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# Graphing rational functions 2

Sal graphs y=(2x)/(x+1). Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

• What if the denominator doesn't factor out? In my example that I am working on, the numerator is "2" and the denominator is "x^5 - 3x^3"
Any tips?
And thanks Salman, your videos are saving me!!
• If the denominator doesn't factor out in the case of finding asymptotes, then you wouldn't have a vertical asymptote or a horizontal asymptote. You would still be able to find points to plot but you wouldn't have any asymptotes.

Hope this helps!
(1 vote)
• From y = (x+1)/(x+1) - no vertical asymptote. Since a vertical asymptote is when f(x) is undefined and that is undefined when the denominator = 0. Since 0/0 has no vertical asymptote - can someone explain the difference between x/0 and 0/0?

Thanks
• The reason why y = (x + 1)/(x + 1) does not have a vertical asymptote is because the equation simplifies to y = 1. Instead of an asymptote, however, there must be a hole at -1.
• If I don't have a horizontal asymptote, but I have a slant asymptote how do I find where, and if, it crosses the x axis?
• Yes, it will cross the x-axis if you have a slant asymptote (oblique asymptote). I am assuming here that you are talking about oblique asymptotes of rational functions and not the oblique asymptotes of hyperbolas (which are not functions).

The rules for determining that you have an oblique asymptote are complicated and then there are some steps for finding the x-intercept. First you determine whether you have a proper rational function or improper one. Only improper rational functions will have an oblique asymptote (and not all of those). A proper one has the degree of the numerator smaller than the degree of the denominator and it will have a horizontal asymptote. An improper rational function has either the same degree or a larger degree in the numerator.

If the degree of the polynomial in the numerator is the same as the degree of the polynomial in the denominator, you will have a horizontal asymptote but no oblique asymptote.

If the degree of the numerator is greater by two or more, then you do not have a horizontal asymptote --nor an oblique one.

Finally! If the degree of the numerator is greater by ONE than the degree of the denominator, then you have an oblique asymptote. You can find the equation of that asymptote by dividing your polynomials to find the quotient, which should be of the form y = mx+b -- the familiar equation of a straight line with slope m. You can find the x-intercept by substituting in a value of zero for the y and solving for x. That is where the oblique asymptote crosses the x-axis.
Note that an oblique asymptote shows how the function behaves, but the function may cross the oblique asymptote before it settles down unlike vertical and horizontal asymptotes which the function approaches but never touches.
• why isnt 0 over 0 just 0?
• It helps to think of division as the inverse of multiplication. In that case, 0/0 would be 0 *(1/0). Before you can complete that multiplication, think about what 1/0 is: the number that, when multiplied by 0, yields a product of 1. Since there is no such number (0*x = 0 for all x, so nothing can be multiplied by 0 to get 1,) 0 has no multiplicative inverse, and therefore 0/0 can not exist. It's undefined.
• Is there any sort of explanation why something like (x+1) in the denominator fails to be a vertical asymptote when it gets canceled out with the same expression in the nominator? I didn`t actually get it from the video. Can`t it still be an asymptote? The graph still approaches it, but never touches.
(1 vote)
• A vertical asymptote happens because as you get closer and closer to the point where you'd divide by zero - in this case, x=-1 - your result is going to keep getting larger and larger (or smaller, if the number is negative. The absolute value becomes huge). That creates the vertical asymptote. The exception is when it cancels out, meaning there's another (x+1) in the nominator... no matter how tiny x+1 is, (x+1)/(x+1) is still going to be 1, so you only get the "hole" in the graph.
• so exactly how do you find the horizontal asymptote?
• Another solution, as Sal mentioned in his first video, is to look at the degree of the numerator and denominator.
If the degrees are equal, the horizontal asymptote is the quotient of the leading coefficient of the numerator and the leading coefficient of the denominator.
If the degree is higher in the denominator, the horizontal asymptote is y = 0 (the x-axis).
If the degree is higher in the numerator, there is no horizontal asymptote.
• Can someone give me an example where there's NO asymptote AT ALL?
• f(x)=(x+3)/(x+3) There are no asymptotes. Just a hole.
• Shouldn't curves be in quadrant 1 and 3 because the graph is positive?I don't understand why the curves became to be on the quadrants 2 and 4...
• Just because "the graph is positive" doesn't mean it needs to be in quadrants 1 and 3. All it will mean is that the curve rises as you move from left to right. You can see that this is the case -- as we move to the right along the top-left-hand curve, the y-value rises, and as we move to the right along the bottom-left-hand curve, the y-value also rises.
• At , is there no horizontal asymptote in this expression? Can anyone explain why? I thought a rational expression must have an asymptote.
• No, not all rational expressions have vertical or horizontal asymptotes. Some have oblique asymptotes instead.

For example, (x³+5x +1) / (x²+2x + 7) has no vertical asymptote because the denominator has no rational zeros. But, because the numerator has a higher degree than the denominator, it does not have a horizontal asymptote either. It does have an oblique asymptote of x − 2.

So, rational expressions may have, but do not necessarily have vertical or horizontal asymptotes. Here is the general pattern:

If there are no rational zeros of the denominator, there are no vertical asymptotes.
If there are rational zeros in the denominator, then those are possible locations of vertical asymptotes, but they must be checked to make sure.
If the degree of the denominator is greater than the degree of the numerator, then the horizontal asymptote is the x axis.
If the degree of the numerator and denominator are equal, then there is a non-zero horizontal asymptote.
If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote but instead there is an oblique asymptote.
(1 vote)
• is horizontal line is undefined slope and vertical slope has zero slope?
(1 vote)
• Slope is defined as change in y divided by change in x.
A horizontal line has no change in y, so its slope is 0 / change in x
which equals zero.
A vertical line has no change in x, so its slope is change in y / 0.
Since division by 0 is not allowed, that means that the slope of the vertical line is undefined.