To find the holes of the function $f(x) = \frac{x+3}{x+2}$, we analyze its numerator and denominator to identify any removable discontinuities. Here’s the step-by-step process:

Steps:

1. Factorize the numerator and denominator:

   • $f(x) = \frac{x+3}{x+2}$.

   The numerator and denominator are already fully simplified.

2. Identify potential holes:

   • A hole occurs where the denominator is zero, but the corresponding factor also cancels in the numerator.
   • The denominator $x+2$ equals zero when $x = -2$.

3. Check for cancellation:

   • There is no $x+2$ term in the numerator to cancel with $x+2$ in the denominator. Thus, there is no hole.

Conclusion:

The function $f(x)$ has no holes. It has a vertical asymptote at $x = -2$, since the denominator becomes zero there, but the numerator does not.

Would you like more details on vertical asymptotes or further clarification?

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Related Questions:

1. What is the domain of $f(x)$?
2. How do vertical asymptotes differ from holes in a rational function?
3. How would the function change if the numerator were $(x+2)(x+3)$?
4. What is the horizontal asymptote of $f(x)$?
5. How do you graph $f(x) = \frac{x+3}{x+2}$?

Tip:

Always simplify the rational function first before analyzing for holes or asymptotes.