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Chapter 9 Exploring Rational Functions Dan Box 9-1 Rational Functions • A function, f(x) is a rational function if it is the division of two polynomial functions, meaning it can be written: • f(x) = p(x) / q(x) , where p(x) and q(x) are polynomial functions, and q(x) ≠ 0. 3x2 + 2x – 4 • Example: f(x) = x3 - 5x2 + 3 • It is important that q(x), the denominator, is not 0. • If a value of x makes the denominator equal to 0, that x value will either be a vertical asymptote or a point of discontinuity. • A point of discontinuity is appears as a whole in the graph at the x value. • A vertical asymptote is an x value that the graph comes very close to, but never actually touches. 9-1 Rational Functions (cont.) A point of discontinuity will look like this: The rational function f(x) = 3x+6 x+2 has a discontinutity at x = -2, illustrated here by the red circle. A vertical asymptote is seen in a graph like this: x2 - 6 The rational function f(x) = x-2 has a vertical asymptote at x = 2, illustrated here by the red dashed line. 9-1 Examples 1. For each equation, find the x values that result in vertical asymptotes or points of discontinuity. Then, use a calculator to check what each x is: X2 + 2x a) f(x) = x+2 x2 + 7x + 6 b) f(x) = c) f(x) = 2. x2 + 4x - 12 x-3 x3 – 6x2 + 9x For each equation, write the equation for the vertical and horizontal asymptotes. a) f(x) = 1 x-3 b) f(x) = 3 (x + 1)(X – 5) c) f(x) = 2 x2 – 2x - 8 9-1 Examples (cont.) 1. Solutions: a) f(x) = b) f(x) = c) f(x) = X2 + 2x X+2 X2 + 7x + 6 X2 + 4x - 12 X-3 X3 a. + 9x b. In the graph, we see a discontinuity at x = -2 Only x = -2 causes the denominator to be 0. = = (X + 6)(x + 1) (X + 6)(x – 2) X-3 X(x-3)(x+3) Here, both x = -6 and x = 2 cause the denominator to be 0. x= 3, x = -3, and x=0 all cause the denominator to be 0. c. In this graph, we see a discontinuity at x = -6, and a vertical asymptote at x=2 In this graph, we see a discontinuity at x = 3, and vertical asymptotes at x=0 and x = -3 9-1 Examples (cont.) 2. For each equation, write the equation for the vertical and horizontal asymptotes. Notice that the denominator = 0 when x = 3. 1 a) f(x) = b) f(x) = c) f(x) = X-3 3 Notice that the denominator = 0 when x = -1 or 5. (X + 1)(X – 5) 2 Notice that the denominator = 0 when x = -2 or 4. X2 – 2x - 8 a. From the graph, the vertical asymptote is x=3. There is also a horizontal asymptote along y=0. b. b. From the graph, the vertical asymptotes are x= -1 and x=5. There is also a horizontal asymptote along y=0. c. From the graph, the vertical asymptotes are x= -2 and x=4. There is also a horizontal asymptote along y=0. 9-1 Problems X+2 1. 2. Find f(-2) and f(4) for f(x)= X2 - x - 6 Determine the points of discontinuity and the vertical asymptotes, first by determining the x values that cause the denominator to x2 + x equal zero, and then by graphing, for f(x) = X2 + 5X + 4 3. Give the equations for the horizontal and vertical asymptotes of f(x) = 2 X2 + 9X -10 x = -1 is a discontinuity f(4) = 1 1) f(2) = undefined (vertical asymptote) x = -4 is vertical asymptote 2)x = -1, x= -4 3) x = -10 (V), x = 1 (v), y = 0 (H) 9-2 Direct, Inverse, and Joint Variation • There are several ways that x and y can be related in a function. 1. We say that x and y are directly related if y is a multiple of x. 1. 2. 2. We say that x and y are inversely related if y is a multiple of 1/ x 1. 2. 3. 3. This means that a direct relationship occurs if y = k * x, where k is any constant, other than zero. For example: y = 4.5*x means y varies directly as x, and the constant of variation (k) is equal to 4.5. This means that an inverse relationship occurs if y = k / x, where k is any constant other than zero. For example: y = 2 / x means y varies inversely as x, with constant of variation 2. Another way of expressing inverse relationships is: x*y = k We say that x and y vary jointly if y is a multiple of two or more variables. 1. 2. This means that joint variation occurs if we say y = k * x * z , where x and z are variables that are not zero, and k is a constant that is not zero. For example, we know Area = base * height on a rectangle. This means A = 1 * B * H. Thus we say the area of a rectangle varies jointly with the width and the height. 9-2 Examples 1. Direct variation: Given y varies directly as x, and when x = 1.5, y = 3 , find y when x = 4. 1. 2. 3. 4. 5. 6. Since y varies directly as x, our equation is y = k * k. Start by using what you know. We know when x = 1.5, y = 3. Plugging this in to y = k*x gives 3 = k * 1.5. Solving for k gives k = 3 / 1.5, or k = 2. Now, set up your second equation, using y = 2 * x. This gives y = 2 * 4, which means when x = 4, y = 8. Note: graphs of direct relationships are straight lines through the origin (0,0). Here is the graph of y = 2x, from the example. 9-2 Examples 1. Inverse Relationships: If y varies inversely as x, and when x = 5, y = 8, what is y when x = -2 ? 1. Remember that in inverse relationships, y = k / x. 2. Use the given information: 8 = k / 5. 3. Cross multiplying gives: 8 * 5 = k * 1, or 40 = k. 4. State the equation: y = 40 / x. 5. Solve using the new equation. y = 40 / -2. This gives y = -20, the solution. 6. Alternatively, you could use the initial equation xy = k. 7. This still gives 8 * 5 = k, k = 40. It would also give -2*y = 40, with the same result : y = -20. Note: Graphs of inverse relationships are curved lines, with vertical and horizontal asymptotes at x = 0, y = 0. The graph of the solution function y = 40 / x is given above. 9-2 Examples 1. Varying Jointly: If y varies jointly as x and z, and when x = 2 and z = 4 , y = 12, what is y when x = 5 and z = 2? 1. Remember that in joint variation, y = k*x*z. 2. Use the given information: 12 = k*(-2)*(4). 3. Solving for k gives: 12 = k*(-8) , 12/-8 = k , k = -1.5 4. State the equation: y = (-1.5)*x*z . 5. Solve using the new equation. y = (-1.5)*(5)*(2). This gives y = -15, the solution. 9-2 Problems Use x = 4, y = -12, and z = 5 to solve the following: A. B. 2. For the following, state if the relationship is direct, inverse, or join variation, and find the constant of variation, k: A. B. 3. Y varies directly as x, what is y when x = 3? Y varies jointly as x and z, what is y when x = 9 and z = 2 X*Y = 4 12*Y = X C. X * Z = -2 * Y D. X = 3 / Y Assume suburb population varies jointly with distance from major city and the number of train stations available to the city. The distance from Naperville to Chicago is 27 miles, and Naperville has 2 train stations. The population of Naperville is 129,600. Given this, how many train stations would you expect in Schaumburg, IL, if Schaumburg has a population of 57,600 and is 24 miles from Chicago? 1A)y=-9 1B)y= -10.8 2A) Inverse, k = 4 2B) direct, k = 1/12 2C) joint, k = -1/3 2D) inverse, k = 3 3) k = 2,400 Schaumburg would have 1 train station. 1. 9-3 Multiplying and Dividing Rational Expressions • Often the first step in multiplying or dividing rational expressions involves simplifying. • Just like with a rational number, like 9/15, you can simplify by removing a common factor from the top and bottom of the expression: 9/15 = (3*3)/(3*5) = 3 / 5 • In a rational expression, we look to remove variables as well numbers. For example: x*(x-2)(x+3) x*(x-2)(x+3) = x*(x-2)(x+3) = (x+3) x2(x-2) = x*x*(x-2) x*x*(x-2) x • Just like with a fraction, we can eliminate x/x and (x-2)/(x-2) because anything divided by itself is equal to 1. 9-3 Multiplying and Dividing Rational Expressions • Remember: When multiplying two fractions, you multiply the numerators and multiply the denominators. When dividing, multiply the denominator of the first rational with the denominator of the second, and the denominator of the first with the numerator of the second. • For any rational function, we multiply: A B * C D = AC BD As long as B ≠ 0 and D ≠ 0. For example: 3x (x+1) 3x(x+1) = x2 * (x+2) X2(x+2) • For any rational function, we divide: A B ÷ C D = AD BC As long as B ≠ 0, C ≠ 0 D ≠ 0. For example: 4x (x+7) 4x(x-2) = x3 * (x-2) X3(x+7) 9-3 Example 1. Give the simplified form of each rational equation: x2 * 3x = x+5 x x x2-4 * x-2 3x*x2 x(x+5) x = 3x3 = x(x+5) 3x3 2 x(x+5) = 3x2 x+5 (x+2)(x-2) (x+2)(x-2) x*(x2-4) x*(x2-4) x2-4 = = = = = = x+2 (x-2) (x-2) x*(x-2) x*(x-2) (x-2) x+15 x+15 4x4 2x*(x+15) 2x*(x+15) 2x*(x+15) ÷ = = = = 4 5 5 4 12x4 6x 2x 6x*4x 24x 24x (x+4)(x-3) -3x4 ÷ x-3 2x8 = 2x8(x+4)(x-3) -3x4(x-3) = 2x8(x+4)(x-3) -3x4(x-3) = 2x4(x+4) -3 Note: 2/24 = 1/12 x/x5 = 1/x4 9-3 Problems 3x(x+2) * x+5 x-10 x7 xy3 Divide: y-8 3x2 y+4 y(x+4) x2(x+4) ÷ y4(y-5) x4 2. y3(y+2) 3 3. xy (y+4) Divide and simplify: ÷ 4y2(y+2) 3x2 (y-8) y3(y-5) 4. x2 4. * 21x2(x+2) 3. y5 1. Multiply and simplify: x2 (y+2) 4x5 2. 7x Multiply: (x-10)(x+5) 1. 9-4 Adding and Subtracting Rational Functions • • • Adding and subtracting rational functions is a lot like adding and subtracting fractions. Just like adding or subtracting a fraction, the first step is to form a common denominator between all the functions involved. To do this, find the Least Common Multiple of the denominators, and then add the numerators. 1 3 1*8 3*3 8 9 17 For example: + = + = + = 3 • • 8 3*8 3*8 24 24 24 You can use this same process for rational functions. The goal is to get the same function as the denominator of both rational equations. For example: 1. x+2 3 7(x+2) 4x*3 7x+14 12x 19x+14 + + + = = = 4x2 7x 7*4x2 4x*7x 28x2 28x2 28x2 3. 2a 7b2 + 3abc 2a*11ac 3abc*7b2 = 11ac + 7b2*11ac 11ac*7b2 = 22a2c 77ab2c + 21ab3c 77ab2c 22a2c + 21ab3c = 77ab2c 9-4 Examples 1. Perform the addition or subtraction of the following rational equations. 3xy y 3xy*15x2 y*2(y+1) 2(y+1) 15x2 2(y+1)*15x2 15x2*2(y+1) Since these denominators have nothing in common, we multiply the numerator and denominator of each fraction by the opposite denominator. 45x3y Now we have a common denominator, we can simplify, then subtract. 2y(y+1) 30x2(y+1) 30x2(y+1) 45x3y – 2y(y+1) 30x2(y+1) Sometimes only one portion of the rational equations needs adjusting. 3x x2 3x*2y2 x2 6xy2 x2 6xy2 – x2 2y 4y3 2y*2y2 4y3 4y3 4y3 4y3 Here, the least common denominator is actually 4y3. We can turn 2y into 4y3 by multiplying the top and bottom of the first rational fraction by 2y2. 9-4 Examples 3. It is possible to combine the concepts of rational multiplication and division with the concepts of rational addition and subtraction. 1 y 1*x2 x2 2y*x2 2y 5 + + y*2y 1*x2 x2 *2y 2y*x2 + y*2y 2x2+2y2 x2 *2y 2x2y 2x2+2y2 2x2y 4 5*6y 4 5*6y 4 30y - 4 6y 1*6y 6y 1*6y 6y 6y The numerator of this equation is the sum of two rational functions, and the denominator is the different of two rational equations. Start by finding the common denominators for the top and bottom separately. The LCD for the numerator is 2x2y and the LCD for the denominator is 6y. Now, combine the numerator and denominator equations. Once the numerator rational and denominator rational are taken care of, the equation is a division of two rationals, just like section 9-3. x 6y 30y - 4 9-4 Problems 1. Find the Least Common Denominator y y 2x y2 x+y 4x2 4x2y 3xy3 3x(y+2) 3xy+6x 10x2 5xy 12x2y3 , 3xy+6x , 10x2y Solve the following equations: 19x2 4(x+3) x+4 y 72xy x 4(x+3)x2+19x2y 3. Solve the following, using sections 9-3 and 9-4 material. x 7y3 + x2 5y 3y 7y3(5xy-6y) 4x 2x2 (x+4)-72y2 y 76x2(x+3) x2 72xy 2. 4x2(x+7x2y3) 9-5 Solving Rational Equations • Any equation that involves one or more rational expressions is a rational equation • Rational equations are usually solved by multiplying each side of the equation by the least common denominator (LCD). • It is important to check solutions to rational equations to be sure that you have no multiplied by zero along the way. The best way to check this is to plug your solutions back in to the initial equation and make sure that you get a real solution. 9-5 Solving Rational Inequalities • Solving rational inequalities is much like solving for rational equations, but sets one rational equation as greater or less than another. • Solving so that one side is equal to zero is helpful in solving inequalities. • It may also be helpful to use a number line to test possible solutions. • Remember, when working with inequalities, if you multiply or divide by a negative number, the direction of the inequality changes For example: -3x < 5 x > - 5/3 9-5 Examples 1. Solve the following rational equation for x. 3 2(x+1) 1 15x 3 30x(x+1) * 2(x+1) The LCD is 30x(x+1). Multiply the LCD by both side of the equation. 90x 30(x+1) 2 15 1 15x * 30x(x+1) Here we see some common terms between the numerator and denominator that can be canceled out. Simplify the fractions, and distribute the expression in the numerator in the right equation. 45x 2(x+1) Now solve using algebra. 30x * 3 1 2 15 * 30(x+1) Next, multiply the 30x and 30(x+1) into the rational expressions. 45x 2x+2 45x-2x-2 0 43x-2 0 43x 2 x 2 43 Plugging the solution x=2/43 into the equation, we get the left side is equal to about 1.433, and the right side is equal to 1.433. Since each side is equal our result is true. 9-5 Examples 2. Solve the following rational inequality for x x 7(x+2) < 5 x x x+1 7(x+2) x+1 Start by subtracting the right side to the left. 7(x+2)(x+1) 7(x+2)(x+1) < 0 x > 0 From here, we know that x is zero or undefined at x=0, -2, and -1. Using this information, use a number line to test ranges. -x < 0 7(x+2)(x+1) < 0 Next, simplify. x2+x – x2 – 2x Combine the two parts now that there is a common denominator. (x+2)(x+1) x*7(x+2) First, get a common denominator. The LCD is 7(x+2)(x+1) x(x+1) – x(x+2) 7(x+2)(x+1) < 0 x(x+1) 7(x+2)(x+1) Plugging x=-3 into the inequality gives -1.5 > 0 which is false. x <0 (x+2)(x+1) > 0*-7 Multiply each side by -7 to remove these constants. Don’t forget to change inequality. X= -1.5 gives 6 > 0 which is true X = -0.5 gives -0.66 > 0 which is false. X = 1 gives .166 > 0 which is true. X ☺ X ☺ -2 -1 0 Thus, the final solution is -2 < x < -1 and x > 0. 9-5 Problems 1. Solve the following rational equations: z -1 = 2 z2 z-1 = z3 z=1 x+2 1 3 1 2 5x+5 = x+1 + 5 x = -2 x x=2 2. Solve the following rational inequalities: > 2 x < -2 and -1<x<1 x+1 x x-2 x 4 < x+4 -4 <x< -2 and 0<x<4 1