6 1 Rational Expressions Rational Expression an expression

6. 1 Rational Expressions Rational Expression – an expression in which a polynomial is divided by another nonzero polynomial. Examples of rational expressions 4 x x 2 x – 5 Domain = {x | x 0} Domain = {x | x 5/2} 2 x– 5 Domain = {x | x 5}

Graph of a Rational Function y= 1 x x -2 -1 -1/2 0 ½ 1 2 y -1/2 -1 -2 Undefined 2 1 ½ The graph does not cross the x = 0 line since x the graph is undefined there. . The line x = 0 is called a vertical asymptote. An Application: Modeling a train track curve.

Multiplication and Division of Rational Expressions A • C = A B • C B 9 x 3 x 2 5 y – 10 = 5 (y – 2) 10 y - 20 10 (y – 2) 2 z 2 – 3 z – 9 z 2 + 2 z – 15 A 2 – B 2 A+B 3 x = 5 =1 10 2 = (2 z + 3) (z – 3) (z + 5) (z – 3) = = (A + B)(A – B) (A + B) = 2 z + 3 z+5 = (A – B)

Negation/Multiplying by – 1 -y – 2 4 y + 8 = y+2 4 y + 8 OR -y - 2 -4 y - 8

Check Your Understanding Simplify: x 2 – 6 x – 7 x 2 -1 1 x-2 (x + 1) (x – 7) (x – 1) 1 x– 2 3 x 2 + x - 6 3 • • (x + 3) 3 (x + 3) (x – 2) 3

6. 2 Addition of Rational Expressions Adding rational expressions is like adding fractions With LIKE denominators: 1 8 + x x+2 x 3 x 2 + 4 x - 4 2 8 + + = 3 x - 1 x+2 = 3 8 4 x - 1 x+2 2 (2 + x) = = 3 x 2 + 4 x -4 (3 x 2 + 4 x – 4) (3 x -2)(x + 2) = 1 (3 x – 2)

Adding with UN-Like Denominators 3 4 1 x 2 – 9 + 1 8 (3) (2) 8 + 6 8 1 8 + 7 8 1 8 + 2 x+3 1 (x + 3)(x – 3) + 2 (x + 3) 1 (x + 3)(x – 3) + 2 (x – 3) (x + 3)(x – 3) 1 + 2(x – 3) (x + 3) (x – 3) = 1 + 2 x – 6 (x + 3) (x – 3) = 2 x - 5 (x + 3) (x – 3)

Subtraction of Rational Expressions To subtract rational expressions: Step 1: Get a Common Denominator Step 2: Combine Fractions DISTRIBUTING the ‘negative sign’ BE CAREFUL!! 2 x x 2 – 1 = - x– 1 (x + 1)(x – 1) x+1 x 2 - 1 = = 2 x – (x + 1) x 2 -1 1 (x + 1) = 2 x 2 – x - 1 x -1

Check Your Understanding Simplify: b 2 b - 4 b-1 b-2 - b 2(b – 2) - b-1 b-2 b 2(b – 2) + -b+1 b-2 b 2(b – 2) + 2(-b+1) 2(b – 2) b – 2 b+2 2(b – 2) = -b + 2 2(b – 2) = -1(b – 2) 2(b – 2) = -1 2

6. 3 Complex Fractions A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both. Examples: 1 5 x x 2 – 16 1 x + 2 x 2 4 7 1 x-4 3 x - 1 x 2 7/20 x x+4 x+2 3 x - 1

6. 4 Division by a Monomial 3 x 2 + x x 4 x 2 + 8 x – 12 4 x 2 15 A 2 – 8 A 2 + 12 4 A 5 x 3 – 15 x 2 15 x 5 x 2 y + 10 xy 2 5 xy 12 A 5 – 8 A 2 + 12 4 A

Polynomial Long Division Example: Divide 4 – 5 x – x 2 + 6 x 3 by 3 x – 2. Begin by writing the divisor and dividend in descending powers of x. Then, figure out how many times 3 x divides into 6 x 3. Multiply. 3 x – 2 2 x 2 6 x 3 – x 2 – 5 x + 4 6 x 3 – 4 x 2 3 x 2 – 5 x Divide: 6 x 3/3 x = 2 x 2. Multiply: 2 x 2(3 x – 2) = 6 x 3 – 4 x 2. Subtract 6 x 3 – 4 x 2 from 6 x 3 – x 2 and bring down – 5 x. Now, divide 3 x 2 by 3 x to obtain x, multiply then subtract. Multiply. 3 x – 2 2 x 2 + x -1 6 x 3 – x 2 – 5 x + 4 6 x 3 – 4 x 2 3 x 2 – 5 x 3 x 2 – 2 x -3 x + 4 -3 x +2 Answer: 2 x 2 + x – 1 + 2 3 x - 2 2 Divide: 3 x 2/3 x = x. Multiply: x(3 x – 2) = 3 x 2 – 2 x. Subtract 3 x 2 – 2 x from 3 x 2 – 5 x and bring down 4. Subtract -3 x + 2 from -3 x + 4, leaving a remainder of 2.

More Long Division 3 x -11 3 3 x + 2 9 x + 9 x -11 x 2 - 5 x - 3 -11 x 2 - 33 x - 33 28 x+30

6. 5 -6. 6 Rational Equations (2 x – 1) 3 x 2 x – 1 (x - 2) = 3 3 x = 3(2 x – 1) 3 x = 6 x – 3 -3 x = -3 x+1 x– 2 (x + 1) = 3 x-2 6 x+1 =x x+1=3 6 = x (x + 1) x=2 6 = x 2 + x x=1 x 2 + x – 6 = 0 (x + 3 ) (x - 2 ) = 0 Careful! – What do You notice about the answer? x = -3 or x=2

Rational Equations Cont… To solve a rational equation: Step 1: Factor all polynomials Step 2: Find the common denominator Step 3: Multiply all terms by the common denominator Step 4: Solve (12 x) x+1 2 x - x– 1 4 x = 1 3 = 6 (x + 1) -3(x – 1) = 4 x 6 x + 6 – 3 x + 3 = 4 x 3 x + 9 = 4 x -3 x 9 = x

Other Rational Equation Examples 3 x– 2 (x + 2)(x – 2) 3 x– 2 + 5 x+2 = 12 x 2 - 4 1 x + 5 x+2 = 12 (x + 2) (x – 2) 4 x + 4 + 1 = 3 x 2 4 3 x 2 - 4 x 3(x + 2) + 5(x – 2) = 3 x + 6 + 5 x – 10 = 8 x – 4 = 12 +4 +4 8 x = 16 x = 2 12 12 (4 x 2) = 3 x 2 - 4 =0 (3 x + 2) (x – 2) = 0 3 x + 2 = 0 or x– 2=0 3 x = -2 or x=2 x = -2/3 or x=2

Check Your Understanding Simplify: x 1 + x 2 – 1 1 x– 2 1 x(x – 1) - 1 x-1 3 x + 1 2(x – 3) x(x – 2) x 2 – 1 - 2 x(x + 1) 3 x(x – 1)(x + 1) Solve 6 - 1 x 2 4 =1 3 2 x – 1 = 2 2 x– 1 + 3 5 x+1 x+2 = x x 2 + x - 2 -1/4 Try this one: Solve for p: 1 =1 + 1 F p q

6. 7 Proportions & Variation Proportion equality of 2 ratios. Proportions are used to solve problems in everyday life. 1. If someone earns $100 per day, then how many dollars can the person earn in 5 days? 100 x (x)(1) = (100)(5) = 5 1 x = 500 2. If a car goes 210 miles on 10 gallons of gas, the car can go 420 miles on X gallons 210 10 3. = 420 x (210)(x) = (420)(10) (210)(x) = 4200 x = 4200 / 210 = 20 gallons If a person walks a mile in 16 min. , that person can walk a half mile in x min. 16 1 = x ½ (x)(1) = ½(16) x = 8 minutes

The Shadow Problem Juan is 6 feet tall, but his shadow is only 2 ½ feet long. There is a tree across the street with a shadow of 100 feet. The sun hits the tree and Juan at the same angle to make the shadows. How tall is the tree? personheight treeheight 6 x = 2. 5 100 personshadow treeshadow 6 ft x 2. 5 x = (100)(6) 2. 5 x = 600 2. 5 x = 240 feet 2 ½ ft 100 ft

7. 6 Direct Variation y = kx y is directly proportional to x. y varies directly with x k is the constant of proportionality Example: y = 9 x (9 is the constant of proportionality) Let y = Your pay Let x – Number of Hours worked Your pay is directly proportional to the number of hours worked. Example 1: Example 2: Salary (L) varies directly as the number of hours worked (H). Write an equation that expresses this relationship. Aaron earns $200 after working 15 hours. Salary = k(Hours) L = k. H Find the constant of proportionality using your equation in example 1. . 200 = k(15) So, k = 200/15 = 13. 33

Inverse Variation y= k x y is inversely proportional to x y varies inversely as x Example: y varies inversely with x. If y = 5 when x = 4, find the constant of proportionality (k) 5 = k 4 So, k = 20

Direct Variation with Power y = kxn y is directly proportional to the nth power of x Example: Distance varies directly as the square of the time (t) Distance = kt 2 D = kt 2 Joint Variation y = kxp • y varies jointly as x and p