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Differential Calculus

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3 Jan 2012 18:54:09 IST

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trace the curve y^2=(x+2)^2(x-6)

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trace the curve y^2=(x+2)^2(x-6)

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Shubham Satyam

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Joined: 4 Jan 2012

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4 Jan 2012 09:41:40 IST

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The equation has roots -2 and 6 as for these value, the equation is having 0 value i.e. The graph cuts the x-axis at these points also its the form of some y^2= .... Means symmetrical about the x-axis (for ex: y^2= 4ax : equation of parabola which is symmetrical about x-axis.) Now you can get the curve easily!

Arjun Virmani

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5 Jan 2012 17:41:49 IST

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as we can see..

we have y^2 on LHS.. which can't be -ve.....

thrfr, RHS >= 0,

thrfr, the curve is defined for x>=6 only!!!!

ats its a simple polynomial thrfr it will be continuous.

and at x=6..... y=0........and x--> infinite y--> infinite (both +ve and -ve as it is square)............

and since thr are no roots in between the graph will not cut the x-axis again, i.e. other than at x=6!!!

It start from (6,0) to infinite (both +ve and -ve as it is square)

and from 2yy' = 2(x+2)(x-6) + (x+2)² = (x+2)(2x -12 + x +2 ) = (x+2)(3x -10)..........

y' = (x+2)(3x -10) / 2y............

we can see that derivative is +ve for +ve y and -ve for -ve y..........(as for x>=6 num. is +ve)

for concavity upwards or downwards diff agn and see!! +ve is cup-up. -ve is cup-down!!

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