Question

How do you find the equation of the tangent and normal line to the curve `y=9x^-1` at x-3?

Answer 1

I'm not sure if you meant `x=3` or `x=-3`

If `x=3` Then
Tangent is `y=-x+6` and Normal is `y=x`

If `x=-3` Then
Tangent is `y=-x-6` and Normal is `y=x`

Explanation:

I'm not sure if you meant `x=3` or `x=-3` so let's do both:

First we need the derivative, which will give us the slope of the tangent at any point:

`y=9/x = 9x^-1 => dy/dx = 9(-x^-2)`

`:. dy/dx = -9/x^2`

We will use the formula `y-y_1=m(x-x_1)` to find the tangents and normals, and use the fact that the tangents and normals are [perpendicular (so their product is -1)

When x=3
`x=3 => y=3 ; y'=-9/9=-1`

Tangent:
Passes through `(3,3)` and has slope `m=-1`
`:. y-3=-1(x-3)`
`:. y-3=-x+3`
`:. y=-x+6`

Normal:
Passes through `(3,3)` and has slope `m=1`
`:. y-3=1(x-3)`
`:. y-3=x-3`
`:. y=x`

When x=-3
`x=-3 => y=-3 ; y'=-1`

Tangent:
Passes through `(-3,-3)` and has slope `m=-1`
`:. y-(-3)=-1(x-(-3))`
`:. y+3=-x-3`
`:. y=-x-6`

Normal:
Passes through `(-3,-3)` and has slope `m=1`
`:. y-(-3)=1(x-(-3))`
`:. y+3=x+3`
`:. y=x`