How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (x^2+1)/(x+1)#?
Vertical:
Slant :
Let y = f(x) = x-1+2/(x+1). Then, reorganizing
(x+1)(y-x+1)=2 and this represents a hyperbola having the pair of
asymptotes given by
Graph of the hyperbola is inserted.
graph{x^2+1-(x+2)(y-x+1)=0 [-80, 80, -80, 80]}
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To find the vertical asymptote, set the denominator equal to zero and solve for x. In this case, x + 1 = 0, so x = -1. Therefore, there is a vertical asymptote at x = -1.
To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, divide the leading coefficients of the numerator and denominator. In this case, since the degrees are both 1, the horizontal asymptote is y = 1.
To find the oblique asymptote, divide the numerator by the denominator using polynomial long division or synthetic division. The quotient will be the equation of the oblique asymptote.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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