Question

Find the equation of the normal line to the curve
f(x) = x^3 - 3x + 1
at the point (2,3)

I believe I have to do the derivative which will be the slope correct?

Answer 1

The derivative I got was 3x-3

Answer 2

derivative is

3x^2 - 3

Answer 3

oh right I forgot to subtract 1

Answer 4

Using that point, what is the slope of the TANGENT at that point

Answer 5

is it 9?

Answer 6

~ derivative.
~ plug in the x-coordinate to find the slope.
~ take the perpendicular slope (after step 2)
~ find the equation with the new slope and point (2,3)

Answer 7

m = 3(2)^2 -3 = 9

Answer 8

the slope of a line perpendicular to the TANGENT, will be the slope of the NORMAL at that point

Answer 9

-1/9

Answer 10

with point (2,3) and slope -1/9, form a line

Answer 11

so one of the equations is y-3 = 9 (x-2) and the normal line is
y-3 = -1/9 (x-2) right?

Answer 12

yep

Answer 13

thank you so much you guys !!!

Answer 14

wait till you start doing those with vectors in 3-space, fun messy as all heck derivatives

Answer 15

\(\large\color{black}{ f(x) = x^3 - 3x + 1 }\)
\(\large\color{black}{ f'(x) = 3x^2 - 3 }\)
\(\large\color{black}{ f'(2) = 3(2)^2 - 3 =9 }\)
\(\large\color{black}{ m_{normal}= -1/9 }\)

good!

Answer 16

Tangent, Normal, BiNormal vectors, form a mini coordinate system at any point in 3-space

Answer 17

forgot to put 1/5 spacing here

Answer 18

1.5*