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.

Answer 1

To construct a rational function that has a vertical asymptote at \( x = 3 \) and a removable discontinuity at \( x = -2 \), we can follow these steps:

1. Identify the factors related to the asymptote and the discontinuity:

   • A vertical asymptote at \( x = 3 \) means that the function will tend to infinity as \( x \) approaches 3. This occurs when the denominator has a factor of \( (x - 3) \).
   • A removable discontinuity at \( x = -2 \) means that the function is not defined at \( x = -2 \), but it can be "removed" by canceling factors. This occurs when both the numerator and denominator have a common factor of \( (x + 2) \).

2. Construct the rational function: The simplest rational function that meets these requirements can be constructed as follows: \[ f(x) = \frac{(x + 2)(\text{some polynomial})}{(x + 2)(x - 3)} \] To create a function with a vertical asymptote at \( x = 3 \) and a removable discontinuity at \( x = -2 \), we can choose a simple polynomial (or constant) for the numerator. A straightforward choice is to use the factor \( (x + 2) \) for the numerator and eliminate it, like this: \[ f(x) = \frac{(x + 2)}{(x + 2)(x - 3)} = \frac{1}{x - 3} \] However, since we want the removable discontinuity to be in the function but not part of its essence, we should just consider: \[ f(x) = \frac{(x + 2)(k)}{(x + 2)(x - 3)} \] where \(k\) is a nonzero constant (it could be \(1\) for simplicity).

   Therefore, we can simplify to: \[ f(x) = \frac{1}{x - 3} \]

3. Full function clarification: While the choice seems a bit limited, we might want to elaborate further so we don't just simplify:

   Let’s give \( f(x) \) as: \[ f(x) = \frac{k(x + 2)}{(x + 2)(x - 3)} \] Once simplified (and assuming \(k=1\)), this gives us: \[ f(x) = \frac{1}{x - 3} \]

The final function that meets these criteria is: \[ f(x) = \frac{(x + 2)}{(x + 2)(x - 3)} = \frac{1}{x - 3} \text{ for } x \neq -2 \]

Summary:

• The vertical asymptote is at \( x = 3 \) due to the factor \( (x - 3) \) in the denominator.
• The removable discontinuity occurs at \( x = -2 \) where the \( (x + 2) \) factor exists in both the numerator and denominator, and thus cancels.

So, our rational function fulfills both criteria successfully.