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Homework Help: 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

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    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
     
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