Hi Stephanie,

Suppose that P is a point on the curve and the tangent to the curve at P passes through (2, 8). Let P have coordinates (a, b), then since P is on the curve y = f(x) = x³, b = f(a) = a³. Hence P has coordinates (a, a³). The slope of the tangent line to the curve at P is f'(a) = 3a² and hence the tangent line to y = x³ at P is

y - a³ = 3a²( x - a) or
y = 3a² x - 2 a²

Since (2, 8) is on this line substitute x = 2, y = 8 and simplify to obtain a cubic equation in a.

The challenge now is to solve this equation for a. One fact to notice is that the point (2, 8), which is on the tangent line, is also on the curve since 8 = 2³. Thus a = 2 must be a solution of the cubic. Check to make sure this is true.

Can you find another solution to the cubic?

Penny