find two straight lines tangent to y=x^3 and they pass through point (2,8)
The equation of a tangent line at point a is f(x) = y + y'(x-a) = a^3 + 3a^2(x-a). Substituting point (2,8) we obtain: 8 = a^3 + 3a^2(2-a). Modifying this equation, -2a^3 + 6a^2 = 8; a^3 - 3a^2 + 4 = 0. We need to factorize left-handed side of the equation. It can be easily seen that a=-1 is a solution of this equation. Dividing by a+1 we obtain (a+1)(a^2-4a+4) = 0, or (a+1)(a-2)^2 = 0. So, there are two solutions: a=-1 and a=2. At last, we have f(x) = (-1)^3 + 3(-1)^2(x-(-1)) = -1+3(x+1) = 3x+2 - first tangent line, and f(x) = 2^3 + 3*2^2*(x-2) = 8 + 12(x-2) = 12x-16 - second tangent line.