17Calculus Precalculus - Basics of Rational Functions
Rational functions are functions (so they pass the vertical line test) that are ratios of polynomials. In addition to the basics of rational functions, we also discuss holes and zeroes of rational functions on this page. Asymptotes and graphing of rational functions are covered on separate pages.
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What Are Rational Functions?
Put another way, rational functions are fractions that have polynomials in the numerator and denominator. They consist of one polynomial divided by another polynomial. You can think of them like this.
\(\displaystyle{ \text{rational function} = \frac{\text{a polynomial}}{\text{another polynomial}}}\) |
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Domain of Rational Functions
When working with rational functions, you need to be very aware of the domain. Since polynomials include all real numbers in their domain, the only situation you need to think about is where the denominator is zero. Here is a video explaining why we need to be careful with zero in the denominator.
Numberphile - Problems With Zero [12min-59secs]
This video includes some limits. If you haven't studied limits yet, no worries. You will still be able to follow this video and maybe even get a head start on limits.
video by Numberphile |
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So your first task when investigating a rational function is determine where the denominator is zero. The page discussing the zeroes of a polynomial will help you with this. The domain then is all the set of all real numbers except for those values where the denominator is zero.
Vertical Asymptotes, Holes and Zeroes
Okay, so you found those values where the denominator is zero and excluded them from the set of real numbers to get the domain. So, what now? Well, the investigation of what is going on where the denominator is zero doesn't stop there. Even though those values are not in the domain, we often want to know what the rational function is doing very near those points. You will look into this in more detail when you get to limits but let's take a look and see what we can find out. We have four cases, three of which involve zero. Here is a table of the values. After the table we discuss each case.
Let the rational function be \(\displaystyle{r(x)=\frac{n(x)}{d(x)}}\)
\(r(x)\) |
\(n(x) \neq 0\) |
\(n(x) = 0\) |
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\(d(x) \neq 0\) |
Zero | |
\(d(x) = 0\) |
Vertical Asymptote |
Hole |
Let's start in the upper left corner (empty cell) and work across.
1. Of course, when the numerator is not zero (\(n(x)\neq0\)) and the denominator is not zero (\(d(x)\neq0\)), we just have a number as our answer. For example, \(1/2\). No surprises here. And, since no zeroes are involved, there is not much to say.
2. Now let's look to the right where numerator is zero (\(n(x)=0\)) and the denominator is not zero (\(d(x)\neq0\)). In this case, since the denominator is not zero, the value of x where this occurs is part of the domain with the result being zero. Again, no surprises here. An example is \(f(x)=(x-1)/x\). When \(x=1\), \(f(1)=0/1=0\). These are called zeroes of the function.
3. The first cell in the last row is a bit more interesting. In this case, we have the numerator being non-zero (\(n(x)\neq0\)) but this is our first case where denominator is zero (\(d(x)=0\)). An example of this is \(g(x)=1/(x-3)\) and when \(x=3\) we have \(g(3)=1/0\). Of course, you know from the video above that this is undefined. But what does the graph look like? Below is the graph of \(g(x)=1/(x-3)\). You can see that as x gets closer and closer to 3 (look at each side separately) the graph goes off to \(\pm \infty\) depending on which side of \(x=3\) you are looking at. We call \(x=3\) a vertical asymptote.
4. The last case is when both the numerator and denominator are zero. In this case, we have something called a hole. An example is \(h(x)=(x^2+x)/(x+1)\). Notice that when \(x=-1\), we get \(h(-1)=0/0\). The graph of this function is shown below.
\(\displaystyle{f(x)=\frac{1}{x-3}}\)
Vertical Asymptote at \(x=3\)
\(\displaystyle{h(x)=\frac{x(x+1)}{x+1}}\)
Hole at \((-1,-1)\)
A few more comments are in order.
1. In the case of a vertical asymptote, it doesn't matter if the graph goes off to positive or negative infinity. The x-value where the denominator is zero is called an asymptote in both cases.
2. In the case of a hole, you need to be careful when graphing on your calculator. The hole will probably not show up in the graph and you may assume nothing is going on there. Always check the results of your calculator to see if they make sense.
A Caution About Simplifying
On the page discussing the domain and range of a function, we go out of our way to explain why the domain is important in calculus. When you are working with rational functions, you need to be extra careful when simplifying. Let's look at the graph above that contains a hole, \(\displaystyle{f(x)=\frac{x^2+x}{x+1}=\frac{x(x+1)}{x+1}}\). If you remember your algebra, your first instinct would be to cancel the \(x+1\) terms from the numerator and denominator to get \(g(x)=x\). That was correct in algebra since you were not really concerned with the domain and functions.
However, doing this cancellation now introduces an error and you need to watch for it. Let's look more closely at what is going on by comparing the two functions \(f(x)\) and \(g(x)\).
\(\displaystyle{f(x)=\frac{x(x+1)}{x+1}}\) |
domain: all real numbers except \(x=-1\) |
\(g(x)=x\) |
domain: all real numbers |
Notice in the table above, \(f(x)\neq g(x)\) because the domains are different, i.e. \(g(-1)=-1\) but \(f(-1)\) is undefined. At all other points, the functions are equal. Even though this is only one point, we must write \(f(x) \neq g(x)\).
However, one way to get around this is to specify a domain that excludes the problem points, in this case \(x=-1\). So it would be correct to say
\(f(x)=g(x); ~~ x \neq -1 \)
We have removed the problem point from the domain and so the functions are now equal for all x except \(x=-1\).
Okay, use what you learned in this tutorial to work these practice problems involving rational functions, zero and holes. After that, your next logical step is to learn more about asymptotes of rational functions.
Practice
Simplify the rational function \(\displaystyle{ \frac{5x-x^2}{x^2-x-20} }\) making sure to specify the domain.
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Simplify the rational function \(\displaystyle{ \frac{5x-x^2}{x^2-x-20} }\) making sure to specify the domain.
Solution |
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2983 video
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Find the zeroes of the rational function \(\displaystyle{\frac{x-2}{x+6}}\).
Problem Statement |
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Find the zeroes of the rational function \(\displaystyle{\frac{x-2}{x+6}}\).
Solution |
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2984 video
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Find the zeroes and holes of the rational function \(\displaystyle{\frac{x-2}{x^2-4}}\).
Problem Statement |
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Find the zeroes and holes of the rational function \(\displaystyle{\frac{x-2}{x^2-4}}\).
Final Answer |
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There is no x-intercept but there is one hole at \(x=2\).
Problem Statement |
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Find the zeroes and holes of the rational function \(\displaystyle{\frac{x-2}{x^2-4}}\).
Solution |
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2985 video
Final Answer |
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There is no x-intercept but there is one hole at \(x=2\). |
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Find the zeroes and holes of the rational function \(\displaystyle{\frac{x^2-x-12}{x^2-9}}\).
Problem Statement |
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Find the zeroes and holes of the rational function \(\displaystyle{\frac{x^2-x-12}{x^2-9}}\).
Final Answer |
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There is a zero at \(x=4\) and there is a hole at \(x=-3\). Although not asked for in this problem, there is also a vertical asymptote at \(x=3\).
Problem Statement |
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Find the zeroes and holes of the rational function \(\displaystyle{\frac{x^2-x-12}{x^2-9}}\).
Solution |
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2986 video
Final Answer |
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There is a zero at \(x=4\) and there is a hole at \(x=-3\). Although not asked for in this problem, there is also a vertical asymptote at \(x=3\). |
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Find the coordinates of all the holes in the graph of \(\displaystyle{h(x)=\frac{x^3-125}{x^2-25}}\).
Problem Statement |
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Find the coordinates of all the holes in the graph of \(\displaystyle{h(x)=\frac{x^3-125}{x^2-25}}\).
Solution |
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1672 video
video by PatrickJMT |
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Write an equation for a rational function with vertical asymptotes \(-4, 8\), x-intercepts at \(-2, 4\) and y-intercept at \(2\).
Problem Statement |
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Write an equation for a rational function with vertical asymptotes \(-4, 8\), x-intercepts at \(-2, 4\) and y-intercept at \(2\).
Solution |
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1660 video
video by MIP4U |
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Find a rational function with x-intercepts \(0, 1\), vertical asymptotes \(2, 4\) and horizontal asymptote \(3/5\).
Problem Statement |
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Find a rational function with x-intercepts \(0, 1\), vertical asymptotes \(2, 4\) and horizontal asymptote \(3/5\).
Solution |
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video by Dr Phil Clark |
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Really UNDERSTAND Precalculus
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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