Rational Functions
Definition of Rational Function
A rational function is any function which can be specified by a rule written as $ f(x) = $ , where $ P(x) $ and $ Q(x) $ are polynomial functions. The domain of a rational function is the set of all real numbers for which $ Q(x) $ . The assumption is normally made that the rational expression $ P(x)/Q(x) $ is in lowest terms, that is, $ P(x) $ and $ Q(x) $ have no factors in common. (See below for analysis of cases where this assumption is not made.)
EXAMPLE 15.1 $ f(x)= $ , $ g(x)= $ , $ h(x)= $ , and $ k(x)= $ , are examples of rational functions. The domains are, respectively, for $ f,{xx} $ , for g, $ {xx} $ , for h, $ {xx,2,-3} $ , and for k, R (since the denominator polynomial is never 0).
Graph of a Rational Function
The graph of a rational function is analyzed in terms of the symmetry, intercepts, asymptotes, and sign behavior of the function.
If $ Q(x) $ has no real zeros, the graph of $ P(x)/Q(x) $ is a smooth curve for all real x.
If $ Q(x) $ has real zeros, the graph of $ P(x)/Q(x) $ consists of smooth curves on each open interval that does not include a zero. The graph has vertical asymptotes at each zero of $ Q(x) $ .
V ertical Asymptotes
The line x = a is a vertical asymptote for the graph of a function f if, as x approaches a through values that are greater than or less than a, the value of the function grows beyond all bounds, either positive or negative. The cases are shown in the following table, along with the notation generally used:
| NOTATION | MEANING | GRAPH |
| $ _{x a} f(x) = $ | As x approaches a from the left, $ f(x) $ is positive and increases beyond all bounds. | y |
| NOTATION | MEANING | GRAPH |
| $ _{x a} f(x) = -$ | As x approaches a from the left, $ f(x) $ is negative and decreases beyond all bounds. |
|
| $ _{x a^{+}} f(x) = $ | As x approaches a from the right, $ f(x) $ is positive and increases beyond all bounds. |
|
| $ _{x a^{-}} f(x) = -$ | As x approaches a from the right, $ f(x) $ is negative and decreases beyond all bounds. | y |
| x | 1 | 1.9 | 1.99 | 1.999 | 3 | 2.1 | 2.01 | 2.001 |
| y | -3 | -30 | -300 | -3000 | 3 | 30 | 300 | 3000 |
Clearly, as x approaches 2 from the left, $ f(x) $ is negative and decreases beyond all bounds, and, as x approaches 2 from the right, $ f(x) $ is positive and increases beyond all bounds, that is, $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=$ . Thus x=2 is a vertical asymptote for the graph.
Horizontal Asymptotes
The line y = a is a horizontal asymptote for the graph of a function f if, as x grows beyond all bounds, either positive or negative, $ f(x) $ approaches the value a. The cases are shown in the following table, along with the notation generally used:
| NOTATION | MEANING | FIGURE 15-8 |
| $ _{x } f(x) = a $ | As x increases beyond all bounds, $ f(x) $ approaches the value a. [In the figure, $ f(x) < a $ for large positive values of x.] |
|
| $ _{x } f(x) = a $ | As x increases beyond all bounds, $ f(x) $ approaches the value a. [In the figure, $ f(x) > a $ for large positive values of x.] |
|
| $ _{x } f(x) = a $ | As x decreases beyond all bounds, $ f(x) $ approaches the value a. [In the figure, $ f(x) < a $ for large negative values of x.] |
|
| $ _{x } f(x) = a $ | As x decreases beyond all bounds, $ f(x) $ approaches the value a. [In the figure, $ f(x) > a $ for large negative values of x.] | [F1] |
Finding Horizontal Asymptotes
Let
\[ f(x)=\frac{P(x)}{Q(x)}=\frac{a_{n}x^{n}+\cdots+a_{1}x+a_{0}}{b_{m}x^{m}+\cdots+b_{1}x+b_{0}} \]
with $ a_{n} $ and $ b_{m} $ . Then
If n < m, the x axis is a horizontal asymptote for the graph of f.
If n = m, the line $ y = a_{n}/b_{m} $ is a horizontal asymptote for the graph of f.
If n > m, there is no horizontal asymptote for the graph of f. Instead, as $ x $ and as $ x -$ , either $ f(x) $ or $ f(x) -$ .
EXAMPLE 15.3 Find the horizontal asymptotes, if any, for $ f(x) = $ .
Since the numerator and denominator both have degree 1, the quotient can be written as
\[ f(x)=\frac{2x+1}{x}\div\frac{x-5}{x}=\frac{2+\frac{1}{x}}{1-\frac{5}{x}} \]
For large positive or negative values of x, this is very close to $ $ , the ratio of the leading coefficients, thus $ f(x) $ . The line y = 2 is a horizontal asymptote.
Oblique Asymptotes
Let
\[ f(x)=\frac{P(x)}{Q(x)}=\frac{a_{n}x^{n}+\cdots+a_{1}x+a_{0}}{b_{m}x^{m}+\cdots+b_{1}x+b_{0}} \]
with $ a_{n} $ and $ b_{m} $ . Then, if $ n = m + 1 $ , $ f(x) $ can be expressed using long division (see Chapter 14) in the form:
\[ f(x)=ax+b+\frac{R(x)}{Q(x)} \]
where the degree of $ R(x) $ is less than the degree of $ Q(x) $ . Then, as $ x $ or $ x -$ , $ f(x) ax + b $ and the line $ y = ax + b $ is an oblique asymptote for the graph of the function.
EXAMPLE 15.4 Find the oblique asymptote for the graph of the function $ f(x) = $ .
Use the long division scheme to write $ f(x) = x - 1 + $ . Hence, as $ x $ or $ x -$ , $ f(x) x - 1 $ , and the line y = x - 1 is an oblique asymptote for the graph of the function.
Graphing a Rational Function
To sketch the graph of a rational function $ y = f(x) = $ :
Find any x-intercepts for the graph [the real zeros of $ P(x) $ ] and plot the corresponding points. Find the y-intercept $ [f(0), 0 f] $ and plot the point $ (0, f(0)) $ . Analyze the function for any symmetry with respect to the axes or the origin.
Find any real zeros of $ Q(x) $ and enter any vertical asymptotes for the graph on the sketch.
Find any horizontal or oblique asymptote for the graph and enter this on the sketch.
Determine whether the graph intersects the horizontal or oblique asymptote. The graphs of $ y = f(x) $ and $ y = ax + b $ will intersect at real solutions of $ f(x) = ax + b $ .
Determine, from a sign chart if necessary, the intervals in which the function is positive and negative. Determine the behavior of the function near the asymptotes.
Sketch the graph of f in each of the regions found in step 5.
EXAMPLE 15.5 Sketch the graph of the function $ f(x) = -12/x $
The graph has no x-intercepts or y-intercepts. Since $ f(-x) = -f(x) $ , the function is odd and the graph has origin symmetry.
Since 0 is the only zero of the denominator, the y-axis, x = 0, is the only vertical asymptote.
Since the degree of the denominator is greater than the degree of the numerator, the x-axis, y = 0, is the horizontal asymptote.
Since there is no solution to the equation $ -12/x = 0 $ , the graph does not intersect the horizontal asymptote.
If x is negative, $ f(x) $ is positive. If x is positive, $ f(x) $ is negative. Hence, $ {x ^{-}} f(x) = $ and $ {x ^{+}} f(x) = -$ .
Sketch the graph (Fig. 15-9).
SOLVED PROBLEMS
15.1. Find any vertical asymptotes for the graph of
- $ f(x) = $
\[ f(x)=\frac{2x}{x^{2}+4} \]
\[ f(x)=\frac{2x-1}{x^{2}-x-2} \]
$ f(x) = $
Since the real zeros of x2 − 4 are ±2, the vertical asymptotes are x = ±2.
Since $ x^{2} + 4 $ has no real zeros, there are no vertical asymptotes.
Since the real zeros of $ x^{2}-x-2 $ are 2 and -1, the vertical asymptotes are x=2 and x=-1.
Since the only real zero of $ x^{3} + 8 $ is -2, the only vertical asymptote is x = -2.
15.2. Find any vertical asymptotes for the graph of $ f(x)= $
It would seem as if the graph has vertical asymptotes $ x = $ , since these are the real zeros of the denominator polynomial. However, the expression for the function is not in lowest terms, in fact,
\[ f(x)=\frac{x(x-1)}{(x+1)(x-1)}=\frac{x}{x+1}\mathrm{if}x\neq1 \]
Since, as $ x ^{+} $ or $ x ^{-} $ , the function value does not increase or decrease beyond all bounds, the line x = 1 is not a vertical asymptote, and the only vertical asymptote is x = -1.
15.3. Find any horizontal asymptotes for the graph of
$ f(x) = $ (b) $ f(x) = $ (c) $ f(x) = $ (d) $ f(x) = $
Since numerator and denominator both have degree 2, the quotient can be written as
\[ f(x)=\frac{4x^{2}}{x^{2}}\div\frac{x^{2}+4}{x^{2}}=\frac{4}{1+\frac{4}{x^{2}}} \]
For large positive or negative values of x, this is very close to $ $ , the ratio of the leading coefficients, thus $ f(x) $ . The line y = 4 is a horizontal asymptote.
Since the degree of the numerator is greater than the degree of the denominator, the graph has no horizontal asymptote.
Since the degree of the numerator is less than the degree of the denominator, the x-axis, y = 0, is the horizontal asymptote.
Since numerator and denominator both have degree 2, the quotient can be written as
\[ f(x)=\frac{3x^{2}+5x+2}{x^{2}}\div\frac{4x^{2}+1}{x^{2}}=\frac{3+\frac{5}{x}+\frac{2}{x^{2}}}{4+\frac{1}{x^{2}}} \]
For large positive or negative values of x, this is very close to $ $ , the ratio of the leading coefficients; thus $ f(x) $ . The line $ y = $ is a horizontal asymptote.
15.4. Find any oblique asymptotes for the graph of
$ f(x)= $ (b) $ f(x)= $ (c) $ f(x)= $ (d) $ f(x)= $
Use the synthetic division scheme to write $ f(x) = x - 4 + $ . Hence, as $ x $ or $ x -$ , $ f(x) x - 4 $ , and the line y = x - 4 is an oblique asymptote for the graph of the function.
Since the degree of the numerator is not equal to 1 more than the degree of the denominator, the graph does not have an oblique asymptote. However, if the synthetic division scheme is used to write
\[ f(x)=x^{2}-4x+16+\frac{-64}{x+4} \]
then, as $ x $ or $ x -$ , $ f(x) x^{2} - 4x + 16 $ . The graph of f then is said to approach asymptotically the curve $ y = x^{2} - 4x + 16 $ .
- Use the long division scheme to write $ f(x) = x - + $ . Hence, as $ x $ or $ x -$ ,
$ f(x) x - $ , and the line $ y = x - $ is an oblique asymptote for the graph of the function.
- Use the long division scheme to write $ f(x) = 2x - 4 + $ . Hence, as $ x $ or $ x -$ ,
$ f(x) 2x - 4 $ , and the line y = 2x - 4 is an oblique asymptote for the graph of the function.
15.5. Sketch a graph of $ f(x) = $
Apply the steps listed above for sketching the graph of a rational function.
Since $ f(0) = 2 $ , the y-intercept is 2. Since $ f(x) $ is never 0, there is no x-intercept. The graph has no symmetry with respect to axes or origin.
Since $ x + 2 = 0 $ when x = -2, this line is the only vertical asymptote.
Since the degree of the denominator is greater than the degree of the numerator, the x-axis is the horizontal asymptote.
Since $ f(x) = 0 $ has no solutions, the graph does not cross its horizontal asymptote.
A sign chart shows that the values of the function are negative on $ (-∞, -2) $ and positive on $ (-2, ∞) $ . Thus, $ {x ^{-}} f(x) = -$ and $ {x ^{+}} f(x) = $ .
The graph is shown in Fig. 15-10.
15.6. Sketch a graph of $ f(x) = - $ .
The graph has no x- intercepts or y-intercepts. Since $ f(-x) = f(x) $ , the function is even and the graph has y-axis symmetry.
Since $ x^{2}=0 $ when x=0, the y-axis is the only vertical asymptote.
Since the degree of the denominator is greater than the degree of the numerator, the x-axis is the horizontal asymptote.
Since $ f(x) = 0 $ has no solutions, the graph does not cross its horizontal asymptote.
Since $ x^{2} $ is never negative, the function values are negative throughout the domain. Thus, $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=-$ .
The graph is shown in Fig. 15-11.
15.7. Sketch the graph of $ f(x)= $
Since $ f(0) = - $ , the y-intercept is $ - $ . Since $ f(x) = 0 $ if x = -3, the x-intercept is -3. The graph has no symmetry with respect to axes or origin.
Since x - 2 = 0 when x = 2, this line is the only vertical asymptote.
Since the numerator and denominator both have degree 1, and the ratio of leading coefficients is $ $ , or 1, the line y = 1 is the horizontal asymptote.
Since $ f(x) = 1 $ has no solutions, the graph does not cross its horizontal asymptote.
A sign chart shows that the values of the function are positive on $ (-∞,-3) $ and $ (2,∞) $ and negative on $ (-3,2) $ . Thus, $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=$ .
The graph is shown in Fig. 15-12.
15.8. Sketch the graph of $ f(x)= $
Since $ f(0)=0 $ , and this is the only zero of the function, the x-intercept and the y-intercept are both 0, that is, the graph passes through the origin. Since $ f(-x)=-f(x) $ , the function is odd and the graph has origin symmetry.
Since $ x^{2}-4=0 $ when $ x= $ , these lines are vertical asymptotes for the graph.
Since the degree of the denominator is greater than the degree of the numerator, the x-axis is the horizontal asymptote.
Since $ f(x) = 0 $ has the solution 0, the graph crosses its horizontal asymptote at the origin.
A sign chart shows that the values of the function are positive on both $ (-2,0) $ and $ (2,) $ and negative on both $ (-,-2) $ and $ (0,2) $ . Thus, $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=$ ; also, $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=$ . The graph is shown in Fig. 15-13.
15.9. Sketch the graph of $ f(x)= $
Since $ f(0)=0 $ , the x-intercept and the y-intercept are both 0, that is, the graph passes through the origin. Since $ f(-x)=f(x) $ , the function is even and the graph has y-axis symmetry.
Since $ x^{2}-4=0 $ when $ x= $ , these lines are vertical asymptotes for the graph.
Since the numerator and denominator both have degree 2, and the ratio of leading coefficients is $ - $ , or -2, the line y = -2 is the horizontal asymptote.
Since $ f(x) = -2 $ has no solutions, the graph does not cross its horizontal asymptote.
A sign chart shows that the values of the function are positive on $ (-2,2) $ and negative on $ (-,-2) $ and $ (2,) $ . Thus, $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=$ , also $ {x^{-}}f(x)=$ and $ {x^{+}}f(x)=-$ .
Moreover, since the behavior near the asymptote x = 2 shows that the function values are large and negative for x greater than 2, and since the graph does not cross its horizontal asymptote, the graph must therefore approach the horizontal asymptote from below for large positive x. The behavior for large negative x is the same, since the function is even.
The graph is shown in Fig. 15-14.
15.10. Sketch the graph of $ f(x)= $ .
Since $ f(0)=0 $ , and this is the only zero of the function, the x-intercept and the y-intercept are both 0; that is, the graph passes through the origin. Since $ f(-x)=-f(x) $ , the function is odd and the graph has origin symmetry. Since $ x^{2}-4=0 $ when $ x= $ , these lines are vertical asymptotes for the graph.
Since the degree of the numerator is 1 more than the degree of the denominator, the graph has an oblique asymptote. Long division shows that
\[ f(x)=\frac{x^{3}}{x^{2}-4}=x+\frac{4x}{x^{2}-4} \]
Thus, as \(x \to \infty\), \(f(x) \to x\), and the line \(y = x\) is the oblique asymptote.
Since $ f(x) = x $ has the solution 0, the graph crosses the oblique asymptote at the origin.
A sign chart shows that the values of the function are positive on $ (-2,0) $ and $ (2,) $ and negative on $ (-,-2) $ and $ (0,2) $ : $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=$ ; also, $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=$ .
Moreover, since the behavior near the asymptote x = 2 shows that the function values are large and positive for x greater than 2, and since the graph does not cross its oblique asymptote here, the graph must therefore approach the oblique asymptote from above for large positive x. Since the function is odd, the graph must therefore approach the oblique asymptote from below for large negative x.
The graph is shown in Fig. 15-15.
15.11. Sketch a graph of $ f(x)= $
Since $ f(0)=0 $ , the y-intercept is 0. Since $ x^{2}+x=0 $ when x=0 and -1, these are both x-intercepts. The graph passes through the origin. There is no obvious symmetry.
Since $ x^{2}-3x+2=0 $ when x=1 and 2, these lines are vertical asymptotes.
Since the numerator and denominator both have degree 2, and the ratio of leading coefficients is $ $ , or 1, the line y = 1 is the horizontal asymptote.
Since $ f(x) = 1 $ has the solution $ $ , the graph crosses the horizontal asymptote at $ (, 1) $ .
A sign chart shows that the values of the function are negative on $ (-1,0) $ and $ (1,2) $ , and positive on $ (-,-1) $ , $ (0,1) $ , and $ (2,) $ . Thus, $ {x^{-}}f(x)=$ and $ {x^{+}}f(x)=-$ ;
also, $ {x^{-}}f(x)=-$ and $ {x^{+}}f(x)=$ .
Moreover, since the behavior near the asymptote x = 2 shows that the function values are large and positive for x greater than 2, and since the graph does not cross its horizontal asymptote here, the graph must therefore approach the horizontal asymptote from above for large positive x. Similarly, the graph must approach the horizontal asymptote from below for large negative x.
The graph is shown in Fig. 15-16.
15.12. Sketch a graph of $ f(x)= $
Since $ f(0) = - $ , the y-intercept is $ - $ . Since $ x^{2} - 9 = 0 $ when $ x = $ , these are both x-intercepts. Since $ f(-x) = f(x) $ , the function is even and the graph has y-axis symmetry.
Since $ x^{2} + 4 $ has no real zeros, the graph has no vertical asymptotes.
Since the numerator and denominator both have degree 2, and the ratio of leading coefficients is $ $ , or 1, the line y = 1 is the horizontal asymptote.
Since $ f(x) = 1 $ has no solutions, the graph does not cross the horizontal asymptote.
A sign chart shows that the values of the function are positive on $ (-∞, -3) $ and $ (3, ∞) $ and negative on $ (-3, 3) $ . The graph is shown in Fig. 15-17.
15.13. To understand an example of the special case when the numerator and denominator of a rational expression have factors in common, analyze and sketch a graph of $ f(x) = $ .
Factoring numerator and denominator yields
\[ f(x)=\frac{(x-2)(x+2)}{(x-2)(x-1)}=\frac{x+2}{x-1}\quad for x\neq2 \]
Thus, the graph of the function is identical with the graph of $ g(x) = (x + 2)/(x - 1) $ , except that 2 is not in the domain of f. Draw a graph of $ y = g(x) $ . The graph of $ y = f(x) $ is conventionally shown as the graph of g with a small circle centered at (2,4) to indicate that this point is not on the graph.
Since $ g(0) = -2 $ , the y-intercept is -2. Since $ g(x) = 0 $ if x = -2, the x-intercept is -2. The graph has no symmetry with respect to axes or origin.
Since x - 1 = 0 when x = 1, this line is the only vertical asymptote.
Since the numerator and denominator both have degree 1, and the ratio of leading coefficients is $ $ , or 1, the line y = 1 is the horizontal asymptote.
Since $ g(x) = 1 $ has no solutions, the graph does not cross its horizontal asymptote.
A sign chart shows that the values of the function are positive on $ (-∞, -2) $ and $ (1, ∞) $ and negative on $ (-2, 1) $ . Thus, $ {x ^{-}} g(x) = -$ and $ {x ^{+}} g(x) = $ .
The graph is shown in Fig. 15-18.
SUPPLEMENTARY PROBLEMS
15.14. Find all intercepts for the graphs of the following rational functions:
Ans. (a) x-intercept: 0, y-intercept: 0; (b) x-intercept: $ $ , y-intercept: $ - $ ;
- x-intercept: 1, y-intercept: none; (d) x-intercept: -3, y-intercept: $ $
15.15. Find all horizontal and vertical asymptotes for the graphs in the previous problem.
horizontal: y = 2, vertical: x = -4; (b) horizontal: y = 4, vertical: none;
horizontal: y = 0, vertical: x = 0, x = 4; (d) horizontal: y = 0, vertical: $ x = $ , $ x = $
15.16. (a) State intercepts and asymptotes and sketch the graph of $ f(x) = $
- Show that f is one-to-one on its domain and that $ f(x) = f^{-1}(x) $ .
Ans. (a) Intercepts: the origin.
Asymptotes: x = 2, y = 2.
The graph is shown in Fig. 15-19.
15.17. State intercepts and asymptotes and sketch the graph of $ f(x) = $ .
Ans. Intercepts: the origin.
Asymptotes: x = 2, $ y = 2x + 4 $ .
The graph is shown in Fig. 15-20.
15.18. State intercepts and asymptotes and sketch the graph of $ f(x) = $ .
Ans. Intercepts: $ (0,) $ .
Asymptotes: x = 2, y = 0.
The graph is shown in Fig. 15-21.
15.19. State intercepts and asymptotes and sketch the graph of $ f(x) = $ .
Ans. Intercepts: $ (0,-2) $ .
Asymptotes: $ x = $ , y = 0.
The graph is shown in Fig. 15-22.
15.20. Find all vertical and oblique asymptotes for the graphs of the following rational functions:
\[ f(x)=\frac{x^{2}}{x+2};(b)f(x)=\frac{x^{2}-4x}{x-1};(c)f(x)=\frac{8x^{3}-1}{x^{2}+4};(d)f(x)=\frac{x^{4}-5x^{2}+6}{x^{3}+x^{2}};(e)f(x)=\frac{x^{3}-2x}{x+6} \]
Ans. (a) vertical: x = -2, oblique: y = x - 2; (b) vertical: x = 1, oblique: y = x - 3;
vertical: none, oblique: y = 8x; (d) vertical: x = 0, x = -1, oblique: y = x - 1;
vertical: x = -6, oblique: none, however, the graph approaches asymptotically the graph of $ y = x^{2} - 6x + 34 $
15.21. State intercepts and asymptotes and sketch the graph of $ f(x) = $ .
Ans. Intercepts: the origin.
Asymptotes: $ x = $ , y = x.
The graph is shown in Fig. 15-23.
15.22. State intercepts and asymptotes and sketch the graph of $ f(x)= $
Ans. Intercepts: the origin.
Asymptotes: y = 0.
The graph is shown in Fig. 15-24.
15.23. State intercepts and asymptotes and sketch the graph of $ f(x)= $ .
Ans. Intercepts: (0,1), (1,0), (-1,0).
Asymptotes: y = x - 1.
The graph is shown in Fig. 15-25.
15.24. A field is to be marked off in the shape of a rectangle of area 144 square feet.
Write an expression for the perimeter P as a function of the length x.
Sketch a graph of the perimeter function and determine approximately from the graph the dimensions for which the perimeter is a minimum.
Ans. (a) $ P(x) = 2x + $
- See Fig. 15-26. Dimensions: 12 feet by 12 feet.
