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Ap Calculus help.

  1. Oct 20, 2005 #1
    This somehow seems easy, but I cannot get the jist of it.

    f(x) = x^3 - x^2 - 4x + 4

    The point (a,b) is on the graph of f and the line tangent to the graph at (a,b) passes through the point (0,-8) which is not on the graph of f. Find the values of a and b.

    okay so, ive spent about an hour just on this part, this problem had 3 parts.

    so i though, maybe, set the original with as and bs, so:

    b= a^3 - a ^2 - 4a + 4

    then maybe take the derivative

    b'= 3a^2 - 2a - 4

    so maybe

    y-8=b'x

    from there on it seems to get pretty nasty,:surprised doesnt seem like the type of problem, that should be this nasty.

    maybe someone can shed some light on this.:smile:

    thanks in advanced.:tongue2:
     
  2. jcsd
  3. Oct 20, 2005 #2

    hotvette

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    You are actually very close. You should be able to expand your last equation and get it in terms of only a and b. Same with the first one. You end up with 2 equations in 2 unknowns (I think). Hint: the point (a,b) is on the graph of each function.
     
    Last edited: Oct 20, 2005
  4. Oct 20, 2005 #3

    hotvette

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    Also, if you haven't already, it really helps to draw a picture. You don't have to actually plot the function, just draw a function, pick a point, draw a tangent to it, etc.
     
  5. Oct 20, 2005 #4
    hmmm, so

    b = 3a^3 - 2a^2 - 4a - 8
    b = a^3 - a^2 - 4a + 4

    0=-2a^3 +a^2 +12

    yuck, im sure this wasnt suppose to happen. :(

    edit: somehow, i think im making it harder than it seems, looks like a very simple problem. :\
     
    Last edited: Oct 20, 2005
  6. Oct 20, 2005 #5

    hotvette

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    Can't you solve the last equation for a? I got it just by staring at the equation for a few minutes.
     
    Last edited: Oct 20, 2005
  7. Oct 20, 2005 #6
    a= (-6/a^2) - (1/2)??????

    i dont know, seems like this question could take up a long, long time. this is supposedly an ap free response question. im thinking this question, is a little too long, to be on a timed test.
     
  8. Oct 20, 2005 #7

    hotvette

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    Much easier than that. Just look at the equation and try to imagine what values of a will satisfy the equation. The first one I tried was the right one.:smile:
     
    Last edited: Oct 20, 2005
  9. Oct 20, 2005 #8
    so i put in 2, get 4x^2 + 3x +6, looks like a mr quadradic to me.:grumpy:

    edit: roar!, mr quad come out a bad choice.
    so only one answer? (2,0)?
     
    Last edited: Oct 20, 2005
  10. Oct 20, 2005 #9

    hotvette

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    [tex]0=-2a^3 +a^2 +12[/tex]

    This is the one you are trying to solve for a. I like the choice you made.:smile:
     
  11. Oct 20, 2005 #10
    so (2,0) is the only answer?
     
  12. Oct 20, 2005 #11

    hotvette

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    The question was to find a & b, which you just did. Congratulations!
     
  13. Oct 20, 2005 #12
    [​IMG]
    thank you. very much. :)
    spent 2 hours on this and its late, time for sleep. thanks again.:smile:
     
  14. Oct 20, 2005 #13

    hotvette

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    Sorry I didn't have a more thoughtful answer (I needed my beauty sleep :biggrin:). Since you wanted (a,b) to satisfy the 2 equations, you could claim that you are done since you found that. But, you are curious (which is good) that there might be more solutions to [itex]0=-2a^3 +a^2 +12[/itex] since it is a cubic. Good point. You have several options. My first choice is to go ahead a plot it (don't think that's against the rules). Sure enough, there is only 1 real root. So, the other 2 must be? Actually, if there were more than 1 real root, that would mean that a line passing through (0,-8) would be tangent to the function at more than 1 point (i.e. multiple a,b combinations). See attached thumnail illustrating the situation.
     

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    Last edited: Oct 20, 2005
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