Question

What is the equation of the normal line of `f(x)=(x-3)/(x+4)` at `x=-1`?

Answer 1

`21y+27x+55=0`

Explanation:

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

Recall: the quotient rule is given by `(dy)/(dx)=(vu'-uv')/v^2` if `y=u/v`

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

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

`f'(x)= 7/(x+4)^2`

When `x=-1`

`f'(-1)=7/(-1+4)^2`

`f'(-1)=7/9`

You just found the gradient of the tangent but we want the normal. Recall that there is a rule `m_1m_2=-1` where `m_1` is a gradient and `m_2` is another gradient and when you multiply it together, it equals to `-1`

ie. `7/9m_2=-1`

`m_2=-9/7`

When `x=-1`, y is equal to `f(-1)=(-1-3)/(-1+4)=-4/3`
`(-1,-4/3)`

Using the point gradient formula,

`(y+4/3)=-9/7(x+1)`

`7y+28/3=-9x-9`

`7y+9x+55/3=0`

`21y+27x+55=0`