What is the equation of the normal line of #f(x)=6x^3-12x^2-x-1# at #x=2#?

Question

1244 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-07T09:37:21

#x+23y+67=0#

#f(x)=6x^3-12x^2-x-1#.

When #x=2, y=f(2)=48-48-2-1=-3#.

Thus, normal to #f# at #(2,-3) is reqd.

We know that, the slope of the tgt. at the pt.#(2,-3)" is "f'(2)#.

We have, #f'(x)=18x^2-24x-1 rArr f'(2)=72-48-1=23#.

Since, normal is #bot# to the tgt., its slope at #(2,-3)# is #-1/23#.

Summing up the information about the normal, it has slope#-1/23,#

&, passes thro. #(2,-3)#.

Using, Point-Slope Form for the Normal, its. eqn. is given by,

#y-(-3)=-1/23(x-2), or, 23(y+3)+(x-2)=0#;

i.e., #x+23y+67=0#

Enjoy Maths.!