How do I determine C, so that the straight line joining (0,3) and (5,2) is tangent to the curve y = c/(x+1) ?

Given curve  y=c/(x+1).. (1)

Its slope at any poit is dy/dx = -c/(x+1)^2   (2)

The slope  and equation of the line joining (0,3) and (5,2)  are

(2-3)/(5)= -(1/5) ....(3)  and

y = -(1/5)x+3........(4).

At the tangent point the slope of line and the curve are same, therefore, from(3) and (2): -1/5=-c/(x+1)^2 ==>

5c=(x+1)^2.........(6)

Solving for the line and the curve, from rom ((4) and (1),

-(1/5)x +3 = c/(x+1)==>

5c=(-x+15)(x+1) = -x^2+14x+15.........(7)

Eliminatin c  between  (6) and (7) we get x cordinate of the tangent touching the curve :

(x+1)^2=-x^2+14x+15==>

x^2-12x-14=0==>

(x+1)(x-7)=0.

x=-1 and x=7. But x=-1 is a vertical asymptote.

Putting x=7 in  Eq (6) gives 5c =(7+1)^2

5c=64

c  =  64/5 =12.8.

The tangent touches curve  at

the point. (7, 12/(7+1))=(7,  1.5)