I'm trying to find points of intersection between a cubic function and a linear function...

Suppose I have cubic function y = (x-4)^3 and linear function y = x-4.
The graph shows that there are three points of intersection, one occurring at (4,0) and the other two off of the x and y axes.

How do I find the points of intersection for the ones not lying on the x or y axes?

y=(x-4)^3  y=x-4

set the equations equal to eachother and solve:
(x-4) = (x-4)^3

divide though by (x-4) -- note that this will elimate the solution you already found: x = 4.

1 = (x-4)^2
1 = x^2 - 8x + 16
0 = x^2 - 8x + 15
Use the quadratic equation to determine that
x = 5 or 3.

Plug x back in to find the y coordinates.

Hmm, yeah I see how that works. It's just my teacher confused me when at first he expanded (x-4)^3 and then subtracted x-4 and then used synthetic division on the equation to find a 0 root.

Thanks for the help,
LK.

[quote author=LivedKrad link=topic=10873.msg103125#msg103125 date=1110422613]
Hmm, yeah I see how that works. It's just my teacher confused me when at first he expanded (x-4)^3 and then subtracted x-4 and then used synthetic division on the equation to find a 0 root.

Thanks for the help,
LK.
[/quote]

The way he did it you won't lose any solutions, which is what happens when you divide by a variable in a situation like this.  Although I think it's pretty trivial to look at the equation and see that 4 is a solution.