consider the curve defined by 2y^3+6x^2y-12x^2+6y=1.

[ashley123456789]

consider the curve defined by 2y^3+6x^2y-12x^2+6y=1.

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

b.write an equation of each 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.

 

[o_O]

a.) What are you having problems with?

b.) Any tangent line to the curve can be found using the slope-intercept form y - y[sub:1fu9lzh6]0[/sub:1fu9lzh6] = m(x - x[sub:1fu9lzh6]0[/sub:1fu9lzh6]) where (x[sub:1fu9lzh6]0[/sub:1fu9lzh6], y[sub:1fu9lzh6]0[/sub:1fu9lzh6]) is a coordinate point that satisfies your equation (in this case, the 2y[sup:1fu9lzh6]3[/sup:1fu9lzh6] + 6x[sup:1fu9lzh6]2[/sup:1fu9lzh6]y - 12x[sup:1fu9lzh6]2[/sup:1fu9lzh6] + 6y = 1) and m is dy/dx at that particular point.

c.) If the tangent line goes through the origin and has a slope of -1, then y = -x. You can substitute this into your original equation to see where they intersect, which according to your question, should be at exactly one point: P

 

[ashley123456789]

a. i have no idea what it is even asking or how i am supposed to answer it.

b. what would the horizontal tangent be

 

[o_O]

a.) You have this: 2y[sup:2dhwh548]3[/sup:2dhwh548] + 6x[sup:2dhwh548]2[/sup:2dhwh548]y - 12x[sup:2dhwh548]2[/sup:2dhwh548] + 6y = 1. You're suppose to take the derivative of it (which in Leibniz's notation is dy/dx).

b.) If you have a horizontal tangent, what does its derivative equal to?