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

Question

2209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T04:45:48

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

#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]}