The long division of 4x by (x²-4x-5) gives 0, which means that the asymptote is the x-axis
I am currently stuck on a problem to find the oblique asymptote with long division. The equation is f(x) = 4x/ (x^2 - 4x-5) . Now I using long division you cannot get an answer because the degree of the divisor is greater than the degree of the dividend. Would that mean that there is just a remainder of 4x OR 4x/(x^2-4x-5). That would basically mean that the oblique asymptote of this equation is 4x, right?
Or am I going in the wrong direction? Thanks
A rational function's graph can have an horizontal asymptote (such as the x-axis) or an oblique ("slant") asymptote, but it will not have both! (And, if the degree of the numerator is larger than the degree of the denominator by 2 or more, it won't have a slant asymptote, either.)
To learn how this works, try here.