A normal line is a line perpendicular to a tangent line at the point where the tangent meets the function. Give the equation of the normal line to the graph of y=3xsqrt{x^2+6}+4 at the point (0,4)

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so step one (as usual in calc) find dy/dx

so dy/dx = (3)(1/2(x^2+6)^-1/2(2x)

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