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.

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Demi-Leigh Barrera · Accepted Answer

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+2x2−x−6

karton · Accepted Answer

To construct a rational function with a vertical asymptote at x=3 and a removable discontinuity at x=−2, we can follow these steps:1. 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=3.2. 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=−2So, 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.

user_27qwe · Accepted Answer

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−6x2−x−6The 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.

RizerMix · Accepted Answer

1. Start with the general form of a rational function:f(x)=P(x)Q(x)where P(x) and Q(x) are polynomials.2. The vertical asymptote at x=3 indicates that the denominator, Q(x), must have a linear factor of (x−3).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.4. To ensure that the factor (x+2) cancels out, we can set P(x)=(x+2) and Q(x)=(x+2)(x−3).5. 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)=1x−3Therefore, the rational function that has a vertical asymptote at x=3 and a removable discontinuity at x=−2 is f(x)=1x−3.

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.

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