# How do you find the vertical, horizontal and slant asymptotes of: #f(x)=(2x)/(x^2+16)#?

horizontal asymptote at y = 0

The denominator of f(x) cannot be zero as this would make f(x) 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.

This has no real solutions hence there are no vertical asymptotes.

Horizontal asymptotes occur as

Slant 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 slant asymptotes. graph{(2x)/(x^2+16) [-10, 10, -5, 5]}

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To find the vertical asymptotes, set the denominator equal to zero and solve for ( x ). In this case, ( x^2 + 16 = 0 ) has no real solutions, so there are no vertical asymptotes.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator (which is 1) is less than the degree of the denominator (which is 2), the horizontal asymptote is at ( y = 0 ).

To find the slant asymptote, perform polynomial long division. Divide the numerator by the denominator. The quotient will be the equation of the slant 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.

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