Finding the Point of Intersection: Y=x+6 and Y=x^3

[teng125]

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

 

[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.

 

[teng125]

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

 

[teng125]

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

 

[HallsofIvy]

You TRY SOMETHING!

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

 

[TD]

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

 

[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.

 

[Integral]

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

 

[teng125]

the answer is it = to 6 and zero??

 

[z-component]

the answer is it = to 6 and zero??

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).

 

[teng125]

i have already drawn it but i can't find the point using the eqn above so i ask for help in this forum

 

[ziad1985]

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.

 

[teng125]

okok got it thanx