Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 47

(1) f(x)=x/(x+1) is a rational function. It has a vertical asymptote at x=-1, and a horizontal asymptote of y=1. On the interval [-1/2,2] the graph is increasing towards the limiting value of y=1.
f(-1/2)=-1 and f(2)=2/3
(2) The slope of the secant line through the endpoints of the interval is m=(2/3-(-1))/(2-(-1/2))=2/3 so the equation of the secant line is y=2/3x-2/3
(3) Since the function is continuous and differentiable on the interval the Mean Value theorem guarantees the existence of a c in the interval such that f'(c) is the slope of the secant line.
f'(x)=((x+1)(1)-(x)(1))/(x+1)^2=1/(x+1)^2
1/(x+1)^2=2/3 ==> x=sqrt(3/2)-1
The corresponding point on the graph is (sqrt(3/2)-1,1-sqrt(2/3))
So the tangent line parallel to the secant line is y-1+sqrt(2/3)=2/3(x-sqrt(3/2)+1)
The graph of the function, secant line, and tangent line: