Question

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

Answer 1

I found: `y=1/3x+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'(x)=(2(x+2)-(2x+1))/(x+2)^2=(2x+4-2x-1)/(x+2)^2=3/(x+2)^2`

at `x=1`
`f'(3)=3/(1+2)^2=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(1)=(2+1)/(1+2)=3/3=1`

The equation of a line passing through `x_0=1 and y_0=1` with slope `m=1/3` is given as:
`y-y_0=m(x-x_0)`
`y-1=1/3(x-1)`
`y=1/3x-1/3+1`
`y=1/3x+2/3`

Graphically: