1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find an equation of the tangent line to the curve at the given point

  1. Feb 11, 2007 #1
    y=1+2x-xcubed, (1,2)
     
  2. jcsd
  3. Feb 11, 2007 #2
    first i set up the problem, i put 1+2x-xcubed-1/x-1
    i got stuck because of the xcubed, i forgot how to factor out polynomials that hae xcubed in them, i know this part doesn't have to do with calculus, but i forgot when i did polynomials like this
     
  4. Feb 11, 2007 #3
    hey i set up the problem help me out mane
     
  5. Feb 11, 2007 #4
    the answer isy=-x+3, but i need to figure out the slope
     
  6. Feb 11, 2007 #5
    i figured out my problem, except i don't know how to factor out xcubed polynomials, or put them into parenthesis
     
  7. Feb 12, 2007 #6

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    You've got the slope in this formula. y=-x+3 is the same as saying y=-1x+3. And since y=mx+b, what do you suppose m is?

    I got the answer without factoring anything.

    I'm not sure how you got the slope (even though you don't realize you got it), but what I did was take the derivative of 1+2x-x3 Then I plugged my x-value into that formula.
     
  8. Feb 12, 2007 #7
    Typically speaking, when a professor wants a tangent line to a curve, he probably wants you to find the derivative. so i guess that would be 2x-3x^2

    So since you are given points....
     
  9. Feb 12, 2007 #8
    well i know what the slope is but how do i find it, obviously it's -1, but the equation is confusing
     
  10. Feb 12, 2007 #9
    y = mx + b
    y = -1x +3

    b = 3, so m = ...
     
  11. Feb 12, 2007 #10
    i have to find the slope from y=1+2x-x3
     
  12. Feb 12, 2007 #11
    i factor it out to x(x+1)(x+1) but i'm wrong because i plug in 1 into the x, and i get 4
     
  13. Feb 12, 2007 #12
    here is what the slope form looks like 1+2x-x3-1/x-1 which is f(x)-a/x-a
    the only part i am having trouble with is the 1+2x-x3, so what do i do with this
     
  14. Feb 12, 2007 #13
    [tex]1+2x-x^3[/tex]

    Derivative is [tex]2-3x^2[/tex]

    You know that the slope is -1 and the point, so use the equation

    [tex]y-y_0=m(x-x_0)[/tex]

    where m = -1, [tex]x_0=1[/tex] and [tex]y_0=2[/tex]
     
  15. Feb 12, 2007 #14
    how did u get 2-3x2
     
    Last edited: Feb 12, 2007
  16. Feb 12, 2007 #15
    I took the derivative of the very first equation in my post. Since the derivative is equal to the slope of the tangent, then plug in x=1

    [tex]2-3(1)^2=-1[/tex]
     
  17. Feb 12, 2007 #16
    so this equals negative -1, ahh makes sense, so what do you do about the x-1 in the denominator
     
  18. Feb 12, 2007 #17
    Ah, you're doing it from first principles. Give me a second to see if I can work it out.
     
  19. Feb 12, 2007 #18
    Ok, from first principles, I get

    [tex]\lim_{x\rightarrow a} \frac{-x^3+2x-1}{x-1}[/tex]

    When I plug in x=1 into the equation, the nominator and denominator both go to zero. So I need to use Hopital's rule and take the derivative of the top and the bottom.

    [tex]\lim_{x\rightarrow a} \frac{-3x^2+2}{1}[/tex]

    Plug in x=1 and you get -1.

    Have you learned Hopital's rule yet?
     
  20. Feb 12, 2007 #19
    nope i checked my notes, actually my teacher may have tought us but i probably missed that day or something
     
  21. Feb 12, 2007 #20
    what is this rule, if you don't feel like explaining it, i'll look it up on the internet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?