Find the equation of the line tangent to the graph of y=4/x^2 at the point (5, 4/25).

the slope of the tangent is given by dy/dx

y = 4/x^2
=> dy/dx = 4*(-2)/x^3 = -8/x^3

at the point (5,4/25), the slope of tangent is -8/5^3 = -8/125

so the tangent
1. passes through (5,4/25)
2. has a slope -8/125

So, its equation is given as (y-4/25) = (-8/125)(x-5)
=> 125y - 20 = -8x + 40
=> 125y + 8x = 60