Solutions for Rational Functions

Transcription

1 Solutions for Rational Functions I. Souldatos Problems Problem Let f(x) = x4 9 x 3 8. Find the domain of f(x). Set the denominator equal to 0: x 3 8 = 0 x 3 = 8 x = 3 8 = 2 So, the domain is all real numbers except 2. Write Domain: x 2. Find the y-intercept of f(x) (if any). Plug-in x = 0. f(0) = 9 8 = 9 8. Find the x-intercept(s) of f(x) (if any). Set the numerator equal to 0, but make sure that the denominator is not equal to 0 at the same time (equivalently, the x-intercepts must belong to the domain of f(x)). x 4 9 = 0 x 4 = 9 x = ± 4 9 So, we have two solutions and none of them makes the denominator equal to 0. Therefore, they are both x-intercepts. Grade yourself: One point for each question. Subtotal: / Same question for f(x) = (x+1)(x+2) (x+1)(x+3), but observe that (x + 1) appears twice. 1

2 Find the domain of f(x). Set the denominator equal to 0. (x + 1)(x + 3) = 0 (x + 1) = 0 or (x + 3) = 0 x = 1 or x = 3 So, the domain is all real numbers except 1, 3. Write Domain: x 1, 3. Find the y-intercept of f(x) (if any). Plug-in x = 0. f(0) = = 2 3. Find the x-intercept(s) of f(x) (if any). Set the numerator equal to 0, but make sure that the denominator is not equal to 0 at the same time (equivalently, the x-intercepts must belong to the domain of f(x)). (x + 1)(x + 2) = 0 (x + 1) = 0 or (x + 2) = 0 x = 1 or x = 2 So, we have two solutions, but x = 1 makes the denominator equal to 0. Therefore, x = 1 is not an x-intercept and x = 2 is the only one. Notice: If you want to cancel out (x + 1), you can cancel when you are finding the x-intercept(s), but not when you are finding the domain or the y-intercept. Grade yourself: One point for each question. If you kept x = 1 as an x-intercept, do not mark your answer to the last question correct. Subtotal: /3 Problem 2. Vertical Asymptotes Vertical asymptotes are determined by the numbers that make the denominator equal to zero Fill-in the following table. Use a calculator to find the vertical asymptotes. All x s for which = 0 All vertical asymptotes x 2 +5x+6 x = 2, 3 x = 2, 3 x 2 5 x = ± 5 x = ± 5 Grade yourself: One point for each question. Subtotal: /4 Page 2 of 5

3 2.2. Summarize your findings from problem 2.1 here: There is a vertical asymptote that every x that makes the denominator equal to Now repeat the same problem for the following functions. Observe that some factors appear both in the numerator and the denominator. All x s for which = 0 All vertical asymptotes (x+1)(x+2) (x+1)(x+3) x = 1, 3 x = 3 Grade yourself: (x 1)(x+3) (x 1)(x+2) x = 1, 2 x = 2 One point for each question. Subtotal: /4 Moral To find the domain of a rational function P (x), you set equal to 0. To find the vertical asymptotes, you cancel any common factor first and then you set the denominator equal to 0. Problem 3. Horizontal Asymptotes (also known as End-Behavior) Horizontal asymptotes are determined by the degree of the numerator, the degree of the denominator and their leading coefficients. Case I: Degree(numerator)<Degree(denominator) 3.1. Fill-in the following table. Use a calculator to find the horizontal asymptotes. Degree(P(x)) Degree() Horizontal asymptote(s) x+1 x y = 0 (x-axis) x 3 +3 x 4 +x (x-axis) (x+1)(2 x) x (x-axis) Grade yourself: One point for each question. Subtotal: / Summarize your findings from problem 3.1 here: If degree(numerator)<degree(denominator), then the x-axis is the horizontal asymptote. Case II: Degree(numerator)=Degree(denominator) Page 3 of 5

4 3.3. Fill-in the following table. Use a calculator to find the horizontal asymptotes. x x 4 +3 x 4 +3 x 4 +x 3 x 4 +x 3 2x 4 +x 3 Degree(P(x)) Degree() Leading coefficient of P (x) Leading coefficient of Horizontal asymptote(s) y = 1 y = 2 y = 1 2 Grade yourself: One point for each question. Subtotal: / Summarize your findings from problem 3.3 here: If degree(numerator)=degree(denominator), then the horizontal asymptote equals y = leading coefficient of P (x) leading coefficient of. Case III: Degree(numerator)>Degree(denominator) 3.5. Fill-in the following table. Use a calculator to find the horizontal asymptotes. Degree(P(x)) Degree() Horizontal asymptote(s) x there is no horizontal asymptote x 3 +3 x 2 +x 3 2 there is no horizontal asymptote (x+1)(2 x) x there is no horizontal asymptote Grade yourself: One point for each question. Subtotal: / Summarize your findings from problem 3.5 here: If degree(numerator)>degree(denominator), then there is no horizontal asymptote. Problem 4. Use your findings from the previous problems to find a possible formula for the function f(x) = P (x) graphed below. Page 4 of 5

5 What are the x-intercepts of f(x)? x = 3, 2 What are the factors of P (x)? (x + 3), (x 2) What are the vertical asymptotes of f(x)? x = 2, 1 What are the factors of? (x + 2), (x 1) What is the horizontal asymptote of f(x)? y = 1 Which one is larger? They are equal (Correct Answer: C) What is the y-intercept of f(x)? 3 Use your answers above to find a possible formula for f(x). f(x) will be of the form a (x+3)(x 2) (x+2)(x 1), where a is a constant number. The degree of the numerator is 2 and the same is true for the degree of the denominator. The leading coefficient of the numerator is a 1 1 = a and the leading coefficient of the denominator is 1 1 = 1. The horizontal asymptote equals y = Therefore, a = 1 and f(x) = (x+3)(x 2) (x+2)(x 1). We verify the y-intercept. f(0) = 3 ( 2) answers. leading coefficient of P (x) leading coefficient of = a 1 = 1. 2 ( 1) = 6 2 = 3. So, everything matches the above Grade yourself: One point for each question, except the last one. For the last question, give yourself one point if you found the numerator correctly, one point if you found the denominator correctly and one point if you found the value of a correctly. Subtotal: /10 Page 5 of 5