]> Chapter 09 Notes

9: Rational Functions

9-1: Inverse Variation

Two variables are in direct variation if $y = k x$ for some constant k. We encountered these earlier in the course, since the graph of these variables is a line.

Two variables are in inverse variation if $x y = k$ , or $y = \frac{k}{x}$ for some constant k. The graph of these variables is not a line…more on that in Chapter 10!

Examples

[1.] The variables x and y vary inversely, so that when $x = 7$ , $y = 2$ . Write the equation that relates x and y.

Since they are inversely related, $y = \frac{k}{x} ⇒ 2 = \frac{k}{7} ⇒ 14 = k$ . Thus, $y = \frac{14}{x}$ .

[2.] The variable z varies directly with x and inversely with the cube of y, so that when $x = 8$ and $y = 2$ , $z = 3$ . Write the equation that relates x, y and z.

Read that carefully! We are told that $z = \frac{k x}{y^{3}}$ ; plugging in gives me $3 = \frac{8 k}{2} ⇒ 6 = 8 k ⇒ \frac{3}{4} = k$ . The equation is $z = \frac{3 x}{4 y^{3}}$ .

9-2: Graphing Inverse Variations

If you can graph $y = \frac{1}{x}$ (and you should!), then you can graph $y = \frac{a}{x - c} + d$ . Would you care to take a guess as to what the other parts of that equation do to the graph?

Examples

[3.] Find the asymptote and y-intercept of $y = \frac{11}{x + 3} - 3$ .

Since the graph has been shifted down three units, the asymptote is at $y = - 3$ .

When $x = 0$ , $y = \frac{11}{3} - 3 = \frac{2}{3}$ , so the y-intercept is at $(0, \frac{2}{3})$ .

[4.] Rewrite $y = \frac{1}{2 x - 6}$ in standard form. Identify the asymptote and y-intercept.

$y = \frac{1}{2 x - 6} = \frac{1}{2 (x - 3)} = \frac{\frac{1}{2}}{x - 3}$ . The asymptote is at $y = 0$ , and the y-intercept is at $\frac{\frac{1}{2}}{- 3} = - \frac{1}{6}$ .

9-3: Rational Functions and their Graphs

A rational function is of the form $\frac{N (x)}{D (x)}$ , where $N (x)$ and $D (x)$ are polynomials. The function $y = \frac{1}{x}$ is the simplest of these—it is sometimes called the reciprocal function.

Understanding rational functions—their graphs, in particular—is essential to Calculus.

Domain Issues

First and foremost, you must be able to find the domain of a rational function. As you (should!) know, any value of x that causes the denominator to be zero is not in the domain. The problem is that those values of x will manifest themselves graphically in two very different ways. In order to distinguish between the two, you must first factor the numerator and denominator.

Vertical Asymptotes

As you know, values of x that cause the denominator to equal zero are not in the domain. Since the denominator is factored, you can easily identify those values of x. If a factor only appears in the denominator, then the graph will have a vertical asymptote at that value of x.

Vertical asymptotes are domain issues—thus, they cannot ever be crossed. They must be approached asymptotically (gradually).

Holes

If there is a factor that appears in both the numerator and denominator, then this will appear as a hole (removable discontinuity) in the graph. Yes—a hole. Literally. Go ahead—graph $y = \frac{x^{2} - 4}{x - 2}$ , and zoom in really close to the point $(2, 4)$ . Really close. See it? Okay, here—look:

An Image

(on your calculator, it may not look exactly like this)

Here's why: for almost every value of x, that common factor cancels out, and contributes nothing to the function value. The exception is the single value of x that makes that factor equal to zero—at that value of x, the graph simply disappears.

Consider my example above: $y = \frac{x^{2} - 4}{x - 2} = \frac{(x + 2) (x - 2)}{x - 2}$ . I know that you want to cancel out that $(x - 2)$ —but before you do, remember that you can only do that if the factor is not equal to zero! Thus, $\frac{x^{2} - 4}{x - 2} = x + 2$ only if $x \neq 2$ . In fact, the graphs of the two sides of that equation are identical except for that one point. This is important for locating the y-coordinate of the hole: the one point that gets removed is the hole. Be sure to look at the forthcoming example below.

One final note: you should note the locations of holes before you identify any vertical asymptotes—otherwise, you might accidentally indicate that there is a hole on a vertical asymptote (which is ludicrous). The lone warning point here (and this is really rare) is if the multiplicity of that common factor differs between the numerator and denominator. In that case, if the numerator is of higher multiplicity, then you get a hole; if the denominator is higher, then you get a vertical asymptote.

End Behavior

What a function does for really extreme values of x (very far from zero) is called the end behavior. The end behavior of polynomials is simple: the ends head off towards positive or negative infinity.

The end behavior of exponential functions was a little more interesting—one end had an asymptote, and the other end went towards infinity.

Rational functions are even more interesting—but simpler than the textbook makes it appear. Put simply, a rational function will have as its end behavior the quotient of the division process.

Okay, even simpler: divide the numerator by the denominator (polynomial long division)—the quotient of the division is the equation that the rational function approaches when x gets really extreme.

Intercepts

An x-intercept occurs when $y = 0$ . The only way that a fraction (rational function) can equal zero is if the numerator equals zero. Fortunately, you have factored the numerator—so finding values of x that make the numerator (and thus the fraction) equal to zero should be a piece of cake.

A y-intercept occurs when $x = 0$ —so plug in zero for x!

Sketching the Graph

Place all of the collected information on a coordinate plane, then connect the dots.

Remember that it is fine to cross the end behavior equation!

Examples

[5.] Find any asymptotes, holes, and intercepts of $y = \frac{x + 2}{x - 4}$ . Identify the end behavior and sketch the graph.

Hurrah—this is already factored. I see that there are no common factors, so there are no holes. There is a value of x that makes the denominator equal to zero, so there is a vertical asymptote at $x = 4$ . The only value of x that makes the numerator equal to zero is -2, so that's where the x-intercept is. When $x = 0$ , the function value is $- \frac{2}{4} = - \frac{1}{2}$ , so that's the y-intercept. The quotient of the division is 1, so the end behavior is the line $y = 1$ (we call this a horizontal asymptote).

Now for the sketch:

An Image

[6.] Find any asymptotes, holes, and intercepts of $y = \frac{x + 3}{x^{2} - x - 12}$ . Identify the end behavior and sketch the graph.

Factor: $\frac{x + 3}{(x - 4) (x + 3)}$ . Thus, this has a hole at $x = - 3$ . For anywhere except the hole, the graph is really $y = \frac{1}{x - 4}$ …that means that the hole has a y-value of $\frac{1}{- 3 - 4} = - \frac{1}{7}$ . There is a vertical asymptote at $x = 4$ . There is no x-intercept, since there is no way for the numerator to equal zero. The y-intercept is at $(0, - \frac{1}{4})$ . The quotient of the division is zero—the $(x - 4)$ can't be divided into 1; thus, the end behavior is $y = 0$ .

The graph:

An Image

(it is really hard to see the horizontal asymptote, and the hole…make it clear on your hand-drawn work)

[7.] Find any asymptotes, holes, and intercepts of $y = \frac{2 x^{2} - x - 6}{5 x^{2} - 11 x - 12}$ . Identify the end behavior and sketch the graph.

Factor: $\frac{(2 x + 3) (x - 2)}{(5 x + 4) (x - 3)}$ . No common factors means no holes. There are two vertical asymptotes, at $x = - \frac{4}{5}$ and $x = 3$ . There are two x-intercepts, at $x = - \frac{3}{2}$ and $x = 2$ . The y-intercept is at $(0, \frac{1}{2})$ . The quotient of the division is $\frac{2}{5}$ , so the end behavior is $y = \frac{2}{5}$ .

The graph:

An Image

(notice how the graph crosses the horizontal asymptote)

[8.] Find any asymptotes, holes, and intercepts of $y = \frac{x^{2} - x}{x + 1}$ . Identify the end behavior and sketch the graph.

Factor: $\frac{x (x - 1)}{x + 1}$ . No common factors means no holes. There is one vertical asymptote at $x = - 1$ . There are x-intercepts at $x = 0$ and $x = 1$ (and one of those is also the y-intercept). Do a little division:

$\begin{array}{r} - 1 | & 1 & - 1 & 0 \\ - 1 & 2 \\ \overset{—}{1} & \overset{—}{- 2} & \overset{—}{| 2} \end{array}$

Since the quotient is $x - 2$ , the end behavior is $y = x - 2$ .

The graph:

An Image

9-4: Rational Expressions

To simplify a rational function, you must factor the numerator and denominator, then cancel out any common factors. Ultimately, a simplified rational function should not be factored (factoring is actually opposite of simplifying).

To multiply rational functions, remember to multiply the numerators and denominators.

To divide rational functions, remember to invert and multiply by the divisor.

Examples

[9.] Simplify: $\frac{x^{2} - 5 x}{x^{2} - 25}$ .

$\frac{x^{2} - 5 x}{x^{2} - 25} = \frac{x (x - 5)}{(x + 5) (x - 5)} = \frac{x}{x + 5}$

[10.] Multiply and simplify: $\frac{3 x^{3}}{x^{2} - 25} \cdot \frac{x^{2} + 6 x + 5}{x^{2}}$ .

$\frac{3 x^{3}}{x^{2} - 25} \cdot \frac{x^{2} + 6 x + 5}{x^{2}} = \frac{3 x^{3}}{(x + 5) (x - 5)} \cdot \frac{(x + 5) (x + 1)}{x^{2}} = \frac{3 x (x + 1)}{(x - 5)} = \frac{3 x^{2} + 3 x}{x - 5}$

[11.] Divide and simplify: $\frac{x^{2} - 3 x - 10}{2 x^{2} - 11 x + 5} \div \frac{x^{2} - 5 x + 6}{2 x^{2} - 7 x + 3}$ .

$\frac{x^{2} - 3 x - 10}{2 x^{2} - 11 x + 5} \div \frac{x^{2} - 5 x + 6}{2 x^{2} - 7 x + 3} = \frac{x^{2} - 3 x - 10}{2 x^{2} - 11 x + 5} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

$= \frac{(x - 5) (x + 2)}{(2 x - 1) (x - 5)} \cdot \frac{(2 x - 1) (x - 3)}{(x - 2) (x - 3)} = \frac{(x - 5) (x + 2)}{(x - 2)} = \frac{x^{2} - 3 x - 10}{x - 2}$

9-5: Adding and Subtracting Rational Expressions

To add or subtract rational functions, don't forget that you need a common denominator. To do that, you really need the lowest common multiple of the denominator. This is just like finding the lowest common multiple of two numbers—we'll talk about it in class, just in case you need a refresher.

You may do quite a bit of adding and subtracting if you encounter a complex fraction (fraction within a fraction).

Examples

[12.] Add and simplify: $\frac{3}{7 x^{2} y} + \frac{4}{21 x y^{2}}$ .

$\frac{3}{7 x^{2} y} (\frac{3 y}{3 y}) + \frac{4}{21 x y^{2}} (\frac{x}{x}) = \frac{9 y + 4 x}{21 x^{2} y^{2}}$

[13.] Subtract and simplify: $\frac{x y - y}{x + y} - \frac{y}{x + 2}$ .

$\frac{x y - y}{x + y} (\frac{x + 2}{x + 2}) - \frac{y}{x + 2} (\frac{x + y}{x + y}) = \frac{(x^{2} y + 2 x y - x y - 2 y) - (x y + y^{2})}{x^{2} + 2 x + x y + 2 y} = \frac{x^{2} y - x y - 2 y - y^{2}}{x^{2} + 2 x + x y + 2 y}$

[14.] Simplify: $\frac{\frac{2}{x} + 6}{\frac{3}{y}}$ .

$\frac{\frac{2}{x} + 6 (\frac{x}{x})}{\frac{3}{y}} = \frac{\frac{2 + 6 x}{x}}{\frac{3}{y}} = \frac{2 + 6 x}{x} \cdot \frac{y}{3} = \frac{2 y + 6 x y}{3 x}$

9-6: Solving Rational Equations

Use common denominators so that you have one fraction equal to another; cross-multiply; beware extraneous solutions!

Examples

[15.] Solve: $\frac{3}{1 - x} = \frac{2}{1 + x}$ .

$3 (1 + x) = 2 (1 - x) ⇒ 3 + 3 x = 2 - 2 x ⇒ 5 x = - 1 ⇒ x = - \frac{1}{5}$

(Check to see that this does work)

[16.] Solve: $\frac{2}{x + 3} + \frac{5}{3 - x} = \frac{6}{x^{2} - 9}$ .

$\frac{2}{x + 3} (\frac{3 - x}{3 - x}) + \frac{5}{3 - x} (\frac{3 + x}{3 + x}) = \frac{6}{x^{2} - 9} ⇒ \frac{(6 - 2 x) + (5 x + 15)}{9 - x^{2}} = \frac{6}{x^{2} - 9} ⇒ \frac{3 x + 21}{9 - x^{2}} = \frac{6}{x^{2} - 9}$ . Watch this next step carefully: $\frac{- 3 x - 21}{x^{2} - 9} = \frac{6}{x^{2} - 9}$ . Since the denominators are equal, the numerators must be equal—there is no need to cross multiply! $- 3 x - 21 = 6 ⇒ - 3 x = 27 ⇒ x = - 9$ . Once again, this solution does check.

[17.] Bob estimates that it would take him 5 hours to code a web site for a client. Dave, who works next door, thinks he can do the same job in 4 hours. Zeke, the intern, claims that he can do the job in just 3 hours. Working together, how long would it take all three men to complete the job?

This is just a variation on the old $D = R T$ formula…the trick is that there is 1 job to be completed!

Let b be Bob's rate of work, d be Dave's rate of work, and z be Zeke's rate of work. The values of these are $b = \frac{1}{5}$ (since $1 = b \cdot 5$ ), $d = \frac{1}{4}$ , and $z = \frac{1}{3}$ . When working together, their rates combine…thus, $1 = (b + d + z) T$ , or $1 = (\frac{1}{5} + \frac{1}{4} + \frac{1}{3}) T$ . Solving for T, we get $1 = (\frac{12}{60} + \frac{15}{60} + \frac{20}{60}) T ⇒ 1 = \frac{47}{60} T ⇒ T = \frac{60}{47} \approx 1.277$ hours.

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