# Find the tangent lines to the curve

1. Jul 21, 2012

### 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.

2. Jul 21, 2012

### tiny-tim

hi frosty8688!
because 1 is the value of x at (1,2) ?

noooo

(1,2) isn't on the curve, is it?

3. Jul 21, 2012

### frosty8688

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

4. Jul 21, 2012

### frosty8688

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

5. Jul 21, 2012

### HallsofIvy

Staff Emeritus
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 $y_0$.

6. Jul 21, 2012

### 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±$\sqrt{3}$

7. Jul 21, 2012

### 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.

8. Jul 21, 2012

### frosty8688

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

9. Jul 21, 2012

### frosty8688

I figured it out.

10. Jul 21, 2012

### azizlwl

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.