I was on winter break and forgot how to do this..
An equation of the line normal to the graph of
y= $\displaystyle sqrt(3x^2 + 2x)$ at (2, 4) is?
find $\displaystyle \frac{dy}{dx}$ and evaluate it at x = 2 ... this value is the slope of the tangent to the curve at the point (2,4).
the normal line is perpendicular to the tangent line, so ... the normal slope is the opposite reciprocal of the tangent slope. determine the normal slope, then use the point-slope form with the point (2,4) to get the desired linear equation.