Hi everyone.

I'm new to this forum so if I'm doing anything wrong, im sorry

I'm just doing some study revision, and a question popped up like iv'e never seen before. It's this:

"Find the equation of the tangent line at the given value of x on the curve y^3+xy^2-25=x+2y^2, x=2"

I'm only used to finding tangent lines given both an x AND a y.

Can someone help me solve this?

What I did was:

y^3+2y^2-25 = 2 +2y^2

y^3 + 2y^2 - 27 - 2y^2 = 0

y^3 - 27 = 0

y^3 = 27

y = 3

Is this the right way to start?

Then I differentiated y^3+xy^2-25=x+2y^2 to make:

3y^2 + 2xy = 1 + 4y

I'm lost at what to do here, if this is even right. =/

Do I do:

3y^2 + 2(2)y = 1 + 4y

= 3y^2 + 4y = 1 + 4y

= 3y^2 = 1

= y^2 = 1/3

= y = sqrt1/3

And then

y-y1 = m(x-x1)

y - 3 = sqrt1/3 (x-2)

y - 3 = sqrt1/3x - 1.154700538

y = sqrt1/3x + 1.845299462 ?

And how can I get 1.845299462 as a fraction?

Thank you!