# For f(x)=(2x+1)/(x+2)  what is the equation of the tangent line at x=1?

Nov 2, 2015

I found: $y = \frac{1}{3} x + \frac{2}{3}$

#### Explanation:

First you need to find the slope $m$ of the tangent line. This is found deriving your function and calculating it at $x = 1$:

$f ' \left(x\right) = \frac{2 \left(x + 2\right) - \left(2 x + 1\right)}{x + 2} ^ 2 = \frac{2 x + 4 - 2 x - 1}{x + 2} ^ 2 = \frac{3}{x + 2} ^ 2$

at $x = 1$
$f ' \left(3\right) = \frac{3}{1 + 2} ^ 2 = \frac{1}{3} = m$

Now we need the $y$ value of the tangence point as well; we find it setting $x = 1$ into the original function:
$f \left(1\right) = \frac{2 + 1}{1 + 2} = \frac{3}{3} = 1$

The equation of a line passing through ${x}_{0} = 1 \mathmr{and} {y}_{0} = 1$ with slope $m = \frac{1}{3}$ is given as:
$y - {y}_{0} = m \left(x - {x}_{0}\right)$
$y - 1 = \frac{1}{3} \left(x - 1\right)$
$y = \frac{1}{3} x - \frac{1}{3} + 1$
$y = \frac{1}{3} x + \frac{2}{3}$

Graphically: