** Let f be the function defined by f(x) = x^3 - x^2 -4x + 4. The point (a,b) is on the graph of f, and the line tangent to the graph at (a,b) passes through the point (0, -8), which is not on the graph of f. Find a and b.**

Ok, the line tangent to the graph of f is the derivaitve of f.

f(x) = x^3 - x^2 -4x + 4

f'(x) = 3x^2 -2x - 4

f'(x) is merely the slope of the line. The equation of the line tangent to f is:

y = (3x^2 - 2x - 4)x + b

NOTE! b is the y intercept, not the variable b in the problem. We were told that the line tangent to f at the point (a,b) passes through (0,-8). We have not found precisly what the slope is, but no matter what it is, it will have a value of zero at this point. Therefore:

-8 = (3(0^3) - 2(0) - 4)0 + b

thus b equals -8

So we have:

y = (3x^2 - 2x -4)x - 8

what this eqation represents is somewhat abstract. The graph of this equation and the graph of f will intersect at (a,b).

Therefore:

f(x) = (3x^2 - 2x -4)x - 8

x^3 - x^2 -4x + 4 = 3x^3 - 2x^2 -4x - 8

-2x^3 + x^2 + 12 = 0

2x^3 - x^2 - 12 = 0

as far as I can tell this can be simplified no further.

Sure you can solve this using the rational roots theorem and or synthetic division. Or with a graphing calcuator. And the answer it yields is correct. x = 2, so a =2 and b = 0. But this was a somewhat inelligent method of solving the problem I think. I think there must be a simple way to solve it, one that does not end up in a third degree polynomial that cannot be reduced. I've never seen a problem in my mathbook that required the solving of a third degree polynomial and didn't say "use a graphing calculator as an aid for solving the problem".

So, any idea's?