Determine the equation of the tangent to y=18/(x+2)^2 at the point (1,2).
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
y=-36/27x - 1.5
final answer: 8x+6y +9=0
Back of the book's answer: