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Are any tangents to the graph f(x)=x^2-3x horizontal?

  1. Sep 30, 2007 #1
    "At what points, if any, are the tangents to the graph f(x)=x^2-3x horizontal?"

    I know how to figure this out using a graphing calculator and using the 'short-cut' (nx^n-1), but I can't do it using limits. Help!

    I'd like to use the equation lim x->a = f(x) - f(a) / x - a

  2. jcsd
  3. Sep 30, 2007 #2
    why would you do it like that? one yields the other. just use the "shortcut"
  4. Oct 1, 2007 #3


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    As ice109 said, once you have worked out the general formula from the definition, you can use that formula and not have to go back to the definition every time! That's the whole point of the formula (which you call a "shortcut").

    However, since you ask, it's not particularly difficult: f(x)= x^2- 3x and f(a)= a^2- 3a.
    f(x)- f(a)= x^2- 3x- a^2+ 3a= (x^2- a^2)- (3x-3a)= (x-a)(x+a)- 3(x-a). Now, when you divide that by x-a, what do you get? What is the limit of that as x goes to a?

    I assume you know that "tangent line horizontal" means that the derivative must be 0.
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