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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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