# Point of intersection

1. Feb 14, 2006

### teng125

may i know how to find the point of intersection between y=x+6 and y=x^3 ??

2. Feb 14, 2006

### TD

See for what values of x, the y values are the same, so solve:

$$x^3 = x + 6 \Leftrightarrow x^3 - x - 6 = 0$$

Divisors of the constant term (6) are possible zeroes.

3. Feb 15, 2006

### teng125

ya,but then from x^3 - x - =0 how to solve pls

4. Feb 15, 2006

### teng125

ya ,from x^3 -x - 6 = 0

5. Feb 15, 2006

### HallsofIvy

Staff Emeritus
You TRY SOMETHING!!!

Have you even tried plugging in numbers to get an idea of how large the solutions should be?

6. Feb 15, 2006

### TD

And as I said, you should try the divisors of the constant term, if you're looking for integer zeroes.

7. Feb 15, 2006

### 0rthodontist

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Scroll down to "solving the depressed cubic" because that is the form you have.

Though indeed, it is easier to look for an integer solution as TD said, and you can see graphically from the original problem that there are no other solutions.

Last edited: Feb 15, 2006
8. Feb 15, 2006

### Integral

Staff Emeritus
You can always plot the 2 functions to get an idea where the intersection is.

9. Feb 15, 2006

### teng125

the answer is it = to 6 and zero??

10. Feb 15, 2006

### z-component

Try graphing the two equations as suggested by others. I graphed it on an online graphing calculator and can see that the two curves intersect at only one point. Remember that intersections are points on the x-y plane, so they should have the form (x,y), not just single numbers. Example: The curves intersect at (1,3) and (5,2).

11. Feb 15, 2006

### teng125

i have already drawn it but i cant find the point using the eqn above so i ask for help in this forum

12. Feb 15, 2006

try to put divisors of 6 , 1,2,3,6 and -1,-2,-3,-6 in the equation(x^3 -x - 6 = 0)
1^3 - 1 - 6 = -6 wrong , I continue 2^3 - 2 - 6 = 8 - 8 = 0 Bingo
the 2 graphs intersect at (2,8).usualy when u have such equation you should what TD told u to do.

13. Feb 15, 2006

### teng125

okok got it thanx