.

SammyS and archaic
For rational functions, to check for horizontal asymptotes (the behavior of the function when x gets very large in negative x and positive x direction), we divide every term by the highest power of x that appears. Then we take the limit as x approaches -infinity and the limit as x approaches + infinity.

I am assuming this is for Pre-Calculus. So if you were to sketch these types of functions, using stuff you learned (including what I am stating in the paragraph), then your graph would not be as accurate as what appears on the graphing calculator. Moreover, would be time consuming to do so the more complicated the function.

## 1. What is an asymptote?

An asymptote is a line that a graph approaches but never touches. It can be horizontal, vertical, or oblique.

## 2. How do you find the horizontal asymptote of a rational function?

To find the horizontal asymptote, you need to look at the degrees of the numerator and denominator of the rational function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

## 3. Can a rational function have more than one vertical asymptote?

Yes, a rational function can have multiple vertical asymptotes. The number of vertical asymptotes is equal to the number of factors in the denominator of the rational function.

## 4. How do you graph a rational function with asymptotes?

To graph a rational function with asymptotes, first find the vertical and horizontal asymptotes. Then, plot points on either side of the vertical asymptotes to determine the behavior of the graph. Finally, connect the points and draw the asymptotes as dotted lines.

## 5. Can a rational function have an oblique asymptote?

Yes, a rational function can have an oblique asymptote if the degree of the numerator is exactly one more than the degree of the denominator. To find the equation of the oblique asymptote, use long division to divide the numerator by the denominator and the resulting quotient will be the equation of the oblique asymptote.

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