Question

Find an equation for the line perpendicular to the tangent to the curve y=x^3 -9x+1 at the point (3,1)

Answer 1

Differentiate the equation to find dy/dx.
dy/dx =3x^2 - 9
slope at (3,1) is dy/dx at x=3 and y=1
so, slope of required line=m= 3(3)^2 -9=27-9=18
now using slope point form equation of tangent is
(y-1)=18(x-3)
y-1=18x-54
18x-y-53=0

Answer 2

Is 18x-y-53=0 the answer?

Answer 3

no its the equation of tangent which I mistakenly derived.
Slope of the line perpendicular to the tangent to the curve is given as
m=-dx/dy=-1/(dy/dx)=-1/18
Now find the equation in slope point form as done above..
Got it??

Answer 4

No. I'm sorry but what would be the answer?

Answer 5

By slope point form
(y-1)=-(1/18)(x-3)
18y-18=-x+3
x+18y-21=0

Answer 6

so x+18y-27=0?

Answer 7

x+18y-21=0

Answer 8

oh, right. Thanks

Answer 9

its a wrong answer :(