1. The problem statement, all variables and given/known data Disclaimer: English is not my first language, so i apologize for any wrong math-terms. We look at the function f(x) = x^3. On the graph for f we have a point, P(a,a^3), where a =/= 0. The tangent to f through P cuts through f in another point, Q. Find Q and show, that the tangent to f through Q has a slope which is 4 times as steep as the tangent to f through P. 2. Relevant equations 3. The attempt at a solution Tangent through P cuts the function f in Q, which means in Q tangentP=x^3. 3a^2(x-a)+a^3=x^3 <=> 3a^2=(x^3-a^3)/(x-a) <=> 3a^2=x^2-a^2 <=> 4a^2=x^2 2a=x Which means Q is located in (2a,(2a)^3)<=> Q(2a,8a^3) But this means that the tangent through Q has a slope which is twice as steep as it should be, since: tangentQ=8a^3+24a^2(x-a) I am not sure how i should solve this problem, and i hope you are able to understand my sloppy english/and math.