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# Graphing rational functions according to asymptotes

Learn how to find removable discontinuities, horizontal asymptotes, and vertical asymptotes of rational functions. This video explores the specific example f(x)=(3x^2-18x-81)/(6x^2-54) before generalizing findings to all rational functions. Don't forget that not every zero of the denominator is a vertical asymptote! Created by Sal Khan.

## Want to join the conversation?

• What do you need to know before watching this video? I cant find any asymptotes or limits videos in algebra 2 here on KA.
• Yea. I suppose this is the introduction video to anymptotes.

The introduction video to "End behavior functions" is given in "End behavior of polynomial functions" Algebra 2 section. And more details on anymptotes are given in "Limits and infinity" in Differential calculus section.
• I learned that there are at most two (2) horizontal asymptotes and there can be an arbitrarily large number of vertical asymptotes for a function. But why at most 2 horizontal asymptotes?

Thank you
• A horizontal asymptote is basically the end behavior of a function, and there can only be two end behaviors (as x approaches negative infinity or positive infinity); that's why there can be only two horizontal asymptotes.
• How do you determine whether or not your function will cross your horizontal asymptote??
(1 vote)
• You find whether your function will ever intersect or cross the horizontal asymptote by setting the function equal to the y or f(x) value of the horizontal asymptote. If you get a valid answer, that is where the function intersects the horizontal asymptote, but if you get a nonsense answer, the function never crosses the horizontal asymptote.
For example, f(x) = (10x+7)/(5x-2) has a horizontal asymptote at f(x) = 2, thus:
``(10x+7)/(5x-2) = 210x+7 = 2(5x-2)10x+7 = 10x-47 = -4Since this is nonsense, the function never crosses the horizontal asymptote.``

Now let us look at an example that does cross the horizontal asymptote:
f(x) = (x²+2)/(x²+2x-6) has a horizontal asymptote at f(x) = 1, thus:
``(x²+2)/(x²+2x-6) = 1(x²+2)= (x²+2x-6)2 = 2x-62x = 8x = 4Therefore, this function crosses its horizontal asymptote at x=4``
• Can there be more than 1 vertical asymptotes
• Sure, as many as you like. y=tan(x) even has infinitely many.
• Just to be clear,
1)Is a removable discontinuity when the numerator is a zero and denominator not and a zero?
2) Is a vertical asymptotes when the denominator is a zero and the numerator isn't?
3) Is a zero when both numerator and denominator are zero?
In this skill every time I found they followed this pattern so I got a few right, then I got one wrong it was completely wrong so I want to make sure I am not doing the incorrect thing; because it worked for me I think I just got lucky. Also the proper way to solve them would be great. The hints are just not working for me this time.

• When you cancel, since "(x-a)/(x-a)" = 1 for all x, you don't change the graph at all, except that you need to note that x != a because /0 is undefined. Now when there are no more factors to cancel you can check the simplified expression for /0 to find asymptotes.
• why is there no videos introducing concepts such as asymptotes and limits and sal just dive straight in the topic ? i have a really hard time following with the examples.
• Why does the denominator = 0 when x=3 or -3?
• The denominator is equal to 6*(x-3)*(x+3). If we substitute 3 for x we have 6*(3-3)*(3+3) = 6*0*6 = 0. If we substitute -3 for x, we have 6*((-3)-3)*((-3)+3) = 6*(-6)*0 = 0. So, the denominator will be 0 when x equal 3 or -3.
• So I have the equation f(x)=7x/(10-3x)^4. Now I am trying to find the vertical asymptote of this equation but I do not know what to do with the ^4. I have made (10-3x)^4=0...but that is as far as I go. Any help with this?
• (10-3x)^4=0 means you have 4 factors = 0
(10-3x)(10-3x)(10-3x)(10-3x)=0
Since them match, just use one: (10-3x)=0 and solve for "x" to find the vertical symptotoe..
If the factors were different, then you would need to solve each different one.
Hope this helps.
• limits and continuity are calculus lessons, aren't they ?
• I was taught to simplify first. Does it matter if you do that first or not?
• As long as you keep track of what values aren't allowed simplifying should be fine.