Posted by **Taeyeon** on Sunday, February 26, 2012 at 2:43am.

Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .

a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1)

b. Write an equation of each horizontal tangent line to the curve.

c. The line through the origin with slope -1 is tangent to the curve at point P. Find the x – and y – coordinates of point P.

- calculus -
**Reiny**, Sunday, February 26, 2012 at 8:13am
If I read your equation as

2y^3 + (6x^2)(y) - 12x^2 + 6y = 1

then your dy/dx is not correct

I get:

8y^2 dy/dx + 6x^2 dy/dx + 12xy - 24x + 6dy/dx = 0

dy/dx(8y^2 + 6x^2 + 6) = 24x - 12xy

dy/dx = 12x(2 - y)/(2(4y^2 + 3x^2 + 3) )

= 6x(2-y)/(4y^2 + 3x^2 + 3)

Please check before I proceed

- calculus -
**Taeyeon**, Sunday, February 26, 2012 at 4:08pm
This is what I found in the paper though.

- calculus -
**Anonymous**, Friday, March 7, 2014 at 2:53am
$4000 is invested at 9% compounded quarterly. In how many years will the account have grown to $14,500? Round your answer to the nearest tenth of a year

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