Question

Find the slope of the tangent line at (8,8) for the function (16-x)y^2=x^3

Round your answer to two decimal places.

Answer 1

using implicit differentiation,

(-1)(y^2) + (16-x)(2yy') = 3x^2
y' = (3x^2+y^2)/(2y(16-x))

So, just evaluate y' at (8,8)

Answer 2

Find equation of all lines having slop -4 that are tangent to the curve y=16/(x-2)