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Solving quadratic

  1. Mar 29, 2008 #1
    [SOLVED] Solving quadratic

    1. The problem statement, all variables and given/known data

    1a) Solve f(x) = x^2- 4x+m in the form f(x) = (x-a)^2+ b
    1b) What is the smallest value f(x) can have?


    3. The attempt at a solution

    1a) Seems simple enough. I set f(x) to 0 and used the completing the square method to solve. Ended up with f(x)=(x-2)^2+ m-4.

    I don’t know how to approach 1b) though. I’m assuming I’ve made a mistake somewhere – there is no smallest value f(x) can have without knowing the value of m. Is there a way to find m that I haven't picked up on, or would the answer just be m - 4?

    Thanks for any advice
     
    Last edited: Mar 29, 2008
  2. jcsd
  3. Mar 29, 2008 #2
    Looks fine to me.
     
  4. Mar 29, 2008 #3

    HallsofIvy

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    Science Advisor

    No, what you have given is fine. No matter what x is (x- 2)2 can never be lower than 0 so f(x)= (x-2)2 + m- 4 can never be lower than m- 4. Since you are not given a specific value of m, that is all you can do.
     
  5. Mar 29, 2008 #4
    Okay, thanks for that
     
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