Sorry if this is a bit simple compared to everything here, but I can't really seem to find an answer.

If I have $$f(x) = \frac{(x-2)(x-4)}{x(x-1)}$$ 1) When is the horizontal asymptote is crossed? Apparently to check if/where the horizontal asymptote is crossed I solve for f(x) = A, where A is the limit, is this true?

2)After solving for the vertical asymptotes I get x = 0 and x = 1. How do I know how each part behaves? My textbook made us use the behavior of the function as it got closer to the x intercepts, but that was for polynomial functions.


1) yes. $\lim{f(x)}_{x\to+\infty} = 1$ and $\lim{f(x)}_{x\to-\infty} = 1$

$f(x) = 1$ where $x = \frac{8}{5}$

2) after finding the vertical asymptotes look at the behavior of the function as you approach it from either side.

$\lim{f(x)}_{x\to0^-}= \infty$ and $\lim{f(x)}_{x\to0^+}= -\infty$

and $\lim{f(x)}_{x\to1^-}= -\infty$ and $\lim{f(x)}_{x\to1^+}= \infty$

You can determine those limits by looking at the graph, or by plugging in carefully chosen values of x. e.g. to determine $\lim{f(x)}_{x\to0^-}= \infty$ and $\lim{f(x)}_{x\to0^+}= -\infty$ notice that when $x=0$ the numerator is positive. Now pick two numbers on either side of 0, say $\pm\frac{1}{2}$. $-\frac{1}{2}$ gives a positive denominator, indicating $f(x)$ will take on larger and larger positive values. $\frac{1}{2}$ gives a negative denominator, indicating larger and larger negative values.


1) If either the $\lim_{x\rightarrow -\infty}f(x)=c_1$ or $\lim_{x\rightarrow +\infty}f(x)=c_2$ exists then you can get the horizontal asymptotes by evaluating $\lim_{x \rightarrow -\infty} f(x)$ and $\lim_{x \rightarrow \infty} f(x)$

If you let $y = f(x)$ then you will see the horizontal asymptotes will occur at the lines $y = c_1$ or $y = c_2$ (provided of course that those limits that defined $c_1,c_2$ exist)

Say you have a function with a horizontal asymptote that you found from evaluating the limit, it's possible (but not guaranteed) that the function has an intercept with the line defined by $y=c_1$. If this is the case (which it is for the f(x) in your question) then you can solve the simultaneous equations defined by $y = f(x)$ and $y=c_1$ or alternatively $f(x) = c_1$ to find the point on the graph that intercepts the horizontal asymptote line. (do the same for $c_2$ if it exists)

2) In a hand wavy way I think the easiest thing to do is to look at what happens on either side of the asymptote.

  • $\begingroup$ Thanks for the answer, I knew the first part. But how do I know if/where the function passes that horizontal asymptote. For the second party, well, how would I do that. Also are questions of this level frowned upon here, everything else seems to be calculus and up. $\endgroup$ – Howcan Dec 9 '13 at 7:50
  • $\begingroup$ @Howcan What exactly do you mean by "if/where the function passes that horizontal asymptote"? I edited to try to clarify but I'm entirely sure this is what you meant. $\endgroup$ – shuttle87 Dec 9 '13 at 7:51
  • $\begingroup$ Well after doing some example questions, the line of the function sometimes passed the horizontal asymptote. Just graph the function I gave as an example (the answer key shows it passed through the asymptote.) $\endgroup$ – Howcan Dec 9 '13 at 7:55
  • $\begingroup$ @Howcan, I see what you are saying now, I edited my answer to try to address that. Basically you just need to solve for the point where the line defined with the asymptote intercepts the graph. $\endgroup$ – shuttle87 Dec 9 '13 at 8:06

Start by simplifying the function. This becomes (x^2-6x+8)/(x^2 -x) The horizontal asymptote is y=1 because both polynomials are the same degree, so it is y=(leading coefficient of numerator)/(leading coefficient of the denominator). Now solve for y=1. 1= (x^2 -6x +8)/(x^2 -x) (x^2 -6x +8)=(x^2 -x) -6x+8=-x 8=5x x=1.6 (or 8/5) The asymptote is crossed at (1.6,1) Use the following rules to find the horizontal asymptote: If the degree of the top is greater than the degree of the bottom, then the horizontal asymptote does not exist. If the degree of the top is equal to the degree of the bottom, then the horizontal asymptote is y=(leading coefficient of numerator)/(leading coefficient of the denominator). If the degree of the top is less than the degree of the bottom, then the horizontal asymptote is y=0 (the x-axis). Sorry for the format, I am on a phone currently.

  • $\begingroup$ This question is five years old. Your answer can surely wait until you get to a real computer and can take the time to format it correctly---the question isn't going anywhere. $\endgroup$ – Xander Henderson Jan 1 at 2:02

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