Final answer:
The student has correctly identified the vertical asymptote for y = 1/x and y = (x-2)/(x+1). For y = sqrt(x+3), they've rightly found no vertical asymptotes. However, they've made a mistake with y = (x^2 - 9)/(x-3); there is a hole at x = 3, not a vertical asymptote.
Step-by-step explanation:
When assessing functions for vertical asymptotes and holes, we analyze the behavior of the function as the input values (x) approach certain critical points. A vertical asymptote reflects a value of x at which the function approaches infinity, indicating the function is undefined at that point. A hole is a point on the graph where the function is not defined due to a cancelable discontinuity.
- a) For y = 1/x, there is a vertical asymptote at x = 0 because as x approaches 0, the function approaches infinity, and there is no x-value for which y is undefined after simplifying, so there is no hole.
- b) For y = (x-2)/(x+1), there is a vertical asymptote at x = -1, since the function goes to infinity as x approaches -1, and no holes as the numerator and denominator have no common factors.
- c) For y = sqrt(x+3), there are no vertical asymptotes or holes since the square root function is defined for all values in its domain, which is x >= -3 in this case.
- d) For y = (x^2 - 9)/(x-3), notice that the numerator can be factored into (x+3)(x-3), and the denominator is (x-3), so there is a cancelable discontinuity, which results in a hole at x = 3, not a vertical asymptote.