In this problem, we're asked to determine the values of \( x \), if any, for which this function is discontinuous. The function given here is \( f(x) = \frac{x - 3}{x^2 + 2x - 15} \). Since this is a rational function, to determine where this function is discontinuous, we need to take the denominator and set it equal to 0. This will tell us anywhere that our rational function has holes or asymptotes, meaning that it would be discontinuous at those points.

So here, I take the denominator \( x^2 + 2x - 15 \), and I want to set that equal to 0. The easiest way to solve this quadratic equation and find those values of \( x \) which are 0 is to factor it. We want to look for two numbers that multiply to -15 and add to 2. Therefore, the factors are \( x - 3 \) and \( x + 5 \). These give us \( (x - 3)(x + 5) = 0 \). Since this is equal to 0, we can take each individual factor and set them equal to 0. Solving \( x - 3 = 0 \), by adding 3 to both sides, we find \( x = 3 \). This is one value of \( x \) for which this function is discontinuous. Solving \( x + 5 = 0 \), by subtracting 5 from both sides, we get \( x = -5 \). This is the other value for which the function is discontinuous.

Thus, this function is specifically discontinuous at \( x = 3 \) and \( x = -5 \). If you are curious about the type of discontinuities we have here, we can take a look back at our function. Since our numerator is \( x - 3 \), the same as one of the factors in the denominator, the \( x = 3 \) represents a removable discontinuity or a hole because it cancels out with the factor in the numerator. Meanwhile, the \( x + 5 \) factor that gives us our value of \( x = -5 \) represents an asymptote in our function. In summary, we have two types of discontinuities: one is a hole, and the other is an asymptote.

Let me know if you have any questions. I'll see you in the next one.