Find the equation to the tangent of a line using known points?

Question

I have carried out the implicit differentiation of the original formula ($x-y^3=2xy$) to get the equation

$$\frac{dy}{dx} = - \frac{2y-1}{3y^2-2x}.$$

Now I need to find the equation of the tangent at point $(-1, 1)$, I've plugged the values into the formula to get

$$\frac{dy}{dx} = - \frac15$$

but have a suspicion I may be missing something.

Many thanks!

Answer 1

You have the slope of the line and a point on the line.

Use the slope point form of line equation $$y=mx+c,$$

where $m$ is the slope, then plug in the point to get $c$.