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Chapter 4
Rational Functions
4.1 Introduction to Rational Functions
If we add, subtract or multiply polynomial functions according to the function arithmetic rules defined in Section 1.5, we will produce another polynomial function. If, on the other hand, we divide two polynomial functions, the result may not be a polynomial. In this chapter we study rational functions - functions which are ratios of polynomials.
Definition 4.1. A rational function is a function which is the ratio of polynomial functions. Said differently, r is a rational function if it is of the form
r(x) = p(x) q(x)
where p and q are polynomial functions.a aAccording to this definition, all polynomial functions are also rational functions. (Take q(x) = 1).
As we recall from Section 1.4, we have domain issues anytime the denominator of a fraction is zero. In the example below, we review this concept as well as some of the arithmetic of rational expressions.
Example 4.1.1. Find the domain of the following rational functions. Write them in the form p q((xx)) for polynomial functions p and q and simplify.
- f (x) = 2 x − 1 x + 1 2. g(x) = 2 −
x + 1
- h(x) =
2 x^2 − 1 x^2 − 1
3 x − 2 x^2 − 1
- r(x) =
2 x^2 − 1 x^2 − 1
÷
3 x − 2 x^2 − 1
Solution.
- To find the domain of f , we proceed as we did in Section 1.4: we find the zeros of the denominator and exclude them from the domain. Setting x + 1 = 0 results in x = −1. Hence,
302 Rational Functions
our domain is (−∞, −1) ∪ (− 1 , ∞). The expression f (x) is already in the form requested and when we check for common factors among the numerator and denominator we find none, so we are done.
- Proceeding as before, we determine the domain of g by solving x + 1 = 0. As before, we find the domain of g is (−∞, −1) ∪ (− 1 , ∞). To write g(x) in the form requested, we need to get a common denominator
g(x) = 2 −
x + 1
x + 1
(2)(x + 1) (1)(x + 1)
x + 1
=
(2x + 2) − 3 x + 1
2 x − 1 x + 1
This formula is now completely simplified.
- The denominators in the formula for h(x) are both x^2 − 1 whose zeros are x = ±1. As a result, the domain of h is (−∞, −1) ∪ (− 1 , 1) ∪ (1, ∞). We now proceed to simplify h(x). Since we have the same denominator in both terms, we subtract the numerators. We then factor the resulting numerator and denominator, and cancel out the common factor.
h(x) = 2 x^2 − 1 x^2 − 1
3 x − 2 x^2 − 1
2 x^2 − 1
− (3x − 2) x^2 − 1
=
2 x^2 − 1 − 3 x + 2 x^2 − 1
2 x^2 − 3 x + 1 x^2 − 1
= (2x − 1)(x − 1) (x + 1)(x − 1)
(2x − 1)(x − 1) (x + 1)(x − 1) = 2 x − 1 x + 1
- To find the domain of r, it may help to temporarily rewrite r(x) as
r(x) =
2 x^2 − 1 x^2 − 1 3 x − 2 x^2 − 1
We need to set all of the denominators equal to zero which means we need to solve not only x^2 − 1 = 0, but also (^3) xx 2 −−^21 = 0. We find x = ±1 for the former and x = 23 for the latter. Our domain is (−∞, −1) ∪
3 ,^1
∪ (1, ∞). We simplify r(x) by rewriting the division as multiplication by the reciprocal and then by canceling the common factor
304 Rational Functions
unbounded behavior, we say the graph of y = f (x) has a vertical asymptote of x = −1. Roughly speaking, this means that near x = −1, the graph looks very much like the vertical line x = −1.
The other feature worthy of note about the graph of y = f (x) is that it seems to ‘level off’ on the left and right hand sides of the screen. This is a statement about the end behavior of the function. As we discussed in Section 3.1, the end behavior of a function is its behavior as x as x attains larger^3 and larger negative values without bound, x → −∞, and as x becomes large without bound, x → ∞. Making tables of values, we find
x f (x) (x, f (x)) − 10 ≈ 2. 3333 ≈ (− 10 , 2 .3333) − 100 ≈ 2. 0303 ≈ (− 100 , 2 .0303) − 1000 ≈ 2. 0030 ≈ (− 1000 , 2 .0030) − 10000 ≈ 2. 0003 ≈ (− 10000 , 2 .0003)
x f (x) (x, f (x)) 10 ≈ 1. 7273 ≈ (10, 1 .7273) 100 ≈ 1. 9703 ≈ (100, 1 .9703) 1000 ≈ 1. 9970 ≈ (1000, 1 .9970) 10000 ≈ 1. 9997 ≈ (10000, 1 .9997)
From the tables, we see that as x → −∞, f (x) → 2 +^ and as x → ∞, f (x) → 2 −. Here the ‘+’ means ‘from above’ and the ‘−’ means ‘from below’. In this case, we say the graph of y = f (x) has a horizontal asymptote of y = 2. This means that the end behavior of f resembles the horizontal line y = 2, which explains the ‘leveling off’ behavior we see in the calculator’s graph. We formalize the concepts of vertical and horizontal asymptotes in the following definitions.
Definition 4.2. The line x = c is called a vertical asymptote of the graph of a function y = f (x) if as x → c−^ or as x → c+, either f (x) → ∞ or f (x) → −∞.
Definition 4.3. The line y = c is called a horizontal asymptote of the graph of a function y = f (x) if as x → −∞ or as x → ∞, f (x) → c.
Note that in Definition 4.3, we write f (x) → c (not f (x) → c+^ or f (x) → c−) because we are unconcerned from which direction the values f (x) approach the value c, just as long as they do so.^4 In our discussion following Example 4.1.1, we determined that, despite the fact that the formula for h(x) reduced to the same formula as f (x), the functions f and h are different, since x = 1 is in the domain of f , but x = 1 is not in the domain of h. If we graph h(x) = 2 x (^2) − 1 x^2 − 1 −^
3 x− 2 x^2 − 1 using a graphing calculator, we are surprised to find that the graph looks identical to the graph of y = f (x). There is a vertical asymptote at x = −1, but near x = 1, everything seem fine. Tables of values provide numerical evidence which supports the graphical observation.
(^3) Here, the word ‘larger’ means larger in absolute value. (^4) As we shall see in the next section, the graphs of rational functions may, in fact, cross their horizontal asymptotes. If this happens, however, it does so only a finite number of times, and so for each choice of x → −∞ and x → ∞, f (x) will approach c from either below (in the case f (x) → c−) or above (in the case f (x) → c+.) We leave f (x) → c generic in our definition, however, to allow this concept to apply to less tame specimens in the Precalculus zoo, such as Exercise 50 in Section 10.5.
4.1 Introduction to Rational Functions 305
x h(x) (x, h(x))
- 9 ≈ 0. 4210 ≈ (0. 9 , 0 .4210)
- 99 ≈ 0. 4925 ≈ (0. 99 , 0 .4925)
- 999 ≈ 0. 4992 ≈ (0. 999 , 0 .4992)
- 9999 ≈ 0. 4999 ≈ (0. 9999 , 0 .4999)
x h(x) (x, h(x))
- 1 ≈ 0. 5714 ≈ (1. 1 , 0 .5714)
- 01 ≈ 0. 5075 ≈ (1. 01 , 0 .5075)
- 001 ≈ 0. 5007 ≈ (1. 001 , 0 .5007)
- 0001 ≈ 0. 5001 ≈ (1. 0001 , 0 .5001)
We see that as x → 1 −, h(x) → 0. 5 −^ and as x → 1 +, h(x) → 0. 5 +. In other words, the points on the graph of y = h(x) are approaching (1, 0 .5), but since x = 1 is not in the domain of h, it would be inaccurate to fill in a point at (1, 0 .5). As we’ve done in past sections when something like this occurs,^5 we put an open circle (also called a hole in this case^6 ) at (1, 0 .5). Below is a detailed graph of y = h(x), with the vertical and horizontal asymptotes as dashed lines.
x
y
− 4 − 3 − (^2) − 1 1 2 3 4 − 2 − 3 − 4 − 5 − 6
1
3
4
5
6
7
8
Neither x = −1 nor x = 1 are in the domain of h, yet the behavior of the graph of y = h(x) is drastically different near these x-values. The reason for this lies in the second to last step when we simplified the formula for h(x) in Example 4.1.1, where we had h(x) = (2(xx+1)(−1)(xx−−1)1). The reason x = −1 is not in the domain of h is because the factor (x + 1) appears in the denominator of h(x); similarly, x = 1 is not in the domain of h because of the factor (x − 1) in the denominator of h(x). The major difference between these two factors is that (x − 1) cancels with a factor in the numerator whereas (x + 1) does not. Loosely speaking, the trouble caused by (x − 1) in the denominator is canceled away while the factor (x + 1) remains to cause mischief. This is why the graph of y = h(x) has a vertical asymptote at x = −1 but only a hole at x = 1. These observations are generalized and summarized in the theorem below, whose proof is found in Calculus.
(^5) For instance, graphing piecewise defined functions in Section 1.6. (^6) In Calculus, we will see how these ‘holes’ can be ‘plugged’ when embarking on a more advanced study of continuity.
4.1 Introduction to Rational Functions 307
The graph of y = f (x) The graph of y = g(x)
- The domain of h is all real numbers, since x^2 + 9 = 0 has no real solutions. Accordingly, the graph of y = h(x) is devoid of both vertical asymptotes and holes.
- Setting x^2 + 4x + 4 = 0 gives us x = −2 as the only real number of concern. Simplifying, we see r(x) = x (^2) −x− 6 x^2 +4x+4 =^
(x−3)(x+2) (x+2)^2 =^
x− 3 x+2.^ Since^ x^ =^ −2 continues to produce a 0 in the denominator of the reduced function, we know x = −2 is a vertical asymptote to the graph. The calculator bears this out, and, moreover, we see that as x → − 2 −, r(x) → ∞ and as x → − 2 +, r(x) → −∞.
The graph of y = h(x) The graph of y = r(x)
Our next example gives us a physical interpretation of a vertical asymptote. This type of model arises from a family of equations cheerily named ‘doomsday’ equations.^7
Example 4.1.3. A mathematical model for the population P , in thousands, of a certain species of bacteria, t days after it is introduced to an environment is given by P (t) = (^) (5^100 −t) 2 , 0 ≤ t < 5.
- Find and interpret P (0).
- When will the population reach 100,000?
- Determine the behavior of P as t → 5 −. Interpret this result graphically and within the context of the problem. (^7) These functions arise in Differential Equations. The unfortunate name will make sense shortly.
308 Rational Functions
Solution.
- Substituting t = 0 gives P (0) = (^) (5^100 −0) 2 = 4, which means 4000 bacteria are initially introduced into the environment.
- To find when the population reaches 100,000, we first need to remember that P (t) is measured in thousands. In other words, 100,000 bacteria corresponds to P (t) = 100. Substituting for P (t) gives the equation (^) (5^100 −t) 2 = 100. Clearing denominators and dividing by 100 gives (5 − t)^2 = 1, which, after extracting square roots, produces t = 4 or t = 6. Of these two solutions, only t = 4 in in our domain, so this is the solution we keep. Hence, it takes 4 days for the population of bacteria to reach 100,000.
- To determine the behavior of P as t → 5 −, we can make a table
t P (t)
- 9 10000
- 99 1000000
- 999 100000000
- 9999 10000000000
In other words, as t → 5 −, P (t) → ∞. Graphically, the line t = 5 is a vertical asymptote of the graph of y = P (t). Physically, this means that the population of bacteria is increasing without bound as we near 5 days, which cannot actually happen. For this reason, t = 5 is called the ‘doomsday’ for this population. There is no way any environment can support infinitely many bacteria, so shortly before t = 5 the environment would collapse.
Now that we have thoroughly investigated vertical asymptotes, we can turn our attention to hori- zontal asymptotes. The next theorem tells us when to expect horizontal asymptotes.
Theorem 4.2. Location of Horizontal Asymptotes: Suppose r is a rational function and r(x) = p q((xx)) , where p and q are polynomial functions with leading coefficients a and b, respectively.
- If the degree of p(x) is the same as the degree of q(x), then y = ab is thea^ horizontal asymptote of the graph of y = r(x).
- If the degree of p(x) is less than the degree of q(x), then y = 0 is the horizontal asymptote of the graph of y = r(x).
- If the degree of p(x) is greater than the degree of q(x), then the graph of y = r(x) has no horizontal asymptotes. aThe use of the definite article will be justified momentarily.
Like Theorem 4.1, Theorem 4.2 is proved using Calculus. Nevertheless, we can understand the idea behind it using our example f (x) = (^2) xx+1−^1. If we interpret f (x) as a division problem, (2x−1)÷(x+1),
310 Rational Functions
- The degrees of the numerator and denominator of h(x) are both three, so Theorem 4.2 tells us y = (^) −^62 = −3 is the horizontal asymptote. We see from the calculator’s graph that as x → −∞, h(x) → − 3 +, and as x → ∞, h(x) → − 3 −.
The graph of y = f (x) The graph of y = g(x) The graph of y = h(x)
Our next example of the section gives us a real-world application of a horizontal asymptote.^11
Example 4.1.5. The number of students N at local college who have had the flu t months after the semester begins can be modeled by the formula N (t) = 500 − (^) 1+3^450 t for t ≥ 0.
- Find and interpret N (0).
- How long will it take until 300 students will have had the flu?
- Determine the behavior of N as t → ∞. Interpret this result graphically and within the context of the problem.
Solution.
- N (0) = 500 − (^) 1+3(0)^450 = 50. This means that at the beginning of the semester, 50 students have had the flu.
- We set N (t) = 300 to get 500 − (^) 1+3^450 t = 300 and solve. Isolating the fraction gives (^) 1+3^450 t = 200. Clearing denominators gives 450 = 200(1 + 3t). Finally, we get t = 125. This means it will take 125 months, or about 13 days, for 300 students to have had the flu.
- To determine the behavior of N as t → ∞, we can use a table.
t N (t) 10 ≈ 485. 48 100 ≈ 498. 50 1000 ≈ 499. 85 10000 ≈ 499. 98
The table suggests that as t → ∞, N (t) → 500. (More specifically, 500−.) This means as time goes by, only a total of 500 students will have ever had the flu. (^11) Though the population below is more accurately modeled with the functions in Chapter 6, we approximate it (using Calculus, of course!) using a rational function.
4.1 Introduction to Rational Functions 311
We close this section with a discussion of the third (and final!) kind of asymptote which can be associated with the graphs of rational functions. Let us return to the function g(x) = x (^2) − 4 x+1 in Example 4.1.4. Performing long division,^12 we get g(x) = x (^2) − 4 x+1 =^ x^ −^1 −^
3 x+1.^ Since the term 3 x+1 →^ 0 as^ x^ → ±∞, it stands to reason that as^ x^ becomes unbounded, the function values g(x) = x − 1 − (^) x+1^3 ≈ x − 1. Geometrically, this means that the graph of y = g(x) should resemble the line y = x − 1 as x → ±∞. We see this play out both numerically and graphically below.
x g(x) x − 1 − 10 ≈ − 10. 6667 − 11 − 100 ≈ − 100. 9697 − 101 − 1000 ≈ − 1000. 9970 − 1001 − 10000 ≈ − 10000. 9997 − 10001
x g(x) x − 1 10 ≈ 8. 7273 9 100 ≈ 98. 9703 99 1000 ≈ 998. 9970 999 10000 ≈ 9998. 9997 9999
y = g(x) and y = x − 1 y = g(x) and y = x − 1 as x → −∞ as x → ∞
The way we symbolize the relationship between the end behavior of y = g(x) with that of the line y = x − 1 is to write ‘as x → ±∞, g(x) → x − 1.’ In this case, we say the line y = x − 1 is a slant asymptote^13 to the graph of y = g(x). Informally, the graph of a rational function has a slant asymptote if, as x → ∞ or as x → −∞, the graph resembles a non-horizontal, or ‘slanted’ line. Formally, we define a slant asymptote as follows.
Definition 4.4. The line y = mx + b where m 6 = 0 is called a slant asymptote of the graph of a function y = f (x) if as x → −∞ or as x → ∞, f (x) → mx + b.
A few remarks are in order. First, note that the stipulation m 6 = 0 in Definition 4.4 is what makes the ‘slant’ asymptote ‘slanted’ as opposed to the case when m = 0 in which case we’d have a horizontal asymptote. Secondly, while we have motivated what me mean intuitively by the notation ‘f (x) → mx+b,’ like so many ideas in this section, the formal definition requires Calculus. Another way to express this sentiment, however, is to rephrase ‘f (x) → mx + b’ as ‘f (x) − (mx + b) → 0.’ In other words, the graph of y = f (x) has the slant asymptote y = mx + b if and only if the graph of y = f (x) − (mx + b) has a horizontal asymptote y = 0.
(^12) See the remarks following Theorem 4.2. (^13) Also called an ‘oblique’ asymptote in some, ostensibly higher class (and more expensive), texts.
4.1 Introduction to Rational Functions 313
so we have that the slant asymptote y = x + 2 is identical to the graph of y = g(x) except at x = 2 (where the latter has a ‘hole’ at (2, 4).) The calculator supports this claim.^17
- For h(x) = x (^3) + x^2 − 4 , the degree of the numerator is 3 and the degree of the denominator is 2 so again, we are guaranteed the existence of a slant asymptote. The long division
x^3 + 1
( ÷
x^2 − 4
gives a quotient of just x, so our slant asymptote is the line y = x. The calculator confirms this, and we find that as x → −∞, the graph of y = h(x) approaches the asymptote from below, and as x → ∞, the graph of y = h(x) approaches the asymptote from above.
The graph of y = f (x) The graph of y = g(x) The graph of y = h(x)
The reader may be a bit disappointed with the authors at this point owing to the fact that in Exam- ples 4.1.2, 4.1.4, and 4.1.6, we used the calculator to determine function behavior near asymptotes. We rectify that in the next section where we, in excruciating detail, demonstrate the usefulness of ‘number sense’ to reveal this behavior analytically.
(^17) While the word ‘asymptote’ has the connotation of ‘approaching but not equaling,’ Definitions 4.3 and 4.4 invite the same kind of pathologies we saw with Definitions 1.11 in Section 1.6.
314 Rational Functions
4.1.1 Exercises
In Exercises 1 - 18, for the given rational function f :
- Find the domain of f.
- Identify any vertical asymptotes of the graph of y = f (x).
- Identify any holes in the graph.
- Find the horizontal asymptote, if it exists.
- Find the slant asymptote, if it exists.
- Graph the function using a graphing utility and describe the behavior near the asymptotes.
- f (x) =
x 3 x − 6 2.^ f^ (x) =
3 + 7x 5 − 2 x
- f (x) =
x x^2 + x − 12
- f (x) =
x x^2 + 1 5.^ f^ (x) =^
x + 7 (x + 3)^2 6.^ f^ (x) =^
x^3 + 1 x^2 − 1
- f (x) = 4 x x^2 + 4
- f (x) = 4 x x^2 − 4 9.^ f^ (x) =^
x^2 − x − 12 x^2 + x − 6
- f (x) =
3 x^2 − 5 x − 2 x^2 − 9
- f (x) =
x^3 + 2x^2 + x x^2 − x − 2
- f (x) =
x^3 − 3 x + 1 x^2 + 1
- f (x) = 2 x^2 + 5x − 3 3 x + 2 14. f (x) = −x^3 + 4x x^2 − 9 15. f (x) = − 5 x^4 − 3 x^3 + x^2 − 10 x^3 − 3 x^2 + 3x − 1
- f (x) =
x^3 1 − x
- f (x) =
18 − 2 x^2 x^2 − 9
- f (x) =
x^3 − 4 x^2 − 4 x − 5 x^2 + x + 1
- The cost C in dollars to remove p% of the invasive species of Ippizuti fish from Sasquatch Pond is given by C(p) =
1770 p 100 − p , 0 ≤ p < 100
(a) Find and interpret C(25) and C(95). (b) What does the vertical asymptote at x = 100 mean within the context of the problem? (c) What percentage of the Ippizuti fish can you remove for $40000?
- In Exercise 71 in Section 1.4, the population of Sasquatch in Portage County was modeled by the function P (t) =
150 t t + 15
where t = 0 represents the year 1803. Find the horizontal asymptote of the graph of y = P (t) and explain what it means.
316 Rational Functions
4.1.2 Answers
- f (x) =
x 3 x − 6 Domain: (−∞, 2) ∪ (2, ∞) Vertical asymptote: x = 2 As x → 2 −, f (x) → −∞ As x → 2 +, f (x) → ∞ No holes in the graph Horizontal asymptote: y = (^13) As x → −∞, f (x) → 13 − As x → ∞, f (x) → (^13)
- f (x) =
3 + 7x 5 − 2 x Domain: (−∞, 52 ) ∪ ( 52 , ∞) Vertical asymptote: x = (^52) As x → 52 −, f (x) → ∞ As x → (^52)
, f (x) → −∞ No holes in the graph Horizontal asymptote: y = − (^72) As x → −∞, f (x) → − (^72)
As x → ∞, f (x) → − (^72) −
- f (x) =
x x^2 + x − 12
x (x + 4)(x − 3) Domain: (−∞, −4) ∪ (− 4 , 3) ∪ (3, ∞) Vertical asymptotes: x = − 4 , x = 3 As x → − 4 −, f (x) → −∞ As x → − 4 +, f (x) → ∞ As x → 3 −, f (x) → −∞ As x → 3 +, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0 − As x → ∞, f (x) → 0 +
- f (x) =
x x^2 + 1 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0 − As x → ∞, f (x) → 0 +
- f (x) =
x + 7 (x + 3)^2 Domain: (−∞, −3) ∪ (− 3 , ∞) Vertical asymptote: x = − 3 As x → − 3 −, f (x) → ∞ As x → − 3 +, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 0 (^19) As x → −∞, f (x) → 0 − As x → ∞, f (x) → 0 +
- f (x) =
x^3 + 1 x^2 − 1
x^2 − x + 1 x − 1 Domain: (−∞, −1) ∪ (− 1 , 1) ∪ (1, ∞) Vertical asymptote: x = 1 As x → 1 −, f (x) → −∞ As x → 1 +, f (x) → ∞ Hole at (− 1 , − 32 ) Slant asymptote: y = x As x → −∞, the graph is below y = x As x → ∞, the graph is above y = x
(^19) This is hard to see on the calculator, but trust me, the graph is below the x-axis to the left of x = −7.
4.1 Introduction to Rational Functions 317
- f (x) = 4 x x^2 + 4 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0 − As x → ∞, f (x) → 0 +
- f (x) = 4 x x^2 − 4
4 x (x + 2)(x − 2) Domain: (−∞, −2) ∪ (− 2 , 2) ∪ (2, ∞) Vertical asymptotes: x = − 2 , x = 2 As x → − 2 −, f (x) → −∞ As x → − 2 +, f (x) → ∞ As x → 2 −, f (x) → −∞ As x → 2 +, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0 − As x → ∞, f (x) → 0 +
- f (x) =
x^2 − x − 12 x^2 + x − 6
x − 4 x − 2 Domain: (−∞, −3) ∪ (− 3 , 2) ∪ (2, ∞) Vertical asymptote: x = 2 As x → 2 −, f (x) → ∞ As x → 2 +, f (x) → −∞ Hole at
Horizontal asymptote: y = 1 As x → −∞, f (x) → 1 + As x → ∞, f (x) → 1 −
- f (x) =
3 x^2 − 5 x − 2 x^2 − 9
(3x + 1)(x − 2) (x + 3)(x − 3) Domain: (−∞, −3) ∪ (− 3 , 3) ∪ (3, ∞) Vertical asymptotes: x = − 3 , x = 3 As x → − 3 −, f (x) → ∞ As x → − 3 +, f (x) → −∞ As x → 3 −, f (x) → −∞ As x → 3 +, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 3 As x → −∞, f (x) → 3 + As x → ∞, f (x) → 3 −
- f (x) =
x^3 + 2x^2 + x x^2 − x − 2
x(x + 1) x − 2 Domain: (−∞, −1) ∪ (− 1 , 2) ∪ (2, ∞) Vertical asymptote: x = 2 As x → 2 −, f (x) → −∞ As x → 2 +, f (x) → ∞ Hole at (− 1 , 0) Slant asymptote: y = x + 3 As x → −∞, the graph is below y = x + 3 As x → ∞, the graph is above y = x + 3
- f (x) =
x^3 − 3 x + 1 x^2 + 1 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Slant asymptote: y = x As x → −∞, the graph is above y = x As x → ∞, the graph is below y = x
4.1 Introduction to Rational Functions 319
- The horizontal asymptote of the graph of P (t) = (^) t^150 +15t is y = 150 and it means that the model predicts the population of Sasquatch in Portage County will never exceed 150.
- (a) C(x) = 100 x+2000 x , x > 0. (b) C(1) = 2100 and C(100) = 120. When just 1 dOpi is produced, the cost per dOpi is $2100, but when 100 dOpis are produced, the cost per dOpi is $120. (c) C(x) = 200 when x = 20. So to get the cost per dOpi to $200, 20 dOpis need to be produced. (d) As x → 0 +, C(x) → ∞. This means that as fewer and fewer dOpis are produced, the cost per dOpi becomes unbounded. In this situation, there is a fixed cost of $ (C(0) = 2000), we are trying to spread that $2000 over fewer and fewer dOpis. (e) As x → ∞, C(x) → 100 +. This means that as more and more dOpis are produced, the cost per dOpi approaches $100, but is always a little more than $100. Since $100 is the variable cost per dOpi (C(x) = 100x + 2000), it means that no matter how many dOpis are produced, the average cost per dOpi will always be a bit higher than the variable cost to produce a dOpi. As before, we can attribute this to the $2000 fixed cost, which factors into the average cost per dOpi no matter how many dOpis are produced.
- (a)
(b) The maximum power is approximately 1. 603 mW which corresponds to 3. 9 kΩ. (c) As x → ∞, P (x) → 0 +^ which means as the resistance increases without bound, the power diminishes to zero.
320 Rational Functions
4.2 Graphs of Rational Functions
In this section, we take a closer look at graphing rational functions. In Section 4.1, we learned that the graphs of rational functions may have holes in them and could have vertical, horizontal and slant asymptotes. Theorems 4.1, 4.2 and 4.3 tell us exactly when and where these behaviors will occur, and if we combine these results with what we already know about graphing functions, we will quickly be able to generate reasonable graphs of rational functions.
One of the standard tools we will use is the sign diagram which was first introduced in Section 2.4, and then revisited in Section 3.1. In those sections, we operated under the belief that a function couldn’t change its sign without its graph crossing through the x-axis. The major theorem we used to justify this belief was the Intermediate Value Theorem, Theorem 3.1. It turns out the Intermediate Value Theorem applies to all continuous functions,^1 not just polynomials. Although rational functions are continuous on their domains,^2 Theorem 4.1 tells us that vertical asymptotes and holes occur at the values excluded from their domains. In other words, rational functions aren’t continuous at these excluded values which leaves open the possibility that the function could change sign without crossing through the x-axis. Consider the graph of y = h(x) from Example 4.1.1, recorded below for convenience. We have added its x-intercept at
2 ,^0
for the discussion that follows. Suppose we wish to construct a sign diagram for h(x). Recall that the intervals where h(x) > 0, or (+), correspond to the x-values where the graph of y = h(x) is above the x-axis; the intervals on which h(x) < 0, or (−) correspond to where the graph is below the x-axis.
x
y
1 2 − 4 − 3 − 2 1 2 3 4 − 1 − 2 − 3 − 4 − 5 − 6
1
3
4
5
6
7
8
As we examine the graph of y = h(x), reading from left to right, we note that from (−∞, −1), the graph is above the x-axis, so h(x) is (+) there. At x = −1, we have a vertical asymptote, at which point the graph ‘jumps’ across the x-axis. On the interval
, the graph is below the (^1) Recall that, for our purposes, this means the graphs are devoid of any breaks, jumps or holes (^2) Another result from Calculus.
