Implicit Derivative to find equations of each Horizontal Tangent line.?

Question

Need to use Implicit Derivative to find equations of each Horizontal Tangent line. Equation is 2y^3 + 6x^2y -12x^2 +6y =1.

Answer 1

6y^2*y' + 6(2xy+x^2*y')-24x+6y'=0

Solve for y'

6y^2*y' + 6x^2*y'+6y'=24x-12xy

Factor and solve

y'(6y^2+6x^2+6)=24x-12xy
y'=(24x-12xy)/(6y^2+6x^2+6)
y'=(2x)*(2-y)/(x^2+y^2+1)

Set y'=0 and solve. This implies setting numerator only to zero (since the denominator can never be zero by definition)
So,

2x*(2-y)=0

or

x=0 and y=2
These are individual coordinate values where horizontal tangents occur. Plug them back into the original equation to solve for the other half of each coordinate pair.

When x=0:

2y^3+6y-1=0 (three solutions)

The Solutions are:
y=-0.08258240024527558+i1.7379469377661714,
-0.08258240024527558-i1.7379469377661714, and
0.16516480049055116

When y=2:

16+12x^2-12x^2+12=1
28=1
no solution

So there is only one point on this line at which there is a horizontal tangent, that point being (0,0.1651). Now, since it has a slope of zero, the equation of the line at that point should just be equal to the y-value. Thus, the equation is y=0.16516480049055116

Answer 2

Well, certainly when the tangent line is vertical, dy/dx will not exist. I assume you found: dy/dx = -(2x + y) / (2y + x) right? Well, in order for this not to exist, we must have: x = -2y Substitute this into the original equation: (-2y)^2 + (-2y)y + y^2 = 27 4y^2 - 2y^2 + y^2 = 27 3y^2 = 27 y^2 = 9 y = 3 or -3 When y = 3, we have: x^2 + 3x + 9 = 27 x^2 + 3x - 18 = 0 (x + 6)(x - 3) = 0 x = 3 or -6 When y = -3, similarly, we have x = -3 or x = 6. Thus, our four candidate points are: (3, 3), (3, -6), (-3, -3), (-3, 6) To verify that these points are indeed vertical asymptotes (and not, say, cusps), we can find dx/dy using implicit differentiation. If these points have a vertical asymptote, then we would expect them to have dx/dy = 0. Implicitly differentiating yields similar results: dx/dy = -(2y + x) / (2x + y) Notice that dx/dy = 0 when on the line 2y + x = 0, which all of our points lie on, so this confirms there are vertical asymptotes at each of them.

Answer 3

y' = 24x/(6y^2 + 12x + 6)
Horizontal tangent equals when y' = 0
therefore
solve 0 = 24x/(6y^2 + 12x + 6) to find out equation of tangent horizontal line oh and i think you have to solove for y as well and sub it back in so you can solve for y' = 0

Answer 4

So deriving by x: 6yyy’ +12xy +6xxy’ –24x +6y’=0;
thus y’ = 2x (y-2) / (yy+xx+1) = 0;
if y=2, then 16 +12xx –12xx +12-1 >0; discard y=2!
if x=0, then 2yyy +6y -1=0 or yyy+3y-0.5=0;
y1= {(1 +sqrt(17))^(1/3) +(1 –sqrt(17))^(1/3)}*(4^(-1/3)) =0.165165;
are there more real roots? Deriving by y: (yyy+3y-0.5)’ = 3yy+3 >0; no more real roots!
Horizontal Tangent: y=0.165165;

Answer 5

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

Implicitly differentiating with respect to x, we have

6y^2(dy/dx) + 6[2xy + (x^2)(dy/dx)] - 24x + 6(dy/dx) = 1

Remember that horizontal tangent lines happen when dy/dx = 0, so we set it to 0 and obtain

6y^2(0) + 6[2xy + (x^2)(0)] - 24x + 6(0) = 1

6(2xy) - 24x = 1
12xy - 24x = 1
12xy = 24x + 1

There are infinitely many tangent lines.
I think I'm wrong and may have done this question incorrectly. Did you write it out correctly?