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Math Help - simple intersect question

  1. #1
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    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.?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    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.

    x^2=x+2

    Then solve for x:

    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 (-1,1)\text{ and }(2,4)

    I hope this makes sense!

    --Chris
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  3. #3
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    ok then how would u do:

    y = x^(1/3)

    y= x^(1/2)

    i think i m getting confused.
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  4. #4
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,
    Quote Originally Posted by Legendsn3verdie View Post
    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 n=m^{1/3} and n=m^{1/2}

    ---> m^{1/3}=m^{1/2}

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

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

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

    m^2=m^3

    m^3-m^2=0

    m^2(m-1)=0

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