# Math Help - Tangent

1. ## Tangent

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
Mt= -36/27

y=-36/27x - 1.5

4x+3y-10=0

2. I got the book's answer. I was doing the same thing as you up until you found "b". Something went different after that. Here's what I did:

y = 18/(x+2)^2

y=mx+b

m = dy/dx = (x+2^2)(0) - (18(2(x+2)(1))/(x+2)^4

= 0-36(x+2)/(x+2)^4

= -36/(x+2)^3

= -36/27

m = -4/3

Substitute m into y=mx+b

y = (-4/3)x+b

Substitute (1,2) to find b

2 = (-4/3)(1) + b

b= 10/3

Into the equation:

y=mx+b

y= -4/3x+10/3

Multiply by 3 to get standard form:

3y=-4x+ 10

Set equal to zero:

$
4x+3y-10 = 0
$