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.
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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