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Rational Functions

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Rational Functions Rational Functions A rational function is a function that is a ratio of 2 polynomials (in reduced form), e.g. p(x) f(x) = q(x) where p(x) and q(x) are polynomials The function is defined when the denominator does not equal 0 (this is called a domain of a function). So, in order to find the domain, we must find numbers that are not in the domain (e.g. where the denominator equals to 0), so all numbers excluding these numbers are in the domain of the function. When the denominator of a rational function equals to 0, one of two things can happen. The rational function may have 1) Vertical asymptote(s) 2) hole(s) or discontinuities (this happens when the function is not in lowest terms and can be reduced further, so one or more factors in denominator cancels with one or more factors in the numerator) Interesting fact though, that if denominator never equal to 0 (e.g. sum of squares), then the rational function may not have any vertical asymptotes. First, we’ll practice with finding the domain of a rational function. 1) Find the domain of the function. x 3 f (x) x 4 Solution: Set denominator equal to 0, then solve for x. x + 4 = 0 x = -4 So, the function is defined for all x, x ≠ 4. We can write the domain in set notation or interval notation (we’ll use interval notation after this example). Set notation: Domain: { x |x ≠ 4} Interval notation: (-∞, 4) (4, ∞) 2) Find the domain. x 1 f (x) x2 1 x2 + 1 = 0 Difference of squares can never be 0 (always positive for real numbers). So, there are no numbers that make the function undefined, hence the domain is all real numbers. D: (-∞, ∞) Next, let’s practice finding asymptotes. On details regarding finding asymptotes, refer notes on asymptotes. 3) Find the domain and all asymptotes of the function. x f (x) x 1 Solution: 1) To find the domain and any vertical asymptotes, we’ll set the denominator equal to 0. x – 1 = 0 x = 1 This is a vertical asymptote (where the function is not defined) Domain: (-∞, 1) (1, ∞) Vertical Asymptote (V.A.): x = 1 Equation of vertical line 2) To find Horizontal (or oblique) asymptotes, we must compare degrees of top and bottom. deg top = deg bottom (both equal to 1) So, horizontal asymptote equals to ratio of leading coefficients of top and bottom. H.A. : y = 1/1 = 1 Since both leading coefficients are 1. This is a horizontal line. 4) Find all asymptotes. x2 6x 8 (x 4)(x 2) x 2 f (x) x2 x 12 (x 4)(x 3) x 3 1) Note that x – 4 cancels, so x = 4 is a hole (discontinuity) but not a V.A. To find vertical asymptote, we must set the remaining factor to 0. x + 3 = 0 V.A. x = -3 2) H.A.: y = 1 Since both deg of top and bottom are equal and lead. coeffs. are 1. 5) Find all asymptotes. x 1 f (x) x2 5 1) x2 + 5 ≠ 0 Why? V.A.: none 2) H.A.: y = 0 Since deg top < deg bottom 6) Find all asymptotes. 2x 3 f (x) 3x 5 1) 3x + 5 = 0 3x = -5 V.A.:x = -5/3 2) H.A.: y = 2/3 Why? 7) Find all asymptotes. x2 49 (x 7)(x 7) f (x) x 7 x 7 x 7 Note that the function can be reduced. Since x – 7 is cancelled from denominator, it’s a hole (not V.A.). So, V.A.: none H.A.: none Since deg top > deg bottom The graph is actually a line y = x + 7 with a hole at x = 7 (giving a point (7, 14)). 8) Find all asymptotes: x 2 1 f (x) x 2 V.A.: x = 2 H.A.: none since deg top > deg bottom O.A. (oblique): y = x + 2 since deg top = 1 + deg bottom (top is 1 bigger) To find O.A., we must perform long (or synthetic for linear factors in denominator). x + 2 x 2 x2 0x 1 -(x2 – 2x) 2x + 1 -(2x – 4) 5 The answer y = x + 2 is the oblique asymptote. Now, we must learn how to graph rational functions. Note: graphing calculator may not show asymptotes, but we must still indicate them on the graph. We can graph rational functions in three steps. 1) Find all asymptotes. 2) Find all intercepts. 3) Test additional points (if necessary) in between Vertical asymptote(s). 1) Graph. 2 f (x) x 3 1) V.A.: x = 3 H.A.: y = 0 2) y – int: (x = 0) (0, -2/3) x- int: none since 2 (x – 3) 0 (x – 3) x 3 2 = 0 Contradiction! 3) We know that the graph to the left of x = 3 will be on the below y = 0 (H.A.) (due to y-int.), but we must test a point to the right of x = 3. Let’s pick x = 4 (to plug into f(x)). f(4) = 2/(4 – 3) = 2 Since y is positive, the graph will be above y = 0 (H.A.) We have enough info to sketch a graph: 2) Graph. x 2 f (x) x 3 1) V.A.: x = -3 H.A.: y = 1 2) y-int: (0, -2/3) x-int: x 2 (x + 3) 0(x + 3) x 3 x – 2 = 0 x = 2 (2,0) We, actually have enough info to sketch a graph. First, we must sketch asymptotes. Then, we know that the graph to the left of x = -3 must be above y = 1 (since there are no other x-int except the one we found), and to the right of x = -3, it must be below y = 1 (because of intercepts). Graph: 3) Graph: x f (x) x2 4 1) V.A.: x = -2, x = 2 H.A.: y = 0 2) y-int: (0, 0) x-int: (0, 0) 3) Since one intercept does not tell us much about the graph, we must test more points (look for sign). I pick the following x on either sides of asymptotes to help determine the shape of graph. x = -3: f(-3) < 0 x = -1: f(-1) > 0 x = 1: f(1) < 0 x = 3: f(3) > 0 Graph: 4) Graph. x 2 1 f (x) x 2 V.A.: x = 2 H.A.: none O.A.: y = x + 2 (long division) y-int: (0, -1/2) x-int: none (since the numerator is never 0) To sketch a graph, we must draw asymptotes first, then sketch a graph. Graph: .
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