Find the horizontal asymptote for `f(x)=(3x^2-1)/(2x-1)` :

A rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator. It will have a horizontal asymptote of `y=a_n/b_n` if the degree of the numerator is the same as the...

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Find the horizontal asymptote for `f(x)=(3x^2-1)/(2x-1)` :

A rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator. It will have a horizontal asymptote of `y=a_n/b_n` if the degree of the numerator is the same as the degree of the denominator (where `a_n,b_n` are the leading coefficients of the numerator and denominator respectively when both are in standard form.)

If a rational function has a numerator of greater degree than the denominator, there will be no horizontal asymptote. However, if the degrees are 1 apart, there will be an oblique (slant) asymptote.

**For the given function, there is no horizontal asymptote.**

We can find the slant asymptote by using long division:

`(3x^2-1)/(2x-1)=(2x-1)(3/2x+3/4-(1/4)/(2x-1))`

The slant asymptote is `y=3/2x+3/4`

The graph of the function and the asymptote in red:

The function f(x) = (3x^2 - 1)/(2x - 1). The highest degree of the numerator is 2 and that of the denominator is 1.

This function does not have a horizontal asymptote. As values of x increase and decrease, the value of y also increases or decreases respectively without a limit.