Question

What is the equation of the line normal to `f(x)=-sqrt((x-5)(x-2)` at `x=6`?

Answer 1

`y=4/5x-34/5`. See the normal-inclusive Socratic graph.

Explanation:

`f<=0`.
At `x = 6, f = -2`. So, the point is `P( 6, -2 )`
To make f real,

`(x-5)(x-2)=(x-7/2)^2-(3/2)^2>=0`, giving `x i<=5 and <= 2`

The Socratic graph is inserted, for yet another proof.

`f'=-1/2(2x-7)/sqrt(((x-5)(x-2)(`= -5/4#, at x = 6.

Slope of the normal is `-1/f'(6)=4/5`.
\
So, the equation to the normal at `P(6, -2)` is

`y-(-2)=4/5(x-6)`. Simplifying,

`y=4/5x-34/5`
graph{(y+sqrt((x-5)(x-2))) ( y-4/5x+34/5) = 0 [-10, 10, -5, 5]}