ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Chapter 9 Rational Functions.

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Chapter 9 Rational Functions Section 9.1Section 9.1 Rational Functions Section 9.2Section 9.2 Long-Run Behavior of Rational Functions Section 9.3Section 9.3 Putting a Rational Function in Quotient Form

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 1 Which of the following is a polynomial? Solution (a) We have (because x 4 − 1 is a difference of squares) This is a quadratic polynomial. (b) In this case, since x 4 + 1 cannot be factored, the denominator cannot be divided into the numerator. Thus, the expression cannot be simplified and written as a polynomial. Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. We define a rational expression to be any expression equivalent to the ratio of two polynomials (except where the expressions are undefined). Rational Expressions Section 9.1 If we add, subtract, or multiply two polynomials, the result is always a polynomial. However, when we divide one polynomial by another, the result is not necessarily a polynomial, just as when we divide one integer by another, the result is not necessarily an integer.

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 2ab Write each expression as the ratio of two polynomials. Solution (a) We put each term over a common denominator and add: (b) We put each term over a common denominator and add : Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 2c Write each expression as the ratio of two polynomials. Solution (c) Since both the numerator and denominator involve algebraic fractions with denominator x, we first multiply them by x to simplify them: Section 9.1, X ≠ 1.

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 2d Write each expression as the ratio of two polynomials. Solution (d) We have Here, the denominator is 1 (a constant polynomial). Any polynomial is a rational expression, because we can think of it as a ratio with denominator equal to the constant polynomial 1. Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Just as we define polynomial functions using polynomial expressions, we define rational functions using rational expressions: Rational Functions A rational function is a function that can be put in the form where a(x) and b(x) are polynomials, and b(x) is not the zero polynomial. A rational function is a function that can be put in the form where a(x) and b(x) are polynomials, and b(x) is not the zero polynomial. Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 3a Let f(t) = 50 + 0.1t represent the population, in millions, of a country in year t, and let g(t) = 200 + 0.05t + 0.03t 2 represent the gross domestic product (GDP), in billions of dollars. The per capita GDP, h(t), is the GDP divided by the population. (a) Show that h(t) is a rational function. Solution (a) We have Since h(t) can be defined using a rational expression in t, then h(t) is a rational function of t. Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 3b Let f(t) = 50 + 0.1t represent the population, in millions, of a country in year t, and let g(t) = 200 + 0.05t + 0.03t 2 represent the gross domestic product (GDP), in billions of dollars. The per capita GDP, h(t), is the GDP divided by the population. (b) Evaluate h(0) and h(20). What does this tell us about the country’s economy? Solution (b) We have We see that after 20 years, the GDP, when divided by the population, rises from $4,000 per person to $4,096 per person. This suggests that on average, the people living in the country have become slightly more prosperous over time. Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Figure 9.1: Figure 9.1 shows the graph of the rational function This graph has many features that graphs of polynomials do not have. It levels off to the horizontal line y = 1 as the value of x gets large (either in a positive or negative direction). This line is called a horizontal asymptote. Furthermore, on either side of the two vertical lines, x = −1 and x = 2, the y- values get very large (either in a positive or negative direction). These lines are called vertical asymptotes. The Graph of a Rational Function Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Just as with polynomials, many aspects of the behavior of rational functions are determined by the factors of the numerator and denominator. Using the Factored Form of a Rational Function Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Solution A fraction is zero provided its numerator is zero. Putting the numerator a(x) = x 2 − x − 6 in factored form (x − 3)(x + 2) we see that it has zeros at x = 3, −2. Finding the Zeros of a Rational Function: Factoring the Numerator Using the Factored Form of a Rational Function In Figure 9.1, the two x-intercepts, at x = −2 and x = 3, correspond to the zeros of f. For polynomials, x-intercepts can be found by factoring. For rational functions, they can be found by factoring the numerator. Example 4 Find the zeros of algebraically. Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Finding the Domain of a Rational Function: Factoring the Denominator The dashed vertical lines in Figure 9.1 tell us that the graph has vertical asymptotes at x = −1 and x = 2. The function f is undefined at these values, so they are not in its domain. Since fractions are undefined only when the denominator is zero, we look for such values by factoring the denominator. Section 9.1 Figure 9.1:

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Solution The numerator, a(x) = x 2 − x − 6, and denominator, b(x) = x 2 − x − 2, are defined everywhere. However, their ratio is undefined if b(x) = 0. Putting b(x) = x 2 − x − 2 into factored form (x − 2)(x + 1) we see that its zeros are 2 and −1, so f(x) is undefined when x = 2 and x = −1. Finding the Domain of a Rational Function: Factoring the Denominator (continued) Example 5 Show that the domain of is all values of x except x = −1 and x = 2. Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. What Causes Vertical Asymptotes? The graph in Figure 9.1 is very steep near the vertical asymptotes. The rapid rise (or fall) of the graph of a rational function near a vertical asymptote is because the denominator becomes small but the numerator does not. Table 9.1 shows this for the asymptote at x = 2. As x gets close to 2, two things happen: The value of the numerator gets close to −4, and the value of the denominator gets close to 0. Thus, the value of y grows larger and larger (either positive or negative), since we are dividing a number nearly equal to −4 by a number nearly equal to 0. Table 9.1 Section 9.1 xNumeratorDenominatorf(x)f(x) 1.900-4.290-0.29014.793 1.990-4.030-0.030134.779 1.999-4.003-0.0031334.778 2.000-4.0000.000undefined 2.001-3.9970.003-1331.889 2.010-3.9700.030-131.890 2.100-3.6900.310-11.903

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Solution We have The expression (x 2 − 5x + 6)/(x 2 − 4) has the same value as the expression (x − 3)/(x + 2), except at x = 2. This is because the latter expression is defined at x = 2, whereas the former is not. This means f and g have different domains: domain is all values except x = ±2 domain is all values except x = −2. (Solution continued on next slide.) Holes in the Graph of a Rational Function Example 6 Cancel the common factor in the expression for Is the resulting expression equivalent to the original expression? What does the answer say about the graph of f? Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Thus, these functions are not exactly the same, even though their graphs are almost identical. See Figures 9.2 and 9.3. Notice that graph of f in Figure 9.2 does not have a vertical asymptote at x = 2, even though the denominator is zero there. Rather it has a hole, where f is undefined. Figure 9.2: Graph ofFigure 9.3: Graph of f(x) = (x 2 − 5x + 6)/(x 2 − 4)g(x) = (x − 3)/(x + 2) Example 6 (continued) Holes in the Graph of a Rational Function (continued) Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 7 Cancel common factors in each of the following rational expressions. Is the resulting expression equivalent to the original one? Solution (a) We haveprovided x ≠ 1. The resulting expression is not equivalent to x − 1, because the original expression is undefined at x = 1. (b)We have Canceling the 2 is valid no matter what the value of x, so the two expressions are equivalent. (c)We have Since x 2 + 1 is positive for all values of x, cancelation does not involve possible division by 0, so the two expressions are equivalent. Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Not All Rational Functions Have Vertical Asymptotes It is tempting to assume that the graph of any rational function has a vertical asymptote. However, this is not always the case. For instance, the graph of in Figure 9.4 has no vertical asymptote. Because the denominator is everywhere positive, the function’s value never “blows up” as it would if we were dividing by zero. It is tempting to assume that the graph of any rational function has a vertical asymptote. However, this is not always the case. For instance, the graph of in Figure 9.4 has no vertical asymptote. Because the denominator is everywhere positive, the function’s value never “blows up” as it would if we were dividing by zero. Figure 9.4: Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. 9.1 RATIONAL FUNCTIONS Key Points Rational expressions Zeros and asymptotes of rational functions The graphs of rational functions Holes in the graphs of rational functions Section 9.1

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. The Quotient Form of a Rational Expression A rational number can be thought of as the numerator divided by the denominator. For example, since 4 goes into 13 three times with a remainder of 1, we have The form on the right is useful for estimating the size of the number. Similarly, the rational function can be thought of as dividing x − 2 into x 2 − 5x + 7. Since x 2 − 5x + 7 = x 2 − 5x + 6 + 1 = (x − 3)(x − 2) + 1, we see that the number of times x − 2 goes into x 2 − 5x + 7 is x − 3 with a remainder of 1, so The form on the right is the analog for rational functions of the mixed- fraction form for rational numbers. Just as the remainder has to be less than the divisor when we divide integers, the remainder in polynomial division has to either have degree less than the degree of the divisor or be the zero polynomial. A rational number can be thought of as the numerator divided by the denominator. For example, since 4 goes into 13 three times with a remainder of 1, we have The form on the right is useful for estimating the size of the number. Similarly, the rational function can be thought of as dividing x − 2 into x 2 − 5x + 7. Since x 2 − 5x + 7 = x 2 − 5x + 6 + 1 = (x − 3)(x − 2) + 1, we see that the number of times x − 2 goes into x 2 − 5x + 7 is x − 3 with a remainder of 1, so The form on the right is the analog for rational functions of the mixed- fraction form for rational numbers. Just as the remainder has to be less than the divisor when we divide integers, the remainder in polynomial division has to either have degree less than the degree of the divisor or be the zero polynomial. Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. The Quotient Form of a Rational Expression (continued) Quotient Form of a Rational Expression The quotient form of a rational expression is The polynomial q is the quotient. The polynomial r is called the remainder, and must have smaller degree than q or be the zero polynomial. The term r(x)/b(x) in the quotient form is called the remainder term. We have a(x) = q(x)b(x) + r(x), so we can think of this as saying that the number of times b(x) goes into a(x) is q(x), with remainder r(x). Quotient Form of a Rational Expression The quotient form of a rational expression is The polynomial q is the quotient. The polynomial r is called the remainder, and must have smaller degree than q or be the zero polynomial. The term r(x)/b(x) in the quotient form is called the remainder term. We have a(x) = q(x)b(x) + r(x), so we can think of this as saying that the number of times b(x) goes into a(x) is q(x), with remainder r(x). Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. For example, we saw above that with q(x) = x − 3 and r(x) = 1. Here the degree of r(x) = 1 is 0, and the degree of b(x) = x − 2 is 1, so the remainder has degree less than the divisor, as required. For example, we saw above that with q(x) = x − 3 and r(x) = 1. Here the degree of r(x) = 1 is 0, and the degree of b(x) = x − 2 is 1, so the remainder has degree less than the divisor, as required. Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Quotient Form and Long-Run Behavior Example 1a For the rational function in quotient form (a)Graph f and describe its long-run behavior. (b)Explain the long-run behavior using the quotient form. Solution (a)See Figure 9.5. We see that, in the long-run, the rational function behaves like the constant function y = 5. We see that y = 5 is a horizontal asymptote. Figure 9.5: For large positive values of x, the graph looks Like the horizontal line y = 5 (b) The function is in quotient form with q(x) = 5. As x gets larger and larger, f(x) gets closer and closer to q(x) because the remainder term gets smaller and smaller. For example, in evaluating f(100), we get Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. In general, the quotient form of a rational function tells us its long-run behavior. This is because, for large values of x, the value of q(x) is much larger than the value of r(x)/b(x). For example, in evaluating at x = 1000, we get Notice how close the function’s value at x = 1000 is to the value of the quotient: 997.001 versus 997. In general: In general, the quotient form of a rational function tells us its long-run behavior. This is because, for large values of x, the value of q(x) is much larger than the value of r(x)/b(x). For example, in evaluating at x = 1000, we get Notice how close the function’s value at x = 1000 is to the value of the quotient: 997.001 versus 997. In general: For a rational function expressed in quotient form the long-run behavior of f is the same as the long-run behavior of the polynomial function q. If q(x) = k, a constant, then the graph has a horizontal asymptote at y = k. For a rational function expressed in quotient form the long-run behavior of f is the same as the long-run behavior of the polynomial function q. If q(x) = k, a constant, then the graph has a horizontal asymptote at y = k. Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. To see if a rational function has constant quotient, we try to write the numerator as a constant times the denominator plus a remainder. For example, to write the rational function in quotient form, we want to write the numerator as Comparing coefficients of x on both sides tells us that the constant must be 5, so and then comparing both sides tells us that remainder = 3. (Continued on next slide.) To see if a rational function has constant quotient, we try to write the numerator as a constant times the denominator plus a remainder. For example, to write the rational function in quotient form, we want to write the numerator as Comparing coefficients of x on both sides tells us that the constant must be 5, so and then comparing both sides tells us that remainder = 3. (Continued on next slide.) Putting a Function into Quotient Form When the Quotient is Constant: Horizontal Asymptotes Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. (Continued.) Now we can write: split numerator quotient form. In general, if the numerator and denominator have the same degree, the ratio of the leading terms gives us the constant quotient when the rational function is in quotient form. (Continued.) Now we can write: split numerator quotient form. In general, if the numerator and denominator have the same degree, the ratio of the leading terms gives us the constant quotient when the rational function is in quotient form. Putting a Function into Quotient Form When the Quotient is Constant: Horizontal Asymptotes Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 2 Find the long-run behavior and horizontal asymptote of by putting it in quotient form. Solution Notice that the leading term of the numerator is 3 times the leading term of the denominator. We write Thus, the quotient is q(x) = 3 and the remainder is r(x) = 4. In the long run f(x) behaves like the constant function q(x) = 3. Figure 9.6 shows the graph of f approaching the graph of y = 3. Figure 9.6: Graph of Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 3 Find the long-run behavior and horizontal asymptote of the rational function Solution In this case the function is already in quotient form with quotient zero: The long-run behavior is given by q(x) = 0, so the horizontal asymptote is y = 0, or the x-axis. A special case of horizontal asymptotes occurs when the numerator has degree less than the denominator. Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. The function is in quotient form with quotient q(x) = x, so for large values of x it is close to the graph of the linear function y = x. See Figure 9.7. Figure 9.7: Graph of What if the quotient polynomial q(x) is not a constant? One possibility is that q(x) is a linear function. In this case, we say the graph has a slant asymptote. What if the quotient polynomial q(x) is not a constant? One possibility is that q(x) is a linear function. In this case, we say the graph has a slant asymptote. Slant Asymptotes and Other Types of Long-Run Behavior Example 4 Graph and describe its long-run behavior. Solution Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 5 The rational function has quotient form Graph f and interpret its long-run behavior in terms of the quotient form. Solution The quotient is q(x) = 2x + 3, so the long-run behavior of f is the same as the line y = 2x + 3. The graph of y = 2x + 3 is a slant asymptote for f. See Figure 9.8. Figure 9.8: Graph of Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. In Example 2, the quotient of is q(x) = 3, the ratio of the leading terms in the numerator and denominator, 30x 3 and 10x 3. Taking the ratio of leading terms works in general to give the long-run behavior of a rational function, since the ratio of leading terms is the leading term of the quotient. In Example 2, the quotient of is q(x) = 3, the ratio of the leading terms in the numerator and denominator, 30x 3 and 10x 3. Taking the ratio of leading terms works in general to give the long-run behavior of a rational function, since the ratio of leading terms is the leading term of the quotient. Long-Run Behavior and the Ratio of the Leading Terms Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 6 Graph and describe its long-run behavior. Solution Considering the ratio of leading terms, we see that the long-run behavior of this function is This means that on large scales, its graph resembles the parabola y = 3x 2. See Figure 9.9. Figure 9.9: In the long run, the graph of looks like the graph of the parabola y = 3x 2 Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Graphing Rational Functions Example 7a Graph the following rational functions, showing all the important features. Solution (a) See Figure 9.10. The numerator is zero at x = −3, so the graph crosses the x-axis here. The denominator is zero at x = −2, so the graph has a vertical asymptote here. At x = 0, y = (0 + 3)/(0 + 2) = 1.5, so the y-intercept is y = 1.5. The long-run behavior is given by the ratio of leading terms, y = x/x = 1, so the graph has a horizontal asymptote at y = 1. Figure 9.10: A graph of the Rational function showing its asymptotes and zero Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 7b Graph the following rational functions, showing all the important features. Solution (b) See Figure 9.11. (See next slide for solution.) Figure 9.11: A graph of the rational function showing intercept and asymptotes Graphing Rational Functions (continued) Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 7b (continued) Graph the following rational functions, showing all the important features. Solution (b) (Continued) Since the numerator of this function is never zero, this rational function has no zeros, which means that the graph never crosses the x-axis. The graph has vertical asymptotes at x = −2 and x = 3 because this is where the denominator is zero. At x = 0, y = 25/(0 + 2)(0 − 3) 2 = 25/18, so the y-intercept is y = 25/18. The long-run behavior of this rational function is given by the ratio of the leading term in the numerator to the leading term in the denominator. The numerator is 25, and if we multiply out the denominator, we see that its leading term is x 3. Thus, the long-run behavior is given by y = 25/x 3, which has a horizontal asymptote at y = 0. Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 8 Find a possible formula for the rational function shown in Figure 9.12. Figure 9.12: A rational function Solution The graph has vertical asymptotes at x = −1 and x = 1. At x = 0, we have y = 2, and at y = 0, we have x = 2. The graph of satisfies each of these requirements. In the long run, as the x-values get larger and larger, the y-values get closer and closer to 0. Looking at the equation of the function, we see that the ratio of the leading terms, does get closer and closer to the horizontal line y = 0 as x gets very large. Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 9 Find a possible formula for the rational function shown in Figure 9.13. Figure 9.13: A rational function Solution The graph has a vertical asymptote at x = −1, and a zero at x = 2, so we try This has the right zero and vertical asymptote, but when x = 0 it has the value y = −2, whereas the graph has a vertical intercept at y = −4. We multiply by 2 to get This has horizontal asymptote at y = 2 and y-intercept at y = −4, as required. Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. We now summarize what we have learned about the graphs of rational functions: For a rational function given by where a and b are polynomials with different zeros, then: The long-run behavior is given by the ratio of the leading terms of a(x) and b(x). The zeros are the same as the zeros of the numerator, a(x). A vertical asymptote or hole occurs at each of the zeros of the denominator, b(x). A hole occurs at an x-value where both a(x) and b(x) are zero. For a rational function given by where a and b are polynomials with different zeros, then: The long-run behavior is given by the ratio of the leading terms of a(x) and b(x). The zeros are the same as the zeros of the numerator, a(x). A vertical asymptote or hole occurs at each of the zeros of the denominator, b(x). A hole occurs at an x-value where both a(x) and b(x) are zero. Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Figure 9.14: A rational function can cross its horizontal asymptote The graph of a rational function never crosses a vertical asymptote. However, the graphs of some rational functions cross their horizontal asymptotes. The difference is that a vertical asymptote occurs where the function is undefined, so there can be no y-value there, whereas a horizontal asymptote explains what happens to the y-values as the x-values get larger and larger (in either a positive or negative direction). There is no reason why the function cannot take on this y-value at a finite x-value. For example, the graph of crosses the line y = 1, its horizontal asymptote; the graph does not cross the vertical asymptote, the y-axis. See Figure 9.14. The graph of a rational function never crosses a vertical asymptote. However, the graphs of some rational functions cross their horizontal asymptotes. The difference is that a vertical asymptote occurs where the function is undefined, so there can be no y-value there, whereas a horizontal asymptote explains what happens to the y-values as the x-values get larger and larger (in either a positive or negative direction). There is no reason why the function cannot take on this y-value at a finite x-value. For example, the graph of crosses the line y = 1, its horizontal asymptote; the graph does not cross the vertical asymptote, the y-axis. See Figure 9.14. Can a Graph Cross an Asymptote? Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. 9.2 LONG-RUN BEHAVIOR OF RATIONAL FUNCTIONS Key Points The quotient form of a rational expression Long-run behavior of rational functions Graphing rational functions Slant asymptotes and other long-run behavior Section 9.2

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. We saw that we could rewrite the rational function by rewriting the numerator as a constant times the denominator plus a remainder. In this section we develop an algorithm for dividing one polynomial into another, similar to long division in arithmetic. We saw that we could rewrite the rational function by rewriting the numerator as a constant times the denominator plus a remainder. In this section we develop an algorithm for dividing one polynomial into another, similar to long division in arithmetic. Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. The Division Algorithm In Example 9 on page 304, we saw that 3x 3 + 7x 2 + 2x − 3720 = (x − 10)(Another polynomial), because the left-hand side is zero at x = 10. To find the other polynomial, we divide x − 10 into 3x 3 + 7x 2 + 2x − 3720. We start by looking at the leading term 3x 3. Since 3x 3 = 3x 2 · x, we can write 3x 3 + 7x 2 + 2x − 3720 = (x − 10)(3x 2 ). Notice we have left space on the right for the lower-degree terms in the second factor on the right. We set this up in more traditional format as To find the next term after 3x 2, we see what is left to divide after we subtract 3x 2 (x − 10) = 3x 3 − 30x 2 : (Continued on next slide.) In Example 9 on page 304, we saw that 3x 3 + 7x 2 + 2x − 3720 = (x − 10)(Another polynomial), because the left-hand side is zero at x = 10. To find the other polynomial, we divide x − 10 into 3x 3 + 7x 2 + 2x − 3720. We start by looking at the leading term 3x 3. Since 3x 3 = 3x 2 · x, we can write 3x 3 + 7x 2 + 2x − 3720 = (x − 10)(3x 2 ). Notice we have left space on the right for the lower-degree terms in the second factor on the right. We set this up in more traditional format as To find the next term after 3x 2, we see what is left to divide after we subtract 3x 2 (x − 10) = 3x 3 − 30x 2 : (Continued on next slide.) Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. The Division Algorithm (Continued) Now we find the next term in the quotient by comparing the divisor, x −10, with the new dividend 37x 2 + 2x − 3720. The new leading term is 37x 2 = 37x·x, so the next term is 37x, giving Continuing in this way, Notice that the remainder is zero, as we expected. (Continued) Now we find the next term in the quotient by comparing the divisor, x −10, with the new dividend 37x 2 + 2x − 3720. The new leading term is 37x 2 = 37x·x, so the next term is 37x, giving Continuing in this way, Notice that the remainder is zero, as we expected. Section 9.3 −

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 1 Divide x 2 + x + 1 into 3x 4 − x 3 + x 2 + 2x + 1. Solution We write 3x 4 − x 3 + x 2 + 2x + 1 = (x 2 + x + 1 )(3x 2 ), leaving space for the remainder as well as the quotient. Or, in traditional format We subtract 3x 2 (x 2 + x + 1) = 3x 4 + 3x 3 + 3x 2 : (Solution continued on next slide.) Polynomial Division with a Remainder In general, we have to allow for the possibility that the remainder might not be zero. Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 1 Divide x 2 + x + 1 into 3x 4 − x 3 + x 2 + 2x + 1. Solution (continued) Again, we find the next term in the quotient by comparing the divisor, x 2 + x + 1, with the new dividend −4x 3 − 2x 2 + 2x. The ratio of the leading terms is now −4x 3 /x 2 = −4x, so the next term is −4x, giving (Solution continued on next slide.) Polynomial Division with a Remainder (continued) Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 1 Divide x 2 + x + 1 into 3x 4 − x 3 + x 2 + 2x + 1. Solution (continued) Continuing in this way, we get the completed calculation This means that 4x − 1 is the remainder, so 3x 4 − x 3 + x 2 + 2x + 1 = (x 2 + x + 1 )(3x 2 − 4x + 2) + 4x − 1. Polynomial Division with a Remainder (continued) Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 2a Write in quotient form and state the quotient: Solution (a)Dividing the denominator into the numerator gives Thus, we can write (x 3 − 2x + 1) = (x − 2)(x 2 + 2x + 2) + 5. so q(x) = x 2 + 2x + 2. The Division Algorithm and Quotient Form Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 2b Write in quotient form and state the quotient: Solution (b) Dividing the denominator into the numerator gives Thus, we can write 2x 3 + 3x 2 = (2x + 3)(x 2 + 1) + (−2x − 3) so q(x) = 2x + 3. The Division Algorithm and Quotient Form Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 2c Write in quotient form and state the quotient: Solution (c) Dividing the denominator into the numerator gives Thus, we can write x 3 − 1 = (x)(x 2 + 1) + (−x − 1) so q(x) = x. The Division Algorithm and Quotient Form Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Example 3 Determine the equation of the slant asymptote of the rational function. Solution Dividing x 2 + 2x − 8 into 2x 3 + 6x 2 − 26x − 30 gives We see that Thus, the quotient is q(x) = 2x + 2, and for large values of x the graph is close to the slant asymptote y = 2x + 2. Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. The Remainder Theorem A particularly simple division is when we divide the polynomial x − a, for a constant a, into some other polynomial p(x). In that case the remainder is a constant, since it must have degree less than the degree of x − a. Thus, p(x) = (x − a)q(x) + constant. What is the constant? If we put x = a in the equation above, we get p(a) = 0 · q(a) + constant = constant. Section 9.3 The Remainder Theorem For any polynomial p(x) and any constant a, we have p(x) = (x − a)q(x) + p(a), where q(x) is the quotient of p(x)/(x − a). In particular, if p(a) = 0 then p(x) = (x − a)q(x), so x − a is a factor of p(x). The Remainder Theorem For any polynomial p(x) and any constant a, we have p(x) = (x − a)q(x) + p(a), where q(x) is the quotient of p(x)/(x − a). In particular, if p(a) = 0 then p(x) = (x − a)q(x), so x − a is a factor of p(x).

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. We could answer both these questions by using the division algorithm to see if the remainder is zero. However, the Remainder Theorem gives us a short cut. (a) We have p(4) = 5(4) 3 − 13(4) 2 − 30(4) + 8 = 5(64) − 13(16) − 120 + 8 = 0, so by the Remainder Theorem x − 4 is a factor of p(x). (b) We have p(− 3) = 2(−3) 3 +10(−3) 2 +11(−3)−6 = 2(−27)+10(9)−33−6= −3, so x − 3 cannot be a factor of p(x) = 2x 3 + 10x 2 + 11x − 6. Example 4 (a) Is x − 4 a factor of the polynomial p(x) = 5x 3 − 13x 2 − 30x + 8? (b) Is x + 3 a factor of the polynomial p(x) = 2x 3 + 10x 2 + 11x − 6? Solution Section 9.3 We have shown, if x − a is a factor of p(x), then p(a) = 0. Now we have proved the converse, that if p(a) = 0 then x − a is a factor of p(x). We have shown, if x − a is a factor of p(x), then p(a) = 0. Now we have proved the converse, that if p(a) = 0 then x − a is a factor of p(x).

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. 9.3 PUTTING A RATIONAL FUNCTION IN QUOTIENT FORM Key Points The Division Algorithm Polynomial division with a remainder Quotient form The Remainder Theorem Section 9.3

ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Chapter 9 Rational Functions.

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ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Chapter 9 Rational Functions.

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Presentation on theme: "ALGEBRA: FORM AND FUNCTION 2 nd edition by McCallum, Connally, Hughes-Hallett, et al.,Copyright 2015, John Wiley & Sons, Inc. Chapter 9 Rational Functions."— Presentation transcript:

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