Select a convenient point (obviously not a max/min or discontinuity) on the function (x^2+3x)/(x^3-x) and write the equation for the tangent line in slope intercept form at that point on the curve. Then write an equation of the normal line to the curve at the same chosen point in point-slope form. I've figured out the derivative of the original function: -x^2-1-6x/(x^2-1)^2 ... would the tangent line have a slope that is the opposite reciprocal of the this derivative? And and all help is appreciated.