What is the horizontal asymptote of f(x) = (3x^2 - 1)/(2x - 1)
2 Answers
embizze | High School Teacher | (Level 2) Educator Emeritus
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:
justaguide | College Teacher | (Level 2) Distinguished Educator
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.