To determine the points of discontinuity for the function \( f(x) = \frac{x - 3}{x^2 + 2x - 15} \), we need to analyze the denominator, as discontinuities in rational functions occur where the denominator equals zero. Thus, we set the denominator \( x^2 + 2x - 15 \) to zero.

To solve this quadratic equation, we can factor it. We are looking for two numbers that multiply to \(-15\) (the constant term) and add to \(2\) (the coefficient of the linear term). The numbers that satisfy these conditions are \(-3\) and \(5\). Therefore, we can factor the quadratic as:

\( (x - 3)(x + 5) = 0 \)

Setting each factor equal to zero gives us the potential points of discontinuity:

\( x - 3 = 0 \) leads to \( x = 3 \)

\( x + 5 = 0 \) leads to \( x = -5 \)

Thus, the function is discontinuous at \( x = 3 \) and \( x = -5 \).

Next, we can classify the types of discontinuities. The factor \( x - 3 \) in the numerator cancels with the factor \( x - 3 \) in the denominator, indicating that \( x = 3 \) is a removable discontinuity (or a hole) in the graph of the function. Conversely, the factor \( x + 5 \) in the denominator does not cancel with any factor in the numerator, indicating that \( x = -5 \) is a vertical asymptote.

In summary, the function \( f(x) \) has two points of discontinuity: a removable discontinuity at \( x = 3 \) and a vertical asymptote at \( x = -5 \).