# How to find the intercepts of y=x^3-3x

• Jul 9th 2012, 09:04 AM
kmerr98277
How to find the intercepts of y=x^3-3x
How to find the intercepts of y=x^3-3x. I keep coming up with x intercept= (sq.rt. 1, 0) and y intercept (0,0) but was told a part of that answer or the whole thing is wrong or it is incomplete.
• Jul 9th 2012, 09:20 AM
Plato
Re: How to find the intercepts of y=x^3-3x
Quote:

Originally Posted by kmerr98277
How to find the intercepts of y=x^3-3x. I keep coming up with x intercept= (sq.rt. 1, 0) and y intercept (0,0) but was told a part of that answer or the whole thing is wrong or it is incomplete.

There are three x-intercepts. Solve $\displaystyle x(x^2-3)=0~.$
• Jul 9th 2012, 09:29 AM
kmerr98277
Re: How to find the intercepts of y=x^3-3x
(x^2-3) factors to (x-3) and (x-1). so...x intercepts are (3,0) and (-1,0)?
• Jul 9th 2012, 09:34 AM
Plato
Re: How to find the intercepts of y=x^3-3x
Quote:

Originally Posted by kmerr98277
(x^2-3) factors to (x-3) and (x-1).

No it does not factor that way.

$\displaystyle (x^2-3)=(x-\sqrt3)(x+\sqrt3)$
• Jul 9th 2012, 09:44 AM
kmerr98277
Re: How to find the intercepts of y=x^3-3x
so... x intercepts are (sq.rt.3,0) and (-sq.rt.3, 0) and i have no idea about the third intercept. How do you enter symbols into the forum? Thanks a lot!!
• Jul 9th 2012, 10:01 AM
Plato
Re: How to find the intercepts of y=x^3-3x
Quote:

Originally Posted by kmerr98277
i have no idea about the third intercept. How do you enter symbols into the forum?

The third is $\displaystyle (0,0)$

[TEX](x-\sqrt{3})(x+\sqrt{3})[/TEX] gives $\displaystyle (x-\sqrt{3})(x+\sqrt{3})$