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OpenStudy (anonymous):

find the point where y=x-2x^2 and y=x^3 + 2x

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OpenStudy (anonymous):

point where the equations are tangent:) 2nd one is \[y=x ^{3}+2x\]

OpenStudy (a_clan):

Tangents are straight lines. But These are not the equation of lines

OpenStudy (anonymous):

i think we find the derivative?

OpenStudy (anonymous):

its where the graphs are tengent r thy touch

OpenStudy (anonymous):

a common point on both?

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OpenStudy (a_clan):

A common point on both the equations can be found by equating both the equations.

OpenStudy (anonymous):

how do u equate thm?

OpenStudy (anonymous):

common point: x-2x^2=x^3+2x x^3+2x^2-x=0 x(x^2+2x-1)=0

OpenStudy (a_clan):

x-2x^2 = x^3 + 2x you will get x. put it back in either eqn, you will get y

OpenStudy (anonymous):

when they hav common point: (x1, y1)=(x2 , y2) isn't it?

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OpenStudy (anonymous):

yes

OpenStudy (anonymous):

y2=y1 so x2=x1

OpenStudy (anonymous):

ohh

OpenStudy (anonymous):

def. of function

OpenStudy (anonymous):

i think i got it! thank u soo much!:)

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OpenStudy (anonymous):

:)

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