Question

What is the equation of the line that is normal to `f(x)= sqrt( x^2+3x+2)` at `x=1`?

Answer 1

`y=sqrt6/5(-2x+7)`
See explanation below

Explanation:

The value of derivative of `f(x)` at `x=1` is the slope of tangent line to `f(x)` at `x=1`.

As per, `f'(x)=1/2(2x+3)/sqrt(x^2+3x+2)`

`f'(1)=1/2(2·1+3)/sqrt(1^2+3·1+2)`=`5/(2sqrt6)`

By other hand: if the straigh line general equation is

`y=mx+b` (where `m` is the slope and `b` is the intersection with `y` axis) then, the normal line equation is `y=-1/mx+c`

So: the normal line equation requested is `y=-2sqrt6/5x+b`

The value b can be found by the fact: the value of `f(1)` and by normal line must be the same (because they intercept in `x=1`

Thus: `f(1)=sqrt6= -2sqrt6/5+b` trasposing elements we found

`b=7sqrt6/5`

Finally, the normal line equation is `y=-2sqrt6/5x+7sqrt6/5` or

`y=sqrt6/5(-2x+7)`