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Math Help - finding constant a

  1. #1
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    finding constant a

    Can you give me hint...Where to start ?

    I want to find constant a for function

    ((ax^2 - 6x + 4)/(x^2 - x - 2))

    when lim x to 2...
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  2. #2
    Member anonimnystefy's Avatar
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    Re: finding constant a

    hi

    you should post the whole question for us to help you.from here we cannot do anything.
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  3. #3
    Grand Panjandrum
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    Re: finding constant a

    Quote Originally Posted by Franciscus View Post
    Can you give me hint...Where to start ?

    I want to find constant a for function

    ((ax^2 - 6x + 4)/(x^2 - x - 2))

    when lim x to 2...
    What have you done?

    I presume you want the limit as x approaches 2 to be finite?

    If so you need (x-2) to be a factor of the numerator (it is already a factor of the denominator).

    CB
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  4. #4
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    Re: finding constant a

    Find a value for constant a in a function f(x) = ((ax^2 - 6x + 4) divided (x^2 - x - 2)) so that the limit exists, when the x approaches 2.
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  5. #5
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    Re: finding constant a

    Thanks for quick answer CB. Yes I meant finite. I'm from Finland, so I'm not familiar with all the notions.

    So in denominator I got (x-2)(x+1). Can I use these values for x in numerator or do I have to factor it other way to find out a?
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  6. #6
    Grand Panjandrum
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    Re: finding constant a

    Quote Originally Posted by Franciscus View Post
    Thanks for quick answer CB. Yes I meant finite. I'm from Finland, so I'm not familiar with all the notions.

    So in denominator I got (x-2)(x+1). Can I use these values for x in numerator or do I have to factor it other way to find out a?
    As I said you need x-2 to be a factor of the numerator (to cancel the x-2 factor in the denominator)

    For x-2 to be a factor of the numerator you need a(2)^2-6(2)+4=0

    CB
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  7. #7
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    Re: finding constant a

    Thanks. Great forum!
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