# How do you find the vertical, horizontal and slant asymptotes of: #f(x) = (6x + 6) / (3x^2 + 1)#?

The horizontal asymptote is

No vertical asymptote

No slant asymptote

So, there is no vertical asymptote.

graph{(6x+6)/(3x^2+1) [-18.02, 18.03, -9.01, 9.01]}

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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 roots 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 hence there are no slant asymptotes. graph{(6x+6)/(3x^2+1) [-10, 10, -5, 5]}

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To find the vertical asymptotes, set the denominator equal to zero and solve for ( x ). To find horizontal asymptotes, 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 to find the horizontal asymptote. For slant asymptotes, perform polynomial long division. If the degree of the numerator is exactly one greater than the degree of the denominator after division, there is a 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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