Question

How do you find the equation of a line tangent to the function `y=1/(2-x)` at x=-1?

Answer 1

Using the first derivative `(dy)/(dx)` we can find the gradient of the tangent, and then use point-gradient formula `y-y_1=m(x-x_1)`

First, finding `(y_1,x_1)` for where `x=-1`
`y=1/(2-(-1))=1/3`
So `(x_1,y_1)` is `(-1,1/3)`

Now, deriving the function,
`y=(2-x)^-1->`changing to index form
`(dy)/(dx)=-(2-x)^-2`
Subbing in `x=-1` to `(dy)/(dx)` to find the gradient of the tangent,
`m=-(2-(-1))^-2=-1/9`

Subbing `(x_1,y_1)` and `m` into the point-gradient formula,

Eq of tangent: `y-1/3=-1/9(x+1)`
`9y-3=-x-1`

`:.x+9y-2=0` is the equation of the tangent to the curve at `x=-1`