# How do you find the equations of the tangent lines to the curve y= (x-1)/(x+1) that are parallel to the line x-2y = 2?

May 23, 2015

Find the equations of the tangent lines to the curve $y = \frac{x - 1}{x + 1}$ that are parallel to the line $x - 2 y = 2$.

There is a bit of algebra and arithmetic for this. Let's focus on the reasoning and the calculus.

One
A line parallel to $x - 2 y = 2$ must have the same slope. The slope of this line is $\frac{1}{2}$. So we want the slope of the tangent line to be $\frac{1}{2}$

Two
How do we find the slope of the tangent line? -- The derivative. So, we want the derivative to be $\frac{1}{2}$.

What is the derivative of $y = \frac{x - 1}{x + 1}$ ?

Use the quotient rule:

$y ' = \frac{x \left(x + 1\right) - \left(x - 1\right) \cdot 1}{x + 1} ^ 2 = \frac{2}{x + 1} ^ 2$

Three

Find $x$ to make $y ' = \frac{1}{2}$

$\frac{2}{x + 1} ^ 2 = \frac{1}{2}$ if and only if

${\left(x + 1\right)}^{2} = 4$

So $x + 1 = \pm 2$

and $x = 1 , - 3$

Four

Find the $y$ values at $x = 1$ ($y = 0$)and at x=-3 (y=2)#

Five

Find the equations of the lines:

through $\left(1 , 0\right)$ with slope $m = \frac{1}{2}$

and through $\left(- 3 , 2\right)$ with slope $m = \frac{1}{2}$.

May 23, 2015

Have a look:

which gives yu two equations:
$y = \frac{1}{2} x + \frac{7}{2}$
and
$y = \frac{1}{2} x - \frac{1}{2}$