Question

How do you find the equation of the tangent line to the curve `y=(2x+1)/(x+2)` at the given point (1,1)?

Answer 1

We have, `y = (2x+1)/(x+2)`

Now, the derivative (using quotient rule)

`dy/dx = (2*(x+2) - (2x+1))/(x+2)^2 = 3/(x+2)^2`

Now `dy/dx` at `(1,1)` is the slope of the tangent to the curve at `(1,1)`

`m = (dy/dx)_(1,1) = 3/9 = 1/3`

Now, point-slope form a straight line is given by

`(y-y_1) = m*(x-x_1)`
Now, `(x_1,y_1) = (1,1)` and `m = 1/3`, substituting these values in above equation we get,

`y-1 = 1/3 * (x-1)` => `y = x/3 + 2/3` (Answer)