Let f(x)= 6x+(4/x^2). then the equation of the tangent line to the graph of f(x) at the point (2,13) is given by y=mx+b.

What is m and what is b?

or f(x) = 6x + 4x^-2

f ' (x) = 6 - 8x^-3 or 6 - 8/x^3

at (2,13) , slope = 6 - 8/2^3 = 6-1 = 5

so y = 5x + b
sub in your given point
13 = 5(2) + b
b = 3

in y = 5x + 3 , m=5 and b=3