chemistry1111 so you'd first antidiff ekx which gets you ekx/k + c, which is your curve. You know one of the coordinates on the curve is (1,e2), so lets sub it into and make an equation. c + ek/k = e2. Now let's find the tangent of the curve at the given point. Go to CAS (menu -> 4-> 9) and u should type "tangentLine(f(x), x, 1)" which gets you the tangent line. Since it passes through origin you make it equal to zero and also sub in zero. You are now left with an equation containing only c and k. Take the equation from before and you have 2 equations. Solve using simultaneous equations and you should get k=2 and c=e2/2.
Hope that helps