1. ## simple intersect question

say you have

y= x^2

and y = x+2

yes i know they intersect at 1,1.. its kinda easy to tell from those two functions. but i forgot how to mathmatically do this.. what formula's would u use and how would u do it.?

2. Originally Posted by Legendsn3verdie
say you have

y= x^2

and y = x+2

yes i know they intersect at 1,1.. its kinda easy to tell from those two functions. but i forgot how to mathmatically do this.. what formula's would u use and how would u do it.?
To find the intersection point, you set the two equations equal to each other.

$\displaystyle x^2=x+2$

Then solve for x:

$\displaystyle x^2=x+2\implies x^2-x-2=0\implies (x-2)(x+1)=0\implies {\color{red}\boxed{x=2}}\text{ and }\color{red}\boxed{x=-1}$

Given these zero values, we can determine the points of intersection.

Verify that the points are $\displaystyle (-1,1)\text{ and }(2,4)$

I hope this makes sense!

--Chris

3. ok then how would u do:

y = x^(1/3)

y= x^(1/2)

i think i m getting confused.

4. Hello,
Originally Posted by Legendsn3verdie
ok then how would u do:

y = x^(1/3)

y= x^(1/2)

i think i m getting confused.
Find a point (m,n) such that $\displaystyle n=m^{1/3}$ and $\displaystyle n=m^{1/2}$

---> $\displaystyle m^{1/3}=m^{1/2}$

Take the 6th power of both sides (because 2*3=6) :

$\displaystyle (m^{1/3})^6=(m^{1/2})^6$

Using the rule $\displaystyle (a^b)^c=(a^c)^b=a^{bc}$, we have :

$\displaystyle m^2=m^3$

$\displaystyle m^3-m^2=0$

$\displaystyle m^2(m-1)=0$

Thus $\displaystyle m= \dots \text{ or } m=\dots$