Chapter 6 Rational Expressions and Functions

6.1 Rational Functions and Equations Rational Function Let p(x) and q(x) be polynomials. Then a rational function is given by f(x) = The domain of f includes all x-values such that q(x) = 0 Examples - 4, x, 3x 2 –6x + 1 x x – 5 3x - 7

Identify the domain of rational function (Ex 2 pg 405) b)g(x) = 2x x 2 - 3x + 2 Denominator = x 2 - 3x + 2 = 0 (x – 1)(x –2) = 0 Factor x = 1 or x = 2 Zero product property Thus D = { x / x is any real number except 1 and 2 }

Using technology ( ex 73, pg 414 ) [ -4.7, 4.7, 1 ] by [ -3.1, 3.1, 1 ] (ex 80, pg 414)

Highway curve ( ex 100, page 415 ) R(m) = slope a) R(0.1) = = 457 About 457 : a safe curve with a slope of 0.1 will have a minimum radius of 457 ft b) As the slope of banking increases, the radius of the curve decreases c) 320 =, 320( 15m + 2) = 1600, 4800m +640 = m = 960, m = = 0.2 Radius

Evaluating a rational function Evaluate f(-1), f(1), f(2) Numerical value x y 3/2 4/3 1 0 __ 4 3 f(x) = 2x x f(-1) = 1 f(1) = undefined and f(2) = 4 Vertical asymptote

6.2 Multiplications and Divisions of Rational Expression To multiply two rational expressions, multiply numerators and multiply denominators., B and C not zero  = ., B and D are nonzero. Example To divide two rational expressions, multiply by the reciprocal of the divisor.  ÷ =, B, C, and D are nonzero Example

6.3 Addition and Subtraction of Rational Expressions To add (or subtract) two rational expressions with like denominators, add (or subtract) their numerators. The denominator does not change. Example, C is not zero Example

Finding the Least Common Multiple Step 1: Factor each polynomial completely Step 2: List each factor the greatest number of times that it occurs in either factorization. Step 3: Find the product of this list of factors. The result is the LCM

6.4 Solving rational equations graphically and numerically ( Ex- 3, pg 442 ) Solution- The LCD for 2,3, and 5 is their product, 30. (Multiply by the LCD ) (Distributive property) x = 6x (Reduce) 4x = -15 (Subtract 6x and 15) x = (Solve) Graphically Y1 = Y2 = [ -9, 9, 1] by [ -6, 6, 1]

Determining the time required to empty a pool (Ex 6.4, pg 450, no.88) A pump can empty a pool in 40 hours. It can empty of the pool in 1 hour. In 2 hour, can empty a pool in th of the pool Generally in t hours it can empty a pool in of the pool. Second pump can empty the pool in 70 hours. So it can empty a pool in of the pool in t hours. Together the pumps can empty of the pool in t hours. The job will complete when the fraction of the pool is empty equals 1. The equation is = 1 (40)(70) = 1 (40)(70) Multiply (40)(70) 70t + 40t = t = 2800 t = = hr. Two pumps can empty a pool in hr

Ex 93 Pg 416 A tugboat can travel 15 miles per hour in still water 36 miles upstream ( 15 – x) Total time 5 hours downstream (15 + x) t = So the equation is = 5 The LCD is (15-x)(15 + x) Multiply both sides by LCD, we get (15 – x)(15 + x)[ ] = 5 (15 – x)(15 + x) x x = 1125 – 5x 2 5x 2 – 45 = 0 5x 2 = 45 x = + 9, x = 3 mph

Modeling electrical resistance R1 = 120 ohms R2 = 160 ohms R

6.6 Modelling with Proportion a c is equivalent to ad = bc b d Example x 6x = 40 or x = 6 feet h feet 4 feet 44 feet

Modeling AIDS cases [ 1980, 1997, 2] by [-10000, , ] Y = 1000 ( x – 1981) 2