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Homework Help: Tangent line

  1. Sep 25, 2006 #1
    Hi, I need help on a math problem that asks me to find k such that the line is tangent to the function, given[tex]f(x) = x^2 - kx[/tex] and [tex]y = 4x - 9[/tex].

    I don't know how to solve for k. Is "tangent line" the same thing as the equation that is the derivative of the function?
     
  2. jcsd
  3. Sep 25, 2006 #2

    chroot

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    To find a tangent line:

    1) Find the derivative of your function. The derivative gives you the slope of the function for any given x.

    2) Find a point on the curve: an x0 and its corresponding y0.

    3) Use point-slope form to combine the point and slope into a single equation.

    In this case, the x0 is arbitrary. You're not trying to find a particular tangent line at a particular x0, you're trying to find the tangent line for any arbitrary x0.

    - Warren
     
    Last edited: Sep 25, 2006
  4. Sep 25, 2006 #3
    No, while the slope of a tangent line at a particular value x for a given function f(x) is the derivative of that function for the particular value x, the derivative of a function f(x) is the expression of the slopes of the tangent lines for all values of x.
     
  5. Sep 25, 2006 #4
    How do I find a point of the curve f(x)=x^2-kx where there are 2 variables?
     
  6. Sep 25, 2006 #5

    chroot

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    Every point on that curve is of the form (x, f(x)). k cannot be eliminated until the very end of this problem.

    - Warren
     
  7. Sep 26, 2006 #6

    HallsofIvy

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    There aren't. There is only one variable, x (you know that because it says "f(x)"). The k is an unknown constant. What is the derivative of f(x)= x2- kx, remembering that the variable is x? Knowing that x must give the same "y" value for y= x2- kx and y= 4x- 9 gives you one equation for the two unknown numbers x and k. Knowing that the derivative of f(x)= x2- kx at that x is the same as the slope of y= 4x- 9 gives you another. Now you have two equations for the two unknown numbers, x and k.

    (There are, by the way, two correct solutions.)
     
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