Determine the equation of the tangent to y=18/(x+2)^2 at the point (1,2).

My answer:

I found the derivative of the function and then I substituted 1 into the Mt.

Then I substituted (1,2) into y=Mtx + b to find the y intercept.

I didn't get the right answer with the textbook. Is mine right or theirs?

f'(x)=-36 / (x+2) ^3

Mt= -36/27

y=-36/27x - 1.5

final answer: 8x+6y +9=0

Back of the book's answer:

4x+3y-10=0