Question

find the tangent of the curve y=x^3 -2x^2 +x+1 using (y-y1)=(m(x-x1)) and d/dx derivatives and stuff like that, you should end up with two answers for the x coordinates: 1/3 and 1.

Answer 1

i need help!!!!!!!!

Answer 2

Is this post completed, because you need a point coordinate to find slope?

Answer 3

yes but through two points. i did the derivative until i ended up with y=6x-4 but thats as far as i know how to do...i dont know how to get to the answers(1/3 and 1). (1/3 and 1 are the answers because your trying to find the x-coordinates only)

Answer 4

the to xcoordinates are what your trying to find, which is1/3 and 1. those xcordinates are where the tangent line touches on the curve

Answer 5

no, when im solving the problem the x coordinates are what i do not know. the x- coordinates are what im trying to find. i know the x cordinates because i have the answer sheet. but what i dont know is how i get to those answers...what work do i do to get to those answers?

Answer 6

If you only want to find x coordinate so:
f'(x) = 3x² - 4x + 1
f'(x) = 0 => x1 = 1 , x2 = 1/3

DONE!

Answer 7

so you take 3x^2 -4x+1 and make it (3x-1)(x-1) and then 3x-1=0, 3x=1, x=1/3; x-1=0, x=1........right?

Answer 8

so find the derivative and then do that?

Answer 9

Taking derivative means find slope points!

Answer 10

so d/dx (x^3 -4x^2 +x+2)=3x^2 -8x+1 how do you get the x coordinates of those slope points?

Answer 11

f'(x) = 0

Answer 12

and i just use the (-b+/-sqrt(b^2-4ac))/(2a) formula right? and i get (4(+/-)sqrt(13))/3

Answer 13

Δ = ...?

Answer 14

x=(4+sqrt(13))/3
x=(4-sqrt(13))/3

Answer 15

Yes, you're correct :)

Answer 16

awesome! :D haha no longer frustrated!... my ap cal teacher kinda flies through the lesson without really giving explanations or making sure we understand. im still totally confused on how to do limits and finding limits and stuff though.

Answer 17

Go ahead post your questions! You'll see it's not as difficult as you thought :)