# How do I identify the horizontal asymptote of #f(x) = (7x+1)/(2x-9)#?

We have a horizontal asymptote

graph{(y-(7x+1)/(2x-9))(y-3.5)=0 [-40.42, 39.58, -17.76, 22.24]}

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To identify the horizontal asymptote of (f(x) = \frac{7x+1}{2x-9}), compare the degrees of the polynomials in the numerator and the denominator:

- The degree of the numerator (the highest power of (x) in the numerator) is 1.
- The degree of the denominator (the highest power of (x) in the denominator) is 1.

When the degrees of the numerator and denominator are equal, the horizontal asymptote is found by dividing the leading coefficients of the numerator and the denominator.

The leading coefficient of the numerator is 7, and the leading coefficient of the denominator is 2. Therefore, the horizontal asymptote is (y = \frac{7}{2}).

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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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