Find the tangent lines to the curve

[frosty8688]

1. How many tangent lines to the curve \left(y=\frac{x}{x + 1}\right) pass through the point (1,2)? At which points do these tangent lines touch the curve?
2. \frac{x}{x + 1}
3. I tried to use the quotient rule and came up with the equation \frac{1}{(x + 1)^{2}}. I tried plugging in 1 to get the slope of 1/4 and the equation y = 1/4x + 7/4. I know that there are two lines, because of the square.

 

[tiny-tim]

hi frosty8688!

How many tangent lines to the curve \left(y=\frac{x}{x + 1}\right) pass through the point (1,2)?

I … came up with the equation \frac{1}{(x + 1)^{2}}. I tried plugging in 1 …

because 1 is the value of x at (1,2) ?

noooo …

(1,2) isn't on the curve, is it? ​

 

[frosty8688]

I did figure out the x values -2±\sqrt{3}. I'm just having problems finding the y values.

 

[frosty8688]

Would it be -2±\sqrt{3} + 1 / 2 which is equal to 1/2 * -1±√3

 

[HallsofIvy]

I did figure out the x values -2±\sqrt{3}. I'm just having problems finding the y values.

Those are the x values of what points? The points of tangency? If so, y is given by x/(x+1).

Let (x_0, y_0)= (x_0, x_0/(x_0+1)) be the point of tangency. Then a line through that point and (1, 2) is given by y= [(y_0- 2)/(x_0-1)](x- 1)+ 2. With y= x/(x+1), y'= 1/(x+1)^2 so we must have (y_0- 2)/(x_0-1)= 1/(x_0+1)^2 as well as y_0= x_0/(x_0+1).

Solve those two equations for x_0 and y_0.

 

[frosty8688]

Here's how I found the x values; \frac{1}{(x+1)^{2}} = \frac{y-2}{x-1}, y-2 = \frac{x-1}{(x+1)^{2}}, \frac{x}{x+1}-2 = \frac{x-1}{(x+1)^{2}}, \frac{[x-2(x+1)]}{x+1}=\frac{x-1}{(x+1)^{2}}, \frac{x-2x-2}{x+1} = \frac{x-1}{(x+1)^{2}}, \frac{-(x+2)}{x+1} = \frac{x-1}{(x+1)^{2}}, -(x+2) = \frac{x-1}{x+1}, (x+2)(x+1) = 1-x, x^{2}+4x+1 = 0. Then I used the quadratic formula and got the x values to be -2±\sqrt{3}

 

[frosty8688]

For the y values, my calculator is giving me the answer \frac{1±\sqrt{3}}{2} and I am wondering how do I get that answer without using the calculator.

 

[frosty8688]

Do I have to multiply \frac{-2±\sqrt{3}}{-2±\sqrt{3}+1} by something.

 

[frosty8688]

I figured it out.

 

[azizlwl]

1. How many tangent lines to the curve \left(y=\frac{x}{x + 1}\right) pass through the point (1,2)? At which points do these tangent lines touch the curve?



2. \frac{x}{x + 1}



3. I tried to use the quotient rule and came up with the equation \frac{1}{(x + 1)^{2}}. I tried plugging in 1 to get the slope of 1/4 and the equation y = 1/4x + 7/4. I know that there are two lines, because of the square.

Even though both have equal gradient, the 2 lines might be of parallel tangent lines.
Reference to a point on the curve for gradient will ensure both are intersecting same points.