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# Graphing rational functions 1

Sal graphs f(x)=(2x+10)/(5x-15). Created by Sal Khan.

## Want to join the conversation?

• How do you find the Horizontal Asymptote if x is multiplied to a higher degree in the denominator than in the numerator? For example: 5x-10/x^2+4x-5 •   If the denominator has a higher degree term than the numerator, the horizontal asymptote will be 0. ALWAYS. If the numerator has a higher degree term than the denominator, there is no horizontal asympotote. When the numerator and denominator both have the same highest degree term (the highest in the numerator and denominator) is when you can divide their coefficients (number in front of a variable). There will be just one horizontal asymptote if that is the case. Also, in the video in this section "asymptotes of rational functions" he covers the question you just asked.

http://www.mathops.com/free/a1rf010.php
• How do you find the holes, or as Sal calls them: breaks (see: ), in a rational function? And how do you know if a function even has a hole? • A hole is where a simplified function is undefined. For example, let's say you have:
f(x) = (x + 5)(x + 2) / (x + 5)
You can simplify it by cancelling out the (x + 5) in the numerator and denominator.
f(x) = x + 2
You may think that because this function has no holes at all because there are no points where it is undefined. However, you always have to look at the original equation to find the holes. In the original equation, you had x + 5 in the denominator. To make x + 5 equal to 0, you would need x to be -5. So at -5, f(x) will have a hole there.
• What if f(x) = 1/x, how would you find the asymptotes? • To find the asymptotes of a function, first recognize that there are three types: vertical, horizontal, and oblique. I can't do as good of a job as Sal, but I can work through this example with you. First, observe that what you have is a rational function. These are the easiest to deal with. Next, recognize what an asymptote actually is: it's a line that the function will get very, very close to, but will never reach.

To find all horizontal asymptotes, observe what happens to y as x gets larger and larger (or more and more negative). If y approaches a specific value, then you have a horizontal asymptote. In your example, As x gets really big, y gets really, really small. Y actually gets infinitely close to zero as x gets infinitely larger. So, you have a horizontal asymptote at y = 0. Applying the same logic to x's very negative, you get the same asymptote of y = 0.

Next, we're going to find the vertical asymptotes of y = 1/x. To do this, just find x values where the denominator is zero and the numerator is non-zero. This clearly happens at x = 0 and nowhere else. So, as we get very close to 0 in x, the y values will approach positive and negative infinity. So, we are actually getting close to forming the vertical line x = 0. This is another asymptote of y = 1/x.

To find oblique asymptotes, the rational function must have the numerator's degree be one more than the denominator's, which it is not. So, there are no oblique asymptotes.

Summing this up, the asymptotes are y = 0 and x = 0. To confirm this, try graphing the function y = 1/x and zooming out very, very far. The graph should look like the lines x = 0 and y = 0.

Sal does a much better job than I do. Take a look at his video here: https://www.khanacademy.org/math/algebra2/polynomial_and_rational/asymptotes-graphing-rational/v/asymptotes-of-rational-functions
• How do you find the horizontal asymptote in a problem like (5x^2-45x+100)/(10x^2-10x-20) ? • I am asked to graph y=10/x ... how am I supposed to graph that? Is there any tutorial on that? • 10/x is part of the reciprocal function family and also its an example of inverse variation. well to graph that:
1. make a tale of positive and negative values
2.plot the points from the table
3. (optional) you can also check x/y - intercepts, asymptotes, and end behavior to get a better feel of the graphs shape and continuity/discontinuity
• Hi I have a question about finding the graph of this function: f(x)=((x+1)(x^2-x-2))/((x-1)(x+2))
I've found the x and y-intercept, vertical and horizontal asymptote (in this case there is no horizontal asymptote). What do I need next to be able to graph it fully? • There's no horizontal asymptote, but there is what is called a slant, or oblique asymptote. Basically it just means it's diagonal. how do you determine how diagonal? Let's find out.

First let's factor the entire thing into binomial terms.

((x+1)(x^2-x-2))/((x-1)(x+2))
((x+1)(x+1)(x-2))/((x-1)(x+2))

Now nothing cancels so this is what we hae to work with. The fact nothing cancels means no holes as well.

Now, to find horizontal/ slant asymptotes we look at the degree of the numerator over the degree of the denominator. I assume you know but just to keep things ordered as I would do it to solve the problem, equal degrees would have the horizontal asymptote be the ratio of the leading coefficients, if the degree of the denominator is larger then the HA is 0 and finally we have the degree of the numerator greater.

To find the slant asymptotes you perform the division depicted in the rational function To do this I am going to expand both the numerator and denominator, if you have trouble with polynomial long division let me know.

(x+1)(x^2-x-2) = x^3-x^2-2x+x^2-x-2 = x^3-3x-2
(x-1)(x+2) = x^2+2x-x+2 = x^2+x+2

So now we have (x^3-3x-2)/(x^2+x+2)

Again, if you need help with the division let me know, but in the end you wind up with x-1 - 4x/(x^2+x+2) Now we use this to find the oblique asymptote. Essentially we look at what this heads to as x goes to infinity. The answer is x-1, as the last term heads to 0.

So there is no horizontal asymptote, but the function does head toward the line x-1 as x heads toward infinity.

I hope this helpped, if not let me know though.
• when will there be more than one horizontal asymptote? • In a rational expression, when I have a numerator that is equal to zero (0) but the denominator is not equal to zero (0) what do I have. The expression is equal to zero (0) but what do I have? Example, what is the expression then called?
At 0.59 seconds Sal talks about the function equaling zero (0).   