Let's analyze the problem.

The given function is:

$f(x) = \frac{3x^2 - 5x - 2}{x^2 + 2x - 8}$

Step 1: Factor both the numerator and the denominator

We begin by factoring the denominator and numerator to check for discontinuities.

Denominator:

$x^2 + 2x - 8 = (x - 2)(x + 4)$ This gives two potential discontinuities at $x = 2$ and $x = -4$.

Numerator:

$3x^2 - 5x - 2 = (3x + 1)(x - 2)$ The factor $(x - 2)$ appears in both the numerator and denominator.

Step 2: Simplifying the function

The common factor of $(x - 2)$ in both the numerator and denominator cancels out, but it still indicates a removable discontinuity at $x = 2$.

After canceling $(x - 2)$, the simplified function becomes:

$f(x) = \frac{3x + 1}{x + 4}$

Step 3: Non-removable discontinuities

The remaining denominator $x + 4$ shows that $x = -4$ is a non-removable discontinuity because it causes the denominator to be zero, which makes the function undefined at that point.

Conclusion:

• At $x = 2$, there is a removable discontinuity.
• At $x = -4$, there is a non-removable discontinuity.

Thus, the correct answer is (B) $-4$ only, which matches the boxed answer in the image.

Would you like more details or have any questions about this solution?

Here are 5 related questions to explore:

1. What is the difference between removable and non-removable discontinuities?
2. How does canceling a factor affect the type of discontinuity?
3. Can a function have more than one removable discontinuity?
4. What are some examples of functions with both types of discontinuities?
5. How do you determine if a discontinuity is removable or non-removable from a graph?

Tip: Always factor both the numerator and denominator of a rational function to identify potential discontinuities.