Rational functions #1 - Questions and Answers

Question 1 Give the range of the rational functions given that neither of the graphs crosses its horizontal asymptote. Write your answers in set notation and interval notation. Answer All real numbers are included in a rational function's range, with the exception of the y values that represent holes or horizontal asymptotes. In Example B, part(b), the horizontal asymptote was y = 2. Therefore, the range is all real numbers, except y = 2 The set notation for the range is {y ∈ R | y ≠ 2} The interval notation is y ∈ (-∞,2) U (2,∞) In 'Try These B, part(a), we were given y = (2-x) / (x+4) In a rational function, the horizontal asymptote is determined by if the degree of the numerator and the denominator are equal. y = (ratio of the leading coefficients in the numerator/denominator) In this case, we have y = -1/1 = -1 Therefore, the set notation for the range is {y ∈ R | y ≠ -1} The interval notation is y∈ (-∞,-1) U (-1,∞) Question 2 True or False. Every rational function has at least one vertical asymptote. Answer 1 False It is untrue that every rational function has at least one vertical asymptote. For instance f(x) = (x²-1) / (x+1) although there is no vertical asymptote, is a rational function. Result: False Answer 2 Result: If there exist any values of x that make the denominator of f(x) equal to zero, then f(x) may have vertical asymptotes at those values. If there are no values of x that make the denominator of f(x) equal to zero, then f(x) does not have any vertical asymptotes. Hence, we can conclude that the statement 'Every rational function has at least one vertical asymptote' is False. Solution: To solve the statement 'True or False. Every rational function has at least one vertical asymptote,' we need to consider the definition of a rational function and the conditions for the existence of vertical asymptotes. A rational function is defined as the ratio of two polynomials, where the denominator polynomial is not equal to zero. Let's assume our rational function is denoted as f(x). To determine if a rational function has vertical asymptotes, we need to check the behavior of the function as x approaches certain values. A vertical asymptote occurs when the function approaches positive or negative infinity as x approaches a particular value. This can happen if the denominator of the rational function becomes zero for certain values of x. Therefore, for a rational function f(x) to have a vertical asymptote, we need to find the values of x that make the denominator of f(x) equal to zero. These values are the potential vertical asymptotes. Answer 3 Step 1: A rational function is defined as the ratio of two polynomial functions, where the denominator is not equal to zero. To find vertical asymptotes of a rational function, we need to identify the values of x for which the denominator becomes zero. Let's consider a general rational function of the form: f(x) = P(x)/Q(x) where P(x) and Q(x) are polynomials and Q(x) is not equal to zero. Step 2: The vertical asymptotes occur when the denominator Q(x) becomes zero. Therefore, to find the vertical asymptotes, we need to solve the equation Q(x) = 0. If Q(x) has any real roots, then those roots correspond to vertical asymptotes. However, it is possible for a rational function to have no real roots for the denominator Q(x) = 0. In such cases, there would be no vertical asymptotes. Therefore, the statement Every rational function has at least one vertical asymptote is false, as there are cases where a rational function has no vertical asymptotes. Answer 4 To solve the statement 'True or False. Every rational function has at least one vertical asymptote,' we can provide a counterexample to show that it is false. Consider the rational function f(x) = 1/x. This function does not have any vertical asymptotes. The denominator x can be equal to 0, but as we approach x = 0, the function tends to positive or negative infinity rather than approaching a finite value. Hence, f(x) does not have a vertical asymptote. Therefore, the statement is false, and we have provided a counterexample to prove it. Question 3 Use the Rational Zero Theorem to list all possible rational zeros for the function f(x) = x³ + x² - 4x – 4. Answer possible rational zeros = (Factors of the constant term) / (Factors of the leading coefficient) List the factors of the constant term and the leading coefficient of the given function. Factors of the constant term, -4: ±1, ±2, ±4 Factors of leading co efficient, 1: ±1 possible rational zeros = (±1, ±2, ±4) / (±1) consider each number is numerator an divide it by each number in the denominator. Possible rational zeros= ±1, ±2, ±4 Therefore, the possible rational zeros of f(x) are ±1, ±2 and ±4 Question 4 Construct a rational function that has a vertical asymptote at x = 3 and a removable discontinuity at x = -2. Explain how you determined your answer. Answer 1 Since x=3 is a vertical asymptote, then x-3 is a factor in the denominator. Since there is a removable discontinuity at x=-2, then x+2 is a common factor in the numerator and denominator. So, a possible rational function is: f(x) = (x+2) / ((x-3)(x+2)) or f(x) = (x+2) / (x²-x-6) Answer 2 To construct a rational function with a vertical asymptote at x = 3 and a removable discontinuity at x = -2, we can follow these steps: Vertical Asymptote at x = 3: A vertical asymptote occurs when the denominator of the rational function is equal to zero at a certain value of x. Therefore, we need a factor of (x - 3) in the denominator to achieve the vertical asymptote at x = 3. Removable Discontinuity at x = -2: A removable discontinuity occurs when there is a common factor in both the numerator and the denominator of the rational function. To achieve a removable discontinuity at x = -2, we need a common factor of (x + 2) in both the numerator and the denominator. Combining these two steps, we can construct the rational function as follows: Let f(x) = (x+2)(x-a) / ((x-3)(x+2)), where a is a constant that we can choose to achieve the desired properties. To find the value of a, we need to ensure that the function has a removable discontinuity at x = -2. This means that the factor of (x + 2) should cancel out in both the numerator and the denominator. Therefore, we set (x + 2) equal to zero and solve for x: (x + 2) = 0 --> x = -2 So, a should be equal to -2. Thus, the rational function that satisfies the given conditions is: f(x) = (x+2)(x-2) / ((x-3)(x+2)) In simplified form, the function is: f(x) = (x+2)² / ((x-3)(x+2)) This function has a vertical asymptote at x = 3 and a removable discontinuity at x = -2. Answer 3 We can start by considering the characteristics of each type of asymptote. Vertical asymptote at x = 3: A vertical asymptote occurs when the denominator of the rational function becomes zero at a certain value of x. To achieve this, we can include a factor of (x - 3) in the denominator. Removable discontinuity at x = -2: A removable discontinuity is a hole in the graph of the function that can be filled to create a continuous function. To accomplish this, we can include a factor of (x + 2) in both the numerator and denominator, and cancel them out later. Combining these elements, we can construct a rational function in the following form: f(x) = (x+2)(x-3) / ((x+2)(x-3)) Notice that both the numerator and denominator contain the factors (x + 2) and (x - 3). When we simplify this expression, we will cancel out these common factors, resulting in a function with a removable discontinuity at x = -2 and a vertical asymptote at x = 3. Simplifying the expression: f(x) = (x² - x - 6) / (x² - x - 6) The factors (x + 2) and (x - 3) cancel out, leaving us with the simplified rational function. This function has a vertical asymptote at x = 3 and a removable discontinuity at x = -2. Answer 4 Start with the general form of a rational function: f(x) = P(x) / Q(x) where P(x) and Q(x) are polynomials. The vertical asymptote at x = 3 indicates that the denominator, Q(x), must have a linear factor of (x - 3). Since we want a removable discontinuity at x = -2, we need both the numerator, P(x), and the denominator, Q(x), to have a factor of (x + 2). However, for the removable discontinuity, this factor must cancel out when simplifying the rational function. To ensure that the factor (x + 2) cancels out, we can set P(x) = (x + 2) and Q(x) = (x + 2)(x - 3). Thus, the rational function that satisfies the given conditions is: f(x) = (x+2) / ((x+2)(x-3)) To simplify this function, we can cancel out the common factor of (x + 2): f(x) = 1 / (x-3) Therefore, the rational function that has a vertical asymptote at x = 3 and a removable discontinuity at x = -2 is f(x) = 1 / (x-3) Question 5 Find a function that has a vertical asymptote at x = 7 and a horizontal asymptote at y = 5. Answer A rational function in the form f(x) = a/(x-b) + c has vertical asymptote x=b and horizontal asymptote y=c. So, we substitute h=-7 and k=5. For simplicity, we use a=1: f(x) = 1/(x - (-7)) + 5 or when simplified as a single expression, f(x) = (1 / (x+7)) + 5 f(x) = (1 / (x+7)) + (5(x+7) / (x+7)) f(x) = (1 + 5x + 35) / (x+7) f(x) = (5x +36) / (x+7)