Question

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.