Rational Functions 4-2

A Rational Function is an equation in the form of f(x) = p(x)/q(x), where p(x) and q(x) are polynomial functions, and q(x) does not equal zero.

1 x The parent rational function is f(x) = . Its graph is a hyperbola, which has two separate branches.

Rational functions may have asymptotes Rational functions may have asymptotes. The function f(x) = has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. 1 x

The rational function f(x) = can be transformed by using methods similar to those used to transform other types of functions. 1 x

1 x Using the graph of f(x) = as a guide, describe the transformation and graph each function. 1 x + 2 1 x – 3 A. g(x) = B. g(x) = Because h = –2, translate f 2 units left. Because k = –3, translate f 3 units down.

Identify the zeros and vertical asymptotes of f(x) = . (x2 + 3x – 4) x + 3 Step 1 Find the zeros and vertical asymptotes. (x + 4)(x – 1) x + 3 f(x) = Factor the numerator. The numerator is 0 when x = –4 or x = 1. Zeros: –4 and 1 The denominator is 0 when x = –3. Vertical asymptote: x = –3

Identify the zeros and vertical asymptotes of f(x) = . (x2 + 7x + 6) x + 3 Step 1 Find the zeros and vertical asymptotes. (x + 6)(x + 1) x + 3 f(x) = Factor the numerator. The numerator is 0 when x = –6 or x = –1 . Zeros: –6 and –1 The denominator is 0 when x = –3. Vertical asymptote: x = –3

Some rational functions, including those whose graphs are hyperbolas, have a horizontal asymptote. The existence and location of a horizontal asymptote depends on the degrees of the polynomials that make up the rational function. Note that the graph of a rational function can sometimes cross a horizontal asymptote. However, the graph will approach the asymptote when |x| is large.

To find Horizontal Asymptotes, compare the degrees of the numerator and the denominator. (3 scenarios): Since degree of numerator is less than the degree of the denominator A horizontal asymptote occurs at y = 0 Since degrees of the numerator and denominator are equal, divide the coefficients of the highest degree terms. A horizontal asymptote occurs at y = 9/5 Since the degree of numerator is larger than the degree of the denominator No horizontal asymptote

Identify the zeros and asymptotes of the function. Then graph. x – 2 x2 – 1 f(x) = x – 2 (x – 1)(x + 1) f(x) = Factor the denominator. The numerator is 0 when x = 2. Zero: 2 Vertical asymptote: x = 1, x = –1 The denominator is 0 when x = ±1. Horizontal asymptote: y = 0 Degree of p < degree of q.

Identify the zeros and asymptotes of the function. Then graph. Graph with a graphing calculator or by using a table of values.

Identify the zeros and asymptotes of the function. Then graph. 4x – 12 x – 1 f(x) = 4(x – 3) x – 1 f(x) = Factor the numerator. The numerator is 0 when x = 3. Zero: 3 Vertical asymptote: x = 1 The denominator is 0 when x = 1. The horizontal asymptote is y = = = 4. 4 1 leading coefficient of p leading coefficient of q Horizontal asymptote: y = 4

Identify the zeros and asymptotes of the function. Then graph. Graph with a graphing calculator or by using a table of values.

Identify the zeros and asymptotes of the function. Then graph. x2 + 2x – 15 x – 1 f(x) = (x – 3)(x + 5) x – 1 f(x) = Factor the numerator. The numerator is 0 when x = 3 or x = –5. Zeros: 3 and –5 The denominator is 0 when x = 1. Vertical asymptote: x = 1 Horizontal asymptote: none Degree of p > degree of q.

A Hole occurs at x = a whenever there is a common factor (x – a) in the numerator and denominator of the function. Example of a function with a hole:

Identify holes in the graph of f(x) = . Then graph. (x – 3)(x + 3) x – 3 f(x) = Factor the numerator. There is a hole in the graph at x = 3. The expression x – 3 is a factor of both the numerator and the denominator. (x – 3)(x + 3) (x – 3) For x ≠ 3, f(x) = = x + 3 Divide out common factors.

The graph of f is the same as the graph of y = x + 3, except for the hole at x = 3. On the graph, indicate the hole with an open circle. The domain of f is {x|x ≠ 3}. Hole at x = 3

Identify holes in the graph of f(x) = . Then graph. x2 + x – 6 x – 2 (x – 2)(x + 3) x – 2 f(x) = Factor the numerator. There is a hole in the graph at x = 2. The expression x – 2 is a factor of both the numerator and the denominator. For x ≠ 2, f(x) = = x + 3 (x – 2)(x + 3) (x – 2) Divide out common factors.

The graph of f is the same as the graph of y = x + 3, except for the hole at x = 2. On the graph, indicate the hole with an open circle. The domain of f is {x|x ≠ 2}. Hole at x = 2

1 x 1. Using the graph of f(x) = as a guide, describe the transformation and graph the function g(x) = . 1 x – 4 g is f translated 4 units right. 2. Identify the asymptotes, domain, and range of the function g(x) = + 2. 5 x – 1 asymptotes: x = 1, y = 2; D:{x|x ≠ 1}; R:{y|y ≠ 2}

3. Identify the zeros, asymptotes, and holes in the graph of 3. Identify the zeros, asymptotes, and holes in the graph of . Then graph. x2 – 3x + 2 x2 – x f(x) = zero: 2; asymptotes: x = 0, y = 1; hole at x = 1

Example: Create a function of the form y = f(x) that satisfies the given set of conditions: Vertical asymptote at x = 2, hole at x = -3

If we were looking for horizontal asymptotes, we would use scenario 3 where the degree of the numerator is greater than the degree of the denominator, and therefore no horizontal asymptote. However, when the degree of the numerator is exactly one greater than the degree of the denominator it is possible that there is a Oblique or Slant ASYMPTOTE.

Find the equation of the slant asymptote for the function . Use polynomial long division What happens as So the slant asymptote is the line

To determine the equation of the oblique asymptote, divide the numerator by the denominator,( use polynomial long division or synthetic division) and then see what happens as The equation of the slant asymptote for this example is y = x.

Example: Determine the slant asymptote for the following function:

5-Minute Check Lesson 3-8A

5-Minute Check Lesson 3-8B