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
Hello,
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 $\displaystyle m= \dots \text{ or } m=\dots$