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Beginner(self study) calculus question.

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the slope of the tangent line to the parabola y = 2x - x2 at the point (2,0)

    2. Relevant equations
    [tex]\frac{f(x) -f(a)}{x - a}[/tex]

    [tex]\frac{f(a + h) - f(a)}{h}[/tex]

    3. The attempt at a solution

    [tex] \frac{(2x - x^2 - (2(2) - (2)^2)}{ x - 2} [/tex]

    [tex] \frac{(2x - x^2 - ( term self cancels )}{ x - 2} [/tex]

    [tex] \frac{x(2 - x)}{x - 2} [/tex]

    This is where I'm stopped. As I wrote in the thread title, I'm independently studying the subject before I have to take the course, so I don't have anyone to direct me in this endeavor. Any help would be appreciated.
    Last edited: Apr 12, 2009
  2. jcsd
  3. Apr 12, 2009 #2
    To find the slope you have to take the limit as x approaches a, by the definition of derivative.
  4. Apr 12, 2009 #3
    Here's one way to do it:

    1. Use the definition of derivative to find the derivative function.

    f'(x) = [tex]\stackrel{lim}{h\rightarrow0}[/tex] [tex]\frac{f(x+h)-f(x)}{h}[/tex]

    2. At the point (2,0), x = 2, so plug in 2 for x in the derivative function.
  5. Apr 12, 2009 #4
    What Dunkle said works too, if you'd like to obtain a function that will tell you the slope at every point.

    Of course the way to were doing it works too, you just have to take a limit as x approaches 2 in your case.
  6. Apr 12, 2009 #5


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    The problem is that the difference quotient is a function in a (or h). So think of your first equation as a template. If you want to figure out the slope of the tangent line to (2, 0), plug in 2 for x in the difference quotient to obtain:

    [tex]f\prime(2) = \lim_{a\rightarrow 2}\frac{f(2) - f(a)}{2 - a}[/tex]

    What this equation is telling you is that the derivative of f at 2, i.e., the slope of the tangent at (2, 0), is equal to the limit of [tex]\frac{f(2) - f(a)}{2 - a}[/tex] as a approaches 2. Note that this new function is a function in a, i.e., it takes an input value a and returns the slope of secant line passing though the value of f at 2 and the value of f at a. So the limit of this new function as a approaches 2 is equal to the slope of the tangent line.

    Here's how to work out the answer, so don't read this if you want to finish working it out on your own:

    [tex]\lim_{a\rightarrow 2}\frac{f(2) - f(a)}{2 - a}[/tex]

    [tex]= \lim_{a\rightarrow 2}\frac{4 - 2^{2} - 2a + a^{2}}{2 - a}[/tex]

    [tex]= \lim_{a\rightarrow 2}\frac{-2a + a^{2}}{2 - a}[/tex]

    [tex]= \lim_{a\rightarrow 2}\frac{(2 - a)(-a)}{2 - a}[/tex]

    [tex]= \lim_{a\rightarrow 2}-a[/tex]

    [tex]= -2[/tex]
  7. Apr 12, 2009 #6
    Wow, thank you all so much for all your help. With said help I was able to complete all the exercise problems in the textbook. Thanks again.
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