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.
To construct a rational function with a vertical asymptote at x = 3 and a removable discontinuity at x = -2, we need to consider the characteristics of rational functions.
A rational function can be represented as the ratio of two polynomials: f(x) = P(x) / Q(x), where P(x) and Q(x) are polynomials.
1. Vertical asymptote at x = 3:
For a rational function to have a vertical asymptote at x = 3, the denominator Q(x) should have a factor (x - 3). This means that the vertical asymptote occurs when the function approaches infinity or negative infinity as x approaches 3.
2. Removable discontinuity at x = -2:
For the function to have a removable discontinuity at x = -2, there should be a factor (x + 2) in both the numerator P(x) and denominator Q(x). This means that there is a hole in the graph at x = -2, but we can cancel out the corresponding factors to make the function continuous at that point.
Based on these considerations, a possible rational function could be:
f(x) = (x + 2)(x - 3) / (x + 2)(x - 3)
In this function, the factors (x + 2) and (x - 3) appear in both the numerator and denominator, resulting in a removable discontinuity at x = -2 and a vertical asymptote at x = 3. By canceling out the common factors, we have f(x) = 1/1 = 1, making the function continuous at x = -2.
It's important to note that this is not the only possible rational function with these characteristics. Other forms with different constant factors or additional terms can also satisfy the given conditions.