Question

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

Answer 1

We must first differentiate using the quotient rule.

Explanation:

Let `y= (g(x))/(h(x))`

Then `g(x) = 2x + 1` and `h(x) = x + 2`. The quotient rule states that `y' = (g'(x) xx h(x) - (g(x) xx h'(x)))/(h(x)^2`

By the power rule:

`g'(x) = 2`

`h'(x) = 1`

We now have all the information we need to know:

`y' =(2(x + 2) - 1(2x + 1))/(x + 2)^2`

`y' = (2x + 4 - 2x - 1)/(x^2 + 4x + 4)`

`y' = 3/(x^2 + 4x + 4)`

The slope of the tangent is given by evaluating `f(a)` in the derivative function, `x = a` being the given point.

`y = 3/(1^2 + 4 + 4)`

`y = 1/3`

The slope of the tangent is `1/3`. Now we know the slope of the tangent and a point that lies both on the function and on the tangent.

By point-slope form:

`y - y_1 + m(x - x_1)`

`y - 1 = 1/3(x - 1)`

`y - 1 = 1/3x - 1/3`

`y = 1/3x + 2/3`

The equation of the tangent is `y = 1/3x + 2/3`.

Hopefully this helps!