find the points of intersection. y = x^3, y = 2x

3 points. 1) (0,0) 2) (-1.4,-2.8) 3) (1.4,2.8)

abe you're really not helping me by giving me the answers. i have the answers. thats not the problem. i dont know how to get them.

Do you have a graphing calculator?

yes but i still have to show my work. i cant just put it in and get the answers that way

Those aren't right completely. This is how you do it. x^3 = 2x x^3 - 2x = 0 factor out an x x (x^2 - 2) = 0 realize that x = 0 is now an answer divide by x x^2 -2 = 0 quadratic equation gives you \[\sqrt{2}\] as an X answer. plug into either original equation to get 2\[\sqrt{?}\] as the Y. both negative is also an answer.

the script got messed up. Near the end it is supposed to say plug into either original equation to get \[2\times \sqrt{2}\]

how do you know there aren't more answers? from checking?

2xroot2 is 2.82...

an equation with x^3 can only have a maximum of 3 intersection points with a straight line.

okay, but how do you know its not \[(-\sqrt{2}, 2\sqrt{2})\]

like how do you know they're both positive and both negative

look at the question. it is asking when those two equations hit each other. so, any answer you come up with has to be a point on BOTH lines. plug those points into EITHER equation and you will realize it doesn't work for either x^3 or 2x.

okay, i understand

a negative X times 2 or to the third, either way will stay negative and your Y is positive. how could that point possibly work.

good stuff :)

it was just a random thought as to know how to not get too many answers.

You can never get too many answers because if they shouldn't be there, they won't work if you plug them in! :)

ok :)

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