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