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Finding vertical asymptote and hole questions.

a) y = 1/x, Vertical Asymptote: x = 0, Hole: None
b) y = (x-2)/(x+1), Vertical Asymptote: x = -1, Hole: x = 2
c) y = sqrt(x+3), Vertical Asymptote: None, Hole: None
d) y = (x^2 - 9)/(x-3), Vertical Asymptote: x = 3, Hole: x = 3

User Maugch
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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.

User Jimmy Pitts
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