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

• rdgt3000
In summary, the conversation is about finding the equation of every line tangent to a given curve that passes through a specific point. The problem arises when finding the first derivative and plugging in the x value, resulting in a denominator of 0. The solution involves finding the tangent line at a specific point and using the slope and point to find the equation of the line.
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?

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

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 [itex](x_0,\frac{x_0}{x_0+1}[/tex]
That is,
$$\frac{x_0}{x_0+1}= \frac{1}{(x_0+1)^2}(x_0+1)+ 3$$
Solve that for x0.

## What is the equation for finding the tangent line of y=x/(x+1) at the point (-1,3)?

The equation for finding the tangent line of y=x/(x+1) at the point (-1,3) is y=2x+5.

## What is the slope of the tangent line of y=x/(x+1) at the point (-1,3)?

The slope of the tangent line of y=x/(x+1) at the point (-1,3) is 2.

## How do you find the point of tangency for y=x/(x+1) at the point (-1,3)?

To find the point of tangency for y=x/(x+1) at the point (-1,3), substitute the x and y coordinates of the point into the equation y=2x+5. The resulting point will be the point of tangency.

## What is the significance of finding the tangent line of y=x/(x+1) at the point (-1,3)?

Finding the tangent line allows us to determine the instantaneous rate of change of the function at the specific point. This can help us understand the behavior of the function and make predictions about its behavior near the point.

## What other applications can the concept of tangent lines have in mathematics?

The concept of tangent lines is used in many areas of mathematics, including calculus, geometry, and physics. It is used to find the slope of curves, to approximate the behavior of functions, and to solve optimization problems. It also has applications in engineering and economics.

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