For any problem in which it asks you to find the tangent of a function at a point on that function we use...
Point Slope Form:
In which x1 and y1 is from your given point.
And m is found by taking the derivative of the original equation at the point given, f'(x1).
In this case a tangent to the function y=2/(x+1) at the point (1,1). Note (1,1)=(x1,y1)
So f'(x) = (d/dx)(2/(x+1)) = (-2/(x+1)^2)
f'(x1) = f'(1) = (-2/(1+1)^2) = -1/2 (m value!)
[y-y1=m(x-x1)] -> [y-1=(-1/2)(x-1)] -> [y=(-1/2)(x-3)] This is the tangent to the function at the given point.
To find this tangent's y-intercept, aka where this tangent intersects the y-axis, plug in 0 for x in the tangent function.
[y(0)=(-1/2)(0-3)] -> (0,1.5) or y=1.5
Therefore, your answer is y=1.5