Answer: Solving Tangent Lines of y=x/(x+1) Through (-1,3)

[rdgt3000]

y= x / (x + 1) I need to find the equation of every line tangent to the curve which passs through the point (-1, 3). The problem arises when you find the first derivative then plug in the x value which results in an answer of 0 in the denominator. Any help?

 

[Office_Shredder]

You're looking for all sets of points (x, y), s.t. y=mx+b, 3 = -1*m + b,
y= x / (x + 1) and m = 1/(x+1)^2

 

[HallsofIvy]

You don't plug x= -1 into the equation! Obviously (-1,3) is not on the given curve. You are looking for tangent lines to that curve that pass through the point (-1, 3) off the curve.

You are looking for the tangent line at (x₀,y₀) that passes through (-1, 3). If y= x / (x + 1) , then y'= m= 1/(x+1)². The equation of the line passing through (-1, 3) with slope m is y= m(x+1)+ 3. You are looking for x₀ such that
y= \frac{1}{(x_0+1)^2}(x+1)+ 3
is passes through (x_0,\frac{x_0}{x_0+1}[/tex]<br /> That is,<br /> \frac{x_0}{x_0+1}= \frac{1}{(x_0+1)^2}(x_0+1)+ 3<br /> Solve that for x₀.