# How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x) = (2x - 3)/( x^2 - 1)#?

vertical asymptotes x = ± 1

horizontal asymptote y = 0

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

Horizontal asymptotes occur as

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 1 , denominator-degree 2) Hence there are no oblique asymptotes. graph{(2x-3)/(x^2-1) [-10, 10, -5, 5]}

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The vertical asymptotes occur where the denominator of the function equals zero. Thus, the vertical asymptotes for the given function are x = -1 and x = 1.

For the horizontal asymptote, if the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is y = 0. If the degrees are equal, then the horizontal asymptote is the ratio of the leading coefficients. In this case, since the degree of the numerator is 1 and the degree of the denominator is 2, the horizontal asymptote is y = 0.

To find the oblique asymptote, divide the numerator by the denominator using polynomial long division or synthetic division. The quotient will represent the oblique asymptote. Since the degree of the numerator is less than the degree of the denominator, there is no oblique asymptote for this function.

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