Math 121. Graphing Rational Functions Fall 2016

Transcription

1 Math 121. Graphing Rational Functions Fall Let x2 85 x (a) State the domain of f, and simplify f if possible. (b) Find equations for the vertical asymptotes for the graph of f. (c) For each vertical asymptote found in part (b), determine the behavior of f just to the right and just to the left of the vertical asymptote. Confirm your answer by creating a table of values. (d) Find all values of c for which there is a hole in the graph of f above x = c. Solution: (a) The domain of f is {x : x 2, x 5}, and we simplify f as (x 5) (x 5) = x 3 x 2 x 2, x 5 (b) The zeros in the denominator of the simplified rational function provide the locations of the vertical asymptotes. Therefore, the graph of f has vertical asymptote x = 2. (c) For numbers x slightly bigger than 2 we have 1 = a very large negative number. a small positive number Therefore, f decreases without bound just to the right of the asymptote. For numbers x slightly smaller than 2 we have 1 = a very large positive number. a small negative number Therefore, f increases without bound just to the left of the asymptote. A table of values is as follows x (d) There would be a hole in the graph above x = 5 since 5 is not in the domain of f, but there is no vertical asymptote at x = 5.

2 2. Find all horizontal asymptotes, or explain why there is no horizontal asymptote, for the rational functions F, G and H defined below. (a) F (x) = 14x8 + 62x x x 11 (b) G(x) = 14x8 + 62x x 8 31x (c) H(x) = x5 + 14x x x 6 Solution: (a) The horizontal asymptote is y = 0 because the degree of the numerator of F is 8 and this is less than the degree of the denominator of F which is 11. (b) The degrees of the numerator and denominator of G are both 8, and so the horizontal asymptote is determined by the ratio of the leading coefficients of G, that is the horizontal asymptote has equation y = = 7 10 (c) There is no horizontal asymptote because the degree of the numerator of H is 8 and this is greater than the degree of the denominator of H which is Find all slant and horizontal asymptotes for the rational function F (x) = 4x3 4x 2 3x + 6 2x Solution: There are no horizontal asymptotes because the degree of the numerator is larger than the degree of the denominator. There is a slant asymptote because the degree of the numerator is the degree of the denominator plus 1. The slant asymptote is found by long division as follows. 2x 2 2x x 3 4x 2 3x + 6 (4x 3 + 2x) 4x 2 5x ( 4x 2 2) 5x + 8 Therefore, and the slant asymptote is the line y = 2x 2. F (x) = 2x x 2x Page 2

3 4. Let x2 5x + 6 x 2 4x + 3. (a) Simplify f and find its domain. (b) Find equations for the vertical asymptote(s) for the graph of f. (c) Find the x- and y-intercepts of the graph of f. (d) For each vertical asymptote found in part (b), determine the behavior of f just to the right and just to the left of the vertical asymptote. (e) Find all values of c for which there is a hole in the graph of f above x = c. (f) Find all horizontal asymptote of f. (g) Use the information above and plot additional points a necessary to graph f. Solution: (a) The domain of f is {x : x 1, x 3} (see the simplified form of f below): (x 1) = x 2 x 1 x 1, x 3 (b) The zeros in the denominator of the simplified rational function provide the locations of the vertical asymptotes. Therefore, the graph of f has vertical asymptote x = 1, there will be a hole in the graph at x = 3. (c) The x-intercept is (2, 0) and the y-intercept is (0, 2). (d) When x is close to, but larger than 1 we have (x 1) 1 = large negative number. small positive number When x is close to, but smaller than 1 we have (x 1) = 1 = large positive number. small negative number Thus the graph decreases without bound as x approaches 1 from the right and increases without bound as x approaches 1 from the left. (e) There would be a hole in the graph above x = 3 (since 3 is not in the domain of f, but there is no vertical asymptote at x = 3). (f) The degrees of the numerator and denominator are the same, so the horizontal asymptote is found by setting y equal to the ratio of the leading coefficients, that is y = 1 (g) A graph of f along with its asymptotes is as follows Page 3

4 y x 5. Use the rational function F defined below to answer the following questions. F (x) = (x + 2)(x 1) (a) Find the domain of F. Express answer in interval notation. (b) Find the x-intercept(s) of F, if there are any. (c) Find the y-intercept(s) of F, if there are any. (d) Find the horizontal asymptote(s) of F, if there are any. (e) Find the slant asymptote(s) of F, if there are any. (f) Find all vertical asymptotes of F, if there are any. For each vertical asymptote, determine the behavior of F just to the right, and just to the left of the asymptote. (g) Use the information from (a) through (f), along with plotting some additional points as necessary to sketch the graph of F along with all of its asymptotes. Solution: (a) The domain of f is (, 1) ( 1, ). (b) The x intercepts are all x-values in the domain where the numerator of f is zero. So (x + 2)(x 1) = 0, that means the intercepts are (, 0) and (1, 0). (c) There is a y-intercept since 0 is in the domain of F, and it is (0, F (0)) where F (0) = (0 + 2)(0 1) = That is, the y-intercept is (0, ). (d) There is no horizontal asymptote because the degree of the numerator of F is larger than degree of the denominator of F (i.e. 2 > 1). Page 4

5 (e) There is a slant asymptote, and it is found by performing the division on F (either long, or synthetic in this case), and you can check so the slant asymptote is y = x. F (x) = (x + 2)(x 1) = x + (f) The vertical asymptote is x = 1 (since the denominator of F is 0 when x = 1 but the numerator is not). To check the behavior the graph of F just to the left of 1, notice that when x is slightly smaller than 1 we get F (x) small negative number which is a large positive number and so the graph of F increases without bound to the left of x = 1. To check the behavior the graph of F just to the right of 1, notice that when x is slightly larger than 1 we get F (x) small positive number which is a large negative number and so the graph of F decreases without bound to the right of x = 1. (g) A graph of f is as follows, the asymptotes are the dashed lines. 20 y x Page 5