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Finding two parallel tangents

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data
    I have a cubic function, y=(x-6)(x-1)(x-9) or y=x3-16x2+69x-54
    I then have two tangents, y=7.75x+14.75 and y=-6.25x+56.25
    What I need to find is another two tangents that are parallel.


    2. Relevant equations



    3. The attempt at a solution
    What I know needs to be found is the two x points which the tangents cross, the rest from there on is simple enough. So how do I find the two x points parallel to the other tangents. Any Help is much appriectated.
     
    Last edited: Jun 5, 2010
  2. jcsd
  3. Jun 6, 2010 #2

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    Gold Member

    To find a tangent that is parallel, you need to find one with the same slope. How can you find a tangent line with slope 7.75?
     
  4. Jun 7, 2010 #3
    Sorry, I'm really blank about the whole thing, Please explain
     
  5. Jun 7, 2010 #4

    Mark44

    Staff: Mentor

    One of your tangent lines has a slope of 7.75. This means that at some point on the graph of the cubic, the value of the derivative function is 7.75. Find the derivative of your cubic function and find both values of x for which dy/dx = 7.75.

    Do the same thing for the other tangent line, the one whose slope is -6.25.
     
  6. Jun 7, 2010 #5
    I really don't know how to do that
     
  7. Jun 7, 2010 #6

    Mark44

    Staff: Mentor

    Your book should have some examples showing how to find the derivative of a given function. The derivative can be used to find the slope of a tangent line at any point on the graph of the function whose derivative you take.

    For example, if f(x) = 2x2 + 3x, f'(x) = 4x + 3. At x = 0, f'(0) = 4(0) + 3 = 3. At (0, 0), the slope of the tangent line is 3.
     
  8. Jun 7, 2010 #7
    can you tell me the answer to one of the tangents, and then I will work out the other myself?
     
  9. Jun 7, 2010 #8
    I'm not trying to find the y value, just the x value, the y value is easy to find after getting the x value.
     
  10. Jun 7, 2010 #9

    Mark44

    Staff: Mentor

    No, that goes against the rules in this forum. You should be able to find the derivative of y = x3 - 16x2 + 69x-54, and set the derivative to 7.75 and -6.25, respectively, to find all of the points of tangency for those values.

    You need to make an effort at this yourself. I'm willing to help, but I'm not willing to do the work for you.
     
  11. Jun 7, 2010 #10
    3x^2-32x+69 then what?
     
  12. Jun 7, 2010 #11
    You want the x (or y) coordinate of the points at which the slope of your function is 7.75. To do this, like Mark said, set your derivative function equal to 7.75.

    [tex]f'(x)=3x^2-32x+69=7.75[/tex]

    Now solve that equation. Use the quadratic formula. You will get two values for x, one you have already, and the second being the one you're looking for.
     
  13. Jun 7, 2010 #12

    Mark44

    Staff: Mentor

    And do the same thing for the other slope, setting the derivative to -6.25.
    [itex]3x^2 - 32x + 69= -6.25[/itex]
     
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