Question

For `f(x) =(x-3)/(x-4)`, what is the equation of the line tangent to `x =3`?

Answer 1

`y=-x+3`

Explanation:

In order to find the equation of the tangent line at `x=3`, we first need to find the gradient of this line at `x=3`.

To do this we find the derivative of `f(x)`. In this case we need to use the quotient rule:

The quotient rule states that for functions `f(x),g(x)`:

`(f/g)^'=(f^'g-fg^')/g^2`

Let:

`f(x)=x-3` and `g(x)=x-4`

`:.`

`((x-3)/(x-4))^'=(1(x-4)-(x-3)1)/(x-4)^2=-1/(x-4)^2`

We now plug `x=3` into this derivative.

`-1/((3)-4)^2=-1`

So gradient `m=-1`

Using the original function to calculate a `y` value:

`f(3)=(x-3)/(x-4)=((3)-3)/((3)-4)=0`

Using point slope form of a line:

`(y_2-y_1)=m(x_2-x_1)`

`y-0=-1(x-3)`

`y=-x+3`

This is the equation of the tangent line at `x=3`

Graph: