1. The problem statement, all variables and given/known data I am to explain all intercepts, critical numbers, extrema, inflection points, and asymptotes of the function f(x)=(x-4)/x^3. 2. The attempt at a solution a) The y-intercept does not exist, as the domain of the function is all real numbers except x=0. Solving the numerator for 0 yields an x-intercept at x=4 b) f'(x) is easily calculated to be (12-2x)/x^4; solving the numerator for 0 yields a critical number of x=6, and x is undefined at 0, yielding our second critical number; Using values from the intervals shows f'(x)>0 on (-infinity, 0] and [0, 6); f'(x)<0 on (6, infinity), indicating what appears to be a local maximum at x=6. Plugging this into f(x): f(6)=(6-4)/6^3 f(6)=2/216 or 1/108 I note that this is a positive number, where f(2) is negative, meaning the line has crossed the x axis between 2 and 6 (specifically at x=4, as predicted) f(2)=(2-4)/2^3 f(2)=-2/8 or -1/4 Inflection points and second derivative information aside, I now come to the point where I am trying to predict the behavior of the function as x approaches positive and negative infinity. c) I know there is a vertical asymptote at x=0 due to the fact that the limit of the function does not exist as x approaches 0, but what about horizontal asymptotes? limit as x approaches infinity of (x-4)/x^3 This is indeterminate, infinity divided by infinity, so using L'Hospital's rule I take the derivative of the numerator and denominator for limit as x approaches infinity of 1/(3x^2) I factor out constants: (1/3) limit as x approaches infinity of 1/x^2 And I know that the limit of 1/x is 0. But here's my problem: I already know that my function has crossed the x axis at 4, so how can there be a horizontal asymptote at y=0?