Construct a rational function that has a vertical asymptote at x = 3 and a removable discontinuity at x equals negative 2. Explain how you determined your answer.
(4 points)
To create a rational function with a vertical asymptote at x = 3 and a removable discontinuity at x = -2, we need to include factors of (x - 3) in the denominator for the vertical asymptote, and (x + 2) in both the numerator and denominator for the removable discontinuity.
One possible rational function that meets these criteria is:
f(x) = (x^2 + 5x + 6) / ((x - 3)(x + 2))
Explanation:
1. Vertical Asymptote at x = 3:
To have a vertical asymptote at x = 3, we include the factor (x - 3) in the denominator. This ensures that the function will approach positive or negative infinity as x approaches 3.
2. Removable Discontinuity at x = -2:
To have a removable discontinuity at x = -2, we include the factor (x + 2) in both the numerator and denominator. This creates a common factor that can be canceled out, resulting in a simplified function without a discontinuity at x = -2.
By having (x + 2) in the numerator, we also ensure that the function has a point of discontinuity at x = -2. However, since this factor is canceled out by having (x + 2) in the denominator as well, the discontinuity becomes removable, resulting in a hole in the graph at x = -2.
Overall, the function f(x) = (x^2 + 5x + 6) / ((x - 3)(x + 2)) satisfies both the requirement of a vertical asymptote at x = 3 and a removable discontinuity at x = -2.