Find the equation of the normal at a point

Find the equation of the normal at the point $(2, 1)$ for the function $x^2 + y^3 - 2y = 3$

I'm still struggling a bit with the application of derivatives, but from what I understand I use the derivative to calculate the slope

\begin{align} \frac{dy}{dx}x^2 + y^3 - 2y &= \frac{dy}{dx} 3\\ (3y^2 - 2)\frac{dy}{dx} &= 0 - 2x\\ \frac{dy}{dx} &= \frac{-2x}{(3y^2 - 2)} \end{align}

I then plug the values of the point to get the slope $m = \frac{-2(2)}{(3(1)^2 - 2)} = \frac{-4}{1} = -4$

I then plug all of this into the equation $y - 1 = -4 (x + 2)$

Is this the correct approach?

Everything looks good, except when you use $m = -4$ in the equation of the desired line: What you post in the end is the equation of the line tangent to the curve at $(2, 1)$.
The slope of the normal is the negative reciprocal of $-4$: $\dfrac 14$.
That gives us $$y - 1 = \frac 14(x - 2)$$
• The slope should be $\frac{1}{4}$, and it should be $y-1=\frac{1}{4}(x-2)$... – stochasm Jun 10 '14 at 20:30