CHAPTER 3 Polynomial and Rational Functions

Transcription

1 CHAPTER 3 Polynomial and Rational Functions Section 3.1 Quadratic Functions A quadratic function has the form f (x) =ax 2 + bx + c and has a parabola as its graph. Using the method of completing the square, it can be converted to standard form f (x) =a(x h) 2 + k, where (h, k) is the vertex of the parabola. Example 3.1: Write the function f (x) =3x 2 7x + 2 in the form f (x) =a(x h) 2 + k. Using the method of completing the square we have, 3x 2 7x + 2 = 3 x x + 2! = 3 x x = 3 x Extrema of a Quadratic = 3 x = 3 x 3 x An extrema is a maximum or minimum point on a graph. The quadratic f (x) =a(x h) 2 + k is a minimum point if a > 0 and a maximum point if a < 0. A shortcut for finding the x-coordinate of the vertex is to use the vertex formula x = b 2a 0.02x x and determine if the vertex is a maxi- Example 3.2: Find the vertex of f (x) = mum or a minimum. Since the coefficient of x 2 is negative, we know that the parabola opens downward and therefore, the vertex is a maximum point. The x-coordinate of the vertex is x = b 2a = 7.1 2( 0.02) = Substituting back into the function, the y value is f (177.5) = 0.02(177.5) (177.5)+14.3 = Therefore, the vertex of the parabola is at (177.5, ) and the vertex is a maximum point. 28

2 Bradshaw - Math 188 Chapter 3 Notes 29 Applications There are many applications of quadratic functions. The common feature is that all the problems ask to create a mathematical model of a situation and then find an extrema of that model. Example 3.3: The product of two numbers is P. The sum of the numbers is 250. What are the two numbers that give the maximum value of P and what is the value of P? Let the two numbers be x and y. Since the sum of the numbers is 250, we have the equation x + y = 250. Since the product of the numbers is P, we have another equation xy = P. Solving the first equation for y gives y = 250 x. Substituting into the second equation, we have P = xy = x(250 x) = x x Since this equation is a quadratic, its maximum is determine using the vertex formula. x = b 2a = 250 2( 1) = 125 Therefore, x = 125, y = 250 x = = 125 and P = xy = 15, 625. Example 3.4: You are building a rectangular pen for your pet pig. You have purchased 500 feet of fencing. What is the largest rectangular area that you can enclose? Let the length be L and the width be W. Since the total length of fencing is the same as the perimeter of the rectangle, we have 2L + 2W = 500. Letting the area of the rectangle be A, we have another equation LW = A. As done in the previous example, we can solve for L in the first equation giving L = 250 W. Substituting this into the area equation gives A = W(250 W). Notice that this is the same format as in the previous example. Thus, the maximum area will be 15, 625 ft 2. Example 3.5: The McBurger Restaurant has hired you to help maximize the revenue for their restaurant. They provide you with the following information. Revenue is found by multiplying the average revenue per table by the number of tables. Currently, the average revenue per table is $75/day and the number of tables is 48. For each table added to the restaurant, the daily revenue will decrease by $1.50. What is the total number of tables the restaurant should have in order to maximize the total revenue? If we let x be the number of tables to be added, then the total number of tables is given by number of tables = 48 + x Since the revenue per table decreases by $1.50 for each additional table, we have revenue per table = x

3 Bradshaw - Math 188 Chapter 3 Notes 30 Therefore, the total revenue is given by total revenue = revenue per table number of tables R =(75 1.5x)(48 + x) R = 1.5x 2 + 3x Since the revenue equation is a quadratic equation, the maximum revenue will be determined by the vertex. x = b 2a = 3 2( 1.5) = 1 Therefore, the total number of tables is = 49, the average revenue per table is (1) =$73.50, and the maximum total revenue is $ Section 3.2 Polynomial Functions A polynomial is a function of the form f (x) =c n x n + c n 1 x n 1 + c n 2 x n c 2 x 2 + c 1 x + c 0 where the exponents are whole numbers (0, 1, 2, 3,... ). The degree of a polynomial is the value of the greatest exponent. The leading coefficient of a polynomial is the constant coefficient of the term with the highest degree. Therefore, f (x) = 6x 3 1 is a polynomial of degree 3 and has a leading coefficient of 6. while f (x) =6x 3 6x 1/2 1 is not a polynomial because one of the exponents is not a whole number. The factors of a polynomial P(x) determine the zeroes or roots of a P(x). For example, P(x) = x 2 4 factors into P(x) =(x 2)(x + 2). The zeroes of P(x) are x = 2 and x = 2. The multiplicity of a zero is the same as the power of the factor. For example, P(x) =x 2 6x + 9 =(x 3) 2. The power of (x 3) is 2. Therefore, the root x = 3 has multiplicity 2. Sketching Polynomials A quick sketch of a polynomial can be obtained from the following characteristics of a polynomial. Endpoint behavior refers to the direction, up or down, in which the tails of the polynomial are heading. If the leading coefficient is positive, the right end of the polynomial heads upward. Otherwise it heads downward. If the degree of the polynomial is even, the ends of the polynomial head in the same direction. If the degree of the polynomial is odd, the ends head in opposite direction. The y-intercept is found by setting x = 0. Every polynomial has a y-intercept. The x-intercepts (if any) are obtained from the factored form of a polynomial. Each factor with a real root (zero) gives an x-intercept of the graph. If a root has an even multiplicity, the graph touches the x-axis, but does not pass through. Example 3.6: Sketch the graph of the polynomial f (x) =(2 x)(x + 1)(x 1) 2.

4 Bradshaw - Math 188 Chapter 3 Notes 31 If we multiply the factors of the polynomial, it has a leading term of x 4. Therefore, the ends of the function both point downward. Substituting x = 0 gives f (0) =(2)(1)(1) = 2 so the y-intercept is (0, 2). The factored form of the polynomial gives us x-intercepts (2, 0), ( 1, 0) and (1, 0). The exponent of 2 on the factor (x 1) indicates that the graph will simply touch the x-axis but not pass through at this point Figure 3.1: Graph of f (x) using 1.5 apple x apple 2.5, 4 apple y apple 4 Example 3.7: A dog can swim at 2 miles per hour and can run at 5 mph. See the illustration below. The dog (at point D) is swimming in a lake. If the dog swims straight to shore at point O, the distance is 440 feet. The dog s owner is at point P, 1500 feet down the shore from O. Assume that 6 O is a right angle. If the dog heads for a point S on shore and then runs the remaining distance to P, what is the minimum time required for the dog to reach its owner? (Hint: 1 mile = 5280 feet.) D 440 O x S 1500 x P The speed of the dog can be found by converting miles to feet. Therefore Similarly, 5 miles/hour = 26,400 feet/hour. Using the fact that 2 miles feet feet 5280 = 10, 560 hour mile hour time(t) = distance(d) rate(r) the time it takes the dog to reach its owner is given by t(x) =t swimming + t running = d swimming + d running r swimming r running p 440 = 2 + x x 10, , 400

5 Bradshaw - Math 188 Chapter 3 Notes 32 Since this equation is not a quadratic, find the minimum point by looking at the graph of the function. As shown in Figure 3.2, the minimum point is near x = 200. Changing the viewing ,000 1,500 Figure 3.2: Graph of t(x) using 0 apple x apple 1500, 0 apple t apple 0.15 rectangle, we gives a better view. From Figure 3.3, we have the minimum at x 192 feet. Plugging this into our time Figure 3.3: Graph of t(x) using 180 apple x apple 210, apple t apple equation, we have t = p , , 400 = hours 5.7 minutes.

6 Bradshaw - Math 188 Chapter 3 Notes 33 Section 3.3 Dividing Polynomials There are two ways of dividing polynomials, long division and synthetic division. Long division has the advantage of always working. However it is a long method. Synthetic division only works when dividing by a polynomial of the form x c but it is very fast. Long Division Long division of polynomials works the same as long division of numbers. x 3 + 3x 2 + 9x + 25 x 3 x 4 2x + 5 x 4 + 3x 3 3x 3 3x 3 + 9x 2 9x 2 2x 9x x 25x x Synthetic Division Synthetic division only work when you are dividing by a factor x process to be collapsed into three lines. c. It allows the long division The Remainder Theorem c gives a remain- The remainder theorem states that the synthetic division of a polynomial P(x) by x der that is equal to P(c). In the problem x 4 2x + 5 x 3 above, the remainder in the synthetic division is 80. Similarly, calculating P(3) gives The Factor Theorem P(3) =3 4 2(3)+5 = = 80 The real importance of the remainder theorem is that if division by x c gives a remainder of zero, then x c is a factor of the polynomial. This is known as the Factor Theorem. Therefore, synthetic division can be used as a method of determining the factors of a polynomial. Example 3.8: Is x + 2 a factor of x 4 3x 3 + 2x x 18? Performing synthetic division by x = 2 gives

7 Bradshaw - Math 188 Chapter 3 Notes Since the remainder is zero, we have that x 4 3x 3 + 2x x 18 =(x + 2)(x 3 5x x 9) Section 3.4 Real Zeros of a Polynomial The textbook has three theorems in this section. We will only be discussing one of the theorems. The Rational Root Theorem Given a polynomial P(x) =c n x n + c n 1 x n 1 + c n 2 x n c 2 x 2 + c 1 x + c 0 where c i are integers, the rational root theorem states that all rational roots must be of the form r = p q where p is a factor of c 0 and q is a factor of c n. Example 3.9: List the possible rational roots of f (x) =6x 5 2x Since c 0 = 7, and c n = 6, the possible rational roots are ±1 1, ±7 1, ±1 2, ±7 2, ±1 3, ±7 3, ±1 6, and ±7 6. The purpose of the rational root theorem is to reduce the number of attempts you make with synthetic division. Turning the Zeroes into a Polynomial If you have the roots of a polynomial along with their multiplicities, you can determine the original polynomial by multiplying the factors. Example 3.10: A polynomial P(x) has roots x = 2andx = 3. The root x = 3 has multiplicity 2. It is also known that P(1) =5. Determine a polynomial of degree 3 meeting these criteria. Since we know the roots and the multiplicities, we have the following. P(x) =a(x 2)(x 3) 2 = a(x 2) x 2 6x + 9 = a x 3 6x 2 + 9x 2x x 18 = a x 3 8x x 18

8 Bradshaw - Math 188 Chapter 3 Notes 35 Since we know that P(1) =5, we have P(x) =a x 3 8x x 18 P(1) =a 1 3 8(1) (1) 18 Section 3.5 Complex Zero Theorem 5 = a( ) a = 5 4 P(x) = 5 4 x 3 8x x 18 Complex Zeros of a Polynomial Any polynomial of degree n with real coefficents, P(x) =c n x n + c n 1 x n 1 + c n 2 x n c 2 x 2 + c 1 x + c 0, with n 1, has complex zeroes that come in pairs, a + bi and a An example of the Complex Zero Theorem can be seen in the above example. Both 4 + 7i and its 3 complex conjugate are roots of the polynomial. As shown in the following example, another use of the Complex Zero Theorem is to obtain the polynomial from its roots. Example 3.11: P(x) is a fourth degree polynomial with real coefficients. Two roots are x = 3i and x = 1 + 3i. If P(0) =10, find P(x). 2 Since the polynomial has degree 4, it must be of the form bi. P(x) =a(x r 1 )(x r 2 )(x r 3 )(x r 4 ). Since P(x) has real coefficients, the complex roots come in pairs, x = 3i and x = 1 + 3i and 1 3i. This gives us four factors. 2 2 This gives the polynomial x = 3i! (x 3i) is a factor x = 3i! (x + 3i) is a factor x = 1 + 3i! (2x 1 3i) is a factor 2 x = 1 3i! (2x 1 + 3i) is a factor 2 3i and P(x) =a(x 3i)(x + 3i)(2x 1 + 3i)(2x 1 3i). Multiplying the first pair of factors gives (x 3i)(x + 3i) =x Multiplying the second pair of factors gives (2x 1 3i)(2x 1 + 3i) =(2x 1) = 4x 2 4x + 10.

9 Bradshaw - Math 188 Chapter 3 Notes 36 This gives the polynomial P(x) =a(x 2 + 9)(4x 2 4x + 10). Performing long multiplication gives P(x) =a(x 2 + 9) 4x 2 4x + 10 = a 4x 4 4x x x 2 36x + 90 = a 4x 4 4x x 2 36x + 90 Since P(0) =10, we have P(0) =a(90) =10, giving a = 1. Therefore, the final answer is 9 P(x) = 2 2x 4 2x x 2 18x Section 3.6 A rational function has the form Rational Functions f (x) = P(x) where P(x) and Q(x) are polynomials. Q(x) The purpose of this section is to graph rational functions without using a calculator. Problems can be divided into two types, depending on if the degree of the numerator is less than or equal or greater than the degree of the denominator. The steps are very similar but not identical. Degree of Numerator apple Degree of Denominator Consider the following function f (x) = 2(x 4)(x + 1)(x 2) (x + 3)(x 1)(x 2) = 2x2 6x 8 x 2 + 2x 3 Locate any holes: Holes are caused by factors that cancel from both the numerator and denominator. To find the y-coordinate of the hole, substitute the x-coordinate into the function. In this example, the hole is at x = 2 giving y = 2(2)2 6(2) 8 (2) 2 + 2(2) 3 = Therefore, the hole is at 2,. 5 Determine Vertical Asymptotes: Vertical asymptotes occur when the denominator is zero (except at a hole.) In this example, the vertical asymptotes are at x = 3 and x = 1. Determine the Horizontal Asymptote: There are two cases. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is found by looking at the ratio of the highest order terms in the function. In this example, the ratio is y = 2x2 x 2 = 2.

10 Bradshaw - Math 188 Chapter 3 Notes 37 Determine the x-intercept(s): If there are any x-intercepts, you find them by setting y = 0. For this example, the x-intercepts occur at ( 1, 0) and (4, 0). Determine the y-intercept: If there is a y-intercept, find it by setting x = 0. For this example, the y-intercept occurs at 0, 8. 3 Determine the place where the graph crosses the horizontal asymptote: If the graph crosses the horizontal asymptote, you find the point(s) by setting the function equal to the value of the horizontal asymptote and solving for x. For this example, 2x 2 6x 8 x 2 + 2x 3 = 2 2x 2 6x 8 = 2x 2 + 4x 6 10x = 2 x = 1 5 Therefore, the graph crosses the horizontal asymptote at 1 5,2. Assemble the above information into a graph: will lead to the following graph. After drawing the above characteristics, some thought Figure 3.4: Graph of f (x) = 2(x 4)(x + 1)(x 2), 11 apple x apple 11, 6 apple y apple 10 (x + 3)(x 1)(x 2) Degree of Numerator > Degree of Denominator Consider the following function f (x) = 2(x 4)(x + 1)(x 2) (x + 3)(x 2) = 2x2 6x 8 x + 3

11 Bradshaw - Math 188 Chapter 3 Notes 38 Locate any holes: As in the above example, holes are caused by factors that cancel from both the numerator and denominator. In this example, the hole is at x = 2 giving Therefore, the hole is at 2, y = 2(2)2 6(2) 8 = Determine Vertical Asymptotes: In this example, the vertical asymptote is x = 3. Determine the Other Asymptote: Functions of this type have an asymptote that is found by performing division. For f (x) = 2x2 6x 8, we can use synthetic division. x The results of the synthetic division allows rewriting the function as f (x) = 2x2 6x 8 x + 3 = 2x x + 3. The other asymptote for the graph is given by the non-fraction part of this result. Therefore, the other asymptote is y = 2x 12. Determine the x-intercept(s): Setting y = 0 gives the x-intercepts at ( 1, 0) and (4, 0). 8 Determine the y-intercept: Setting x = 0 gives the y-intercept at 0,. 3 Determine the place where the graph crosses the other asymptote: If the graph crosses the other asymptote, you find the point(s) by setting the function equal to the other asymptote and solving for x. For this example, 2x 2 6x 8 = 2x 12 x + 3 2x 2 6x 8 =(x + 3)(2x 12) 2x 2 6x 8 = 2x 2 6x 36 8 = 36 Since this statement is false, the graph does not cross the asymptote. Assemble the above information into a graph: will lead to the following graph. After drawing the above characteristics, some thought Section 3.7 Polynomial and Rational Inequalities Solving polynomial and rational inequalities uses a method different from solving equalities. The process consists of three parts: Find the critical values Check each interval to see if it works

12 Bradshaw - Math 188 Chapter 3 Notes Figure 3.5: Graph of f (x) = 2(x 4)(x + 1)(x 2), 10 apple x apple 10, 50 apple y apple 10 (x + 3)(x 2) Write the solution For example, solve x2 5x apple 2. x 3 Find the critical values: Start by rewriting the problem so that one side is zero and the other side is completely factored. However, do not multiply or divide by a term containing x. 5x x 3 apple 2 x 2 5x 2 apple 0 x 3 5x 2(x 3) apple 0 x 3 x 3 x 2 7x + 6 apple 0 x 3 (x 6)(x 1) apple 0 x 3 x 2 The critical values are the value that make the resulting expression either zero or undefined. For this example, there are three critical values, x = 1, x = 3, and x = 6. Check each interval: To check the intervals, create a table such as the following. Use the final inequality from above to check the values in each interval to see if the values make the inequality true or false. For example, in the interval (1, 3), checking x = 2 results in x 2 (x 6)(x 1) x 3! (2 6)(2 1) (2 3) = 4 1 = 4 which is not apple 0. Therefore, points in the interval (1, 3) are false. (,1) 1 (1, 3) 3 (3, 6) 6 (6, ) True True False False True True False

13 Bradshaw - Math 188 Chapter 3 Notes 40 Write the solution: The solution consists of any values listed as true. Therefore, the solution is given by x 2 (,1] [ (3, 6] 1 3 6