Rational Functions (Fractions) Rational comes from ratio --ratios are fractions--division. Asymptotes The first degree rational function always has two asymptotes one vertical and one horizontal. The vertical asymptote occurs because there is a value of x that makes the denominator equal zero --and division by zero is undefined in our number system. The horizontal asymptote is best understood with limits in Calculus. It defines the value y approaches as the x-values head for either end of the axis, that is: x gets extremely small or extremely large. First-Degree Rational Function asymptotes are: x = h and y = k a < 0, curve always increases a > 0, curve always decreases x -h f(x) = + k a (standard form) f(x) = a x + b 1 1 (general form) a x + b 2 2 a x + b 2 vertical asymptote: solve 2 = 0 horizontal asymptote: y = a1 a 2 zero: set numerator = 0 y-intercept: set x = 0, get 1 b b2 Note: asymptotes are LINES so we have to write an equation. We cannot write "the vertical asymptote is 4". We must write x = 4 As we see, the domain and range would both be R if there were no asymptotes so we describe them exactly that way. We eliminate the asymptote value like this: domain: x ‘; x ! h range: y ‘; y ! k If reality dictates, the domain and range could be limited to only positive values. . The First Degree Rational Function (fractions)From General to Standard Form Though I much prefer to work with the general form of the rule, we should know how to change from one form to the other. The fraction tells us how --we have to divide --then make a minor modification once we're done. Example 1 Put this rational function rule of correspondence into standard form: f (x) = 6x − 15 2x − 8 We will divide 6x – 15 by 2x – 8. The quotient gives us k and the remainder gives us a --after the modification. 3 6x-15 9 2 8 6 15, so 3 2x-8 2 8 6 249 x x x x − − = + − − Now our only problem is that we don't have x – h in the denominator, we have a multiple of it. So, we factor out 2 from 2x – 8, then we divide the 9 in the numerator by 2. Now, we have standard form: f (x) = 9 2(x − 4) + 3 tf (x) = 9/2 (x − 4) + 3 Since a = 9/2 > 0, the curve is always DECREASING (rational functions are backwards) The vertical asymptote is x = 4 and the horizontal asymptote is y = 3. The domain is x ‘; x ! 4 range: y ‘; y ! 3 The y-intercept is 15/8 (set x = 0) and the zero is 5/2 (set the numerator = 0) Let's graph it. 1 2 3 4 12345 -5 -4 -3 -2 -1 -6 -1 -2 -3 -4 -5 5 6 f(x) = 6x -15 2x -8 f(x) = 9/2 x -4 + 3 The First Degree Rational Function (fractions)Now one from the general form (easiest in my opinion) Example 2 Describe the properties and draw the graph of g(x) = x − 3 2x + 6 Vertical Asymptote: set 2x + 6 = 0 so x = – 3 is VA Horizontal Asymptote: ratio of the coefficients of x, so y = ½ is HA y-intercept: set x = 0, we get (0, – ½); x-intercept: set numerator = 0 we get (3, 0) once we have the intercepts, we know the curve is increasing. f(x) = x -3 2x + 6 -3 ½(0, -½) (3, 0) Finding the Inverse Function In this case, we'll first solve for x, then switch the x's and y's, like this: Example 3 Find the inverse function y = 3x − 5 x + 4 1. cross multiply: y(x + 4) = 3x – 5 2. remove brackets: xy + 4y = 3x – 5 3. transpose all x’s to left side: xy – 3x = – 4y – 5 4. factor out x: x(y – 3) = – (4y + 5) 5. divide both sides by y – 3 6. switch the x's and y's x = − (4 y + 5) y − 3 , so f − 1(x) = − (4 x + 5) x − 3 ; x c ≠, x ! 3 Notice how we put all the x terms together and then factored it out. We would get the same result had we switched the variables first, then solved for y. . The First Degree Rational Function (fractions)Practice 1) Find the zero and the y-intercept for f (x) = 2x + 5 x + 1 2) List the vertical asymptote, horizontal asymptote, y-intercept and zero for: (remember asymptotes are lines -– they have equations!!) d) y = x + 1 x − 1 c) g(x) = 3x + 2 2 − 4x b) f (x) = 4x − 5 2x − 7 a) y = 3x − 1 x − 1 3) Find the inverse function for y = 2x − 5 x + 9 4) John calculates his average hourly profit doing landscape work with the rule P(x) = , 25x − 75 x + 2 where P(x) is the average hourly profit and x is the number of hours he works. a) How many hours must John work to break even? (no profit, no loss) b) What is his hourly profit if he works 8 hours? c) What is his maximum hourly profit? 5) a) Graph and describe the properties of f (x) = . 3x + 7 x + 1 b) Solve f (x) < 2 . Solutions 1) x = – 5/2, y = 5 2) d) y = x + 1 x − 1 VA: x = 1 HA: y = 1 y-int: (0, –1) zero: (–1, 0) c) g(x) = 3x + 2 2 − 4x VA: x = 1/2 HA: y = –3/4 y-int: (0, 1) zero: (–2/3, 0) b) f (x) = 4x − 5 2x − 7 VA: x = 7/2 HA: y = 2 y-int: (0, 5/7) zero: (5/4, 0) a) y = 3x − 1 x − 1 VA: x = 1 HA: y = 3 y-int: (0, 1) zero: (1/3, 0) The First Degree Rational Function (fractions)3) Find the inverse function for y = 2x − 5 x + 9 Switch x and y’s: r x = , then cross-multiply: x(y + 9) = 2y – 5. 2y − 5 y + 9 we’re trying to find y, so we multiply out the bracket and put the y-terms together on one side: xy + 9x = 2y – 5 so we transpose to get xy – 2y = – 9x – 5. Now we factor out y, and divide both sides by it’s coefficient: y(x – 2) = – 9x – 5, so the inverse function is y = − 9x − 5 x − 2 4) c) maximum < $25 since the horizontal asymptote is y = 25 a) P(x) = 0, so x = 3 b) P(8) = $12.50 make numerator = 0 5)a) 3 -1 (0, 7) (-7/3, 0) f (x) < 0 –7/3 < x < –1 f (x) > 0 x < – 7/3, or x > –1 Range: R , y ! 3 Domain: R , x ! –1 b) f (x) < 2 means 3x + 7 < 2x + 2, so x < –5 Functions MathRoom Index The First Degree Rational Function (fractions)