To construct a rational function with a vertical asymptote at \( x = 3 \) and a removable discontinuity at \( x = -2 \), we can use the following approach:
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Vertical Asymptote at \( x = 3 \): A vertical asymptote occurs when the denominator of the function goes to zero but the numerator does not. Therefore, we will include a factor of \( (x - 3) \) in the denominator.
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Removable Discontinuity at \( x = -2 \): A removable discontinuity occurs when both the numerator and denominator have a common factor that cancels out. Thus, we need to include a factor of \( (x + 2) \) in both the numerator and denominator.
Putting this together, we can form the rational function:
\[ f(x) = \frac{(x + 2)(k)}{(x + 2)(x - 3)} \]
Here, \( k \) is any non-zero constant. The \( (x + 2) \) cancels out, resulting in:
\[ f(x) = \frac{k}{x - 3} \]
This function has:
- A vertical asymptote at \( x = 3 \).
- A removable discontinuity at \( x = -2 \) because while the factor \( (x + 2) \) appears in both the numerator and denominator, it cancels out.
Thus, a simple rational function that fulfills the requirements is:
\[ f(x) = \frac{1}{x - 3} \]
(Note: The \( (x + 2) \) factor was included merely to illustrate the removable discontinuity; it cancels out, leaving us with the desired vertical asymptote).
In summary, the key points were identifying where you want vertical asymptotes and removable discontinuities, represented by the factors in the rational function.