Answer to Question #4828 in Abstract Algebra for Nonsikelelo

Question

find two straight lines tangent to y=x^3 and they pass through point (2,8)

Answer 1

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.