Calculus
posted by Anonymous .
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 = 61 = 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