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Math Help - Locate relative extrema is this working correct?

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    Locate relative extrema is this working correct?

    Locate and describe the relative extrema then sketch the graph of



    is the theory correct?
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    review your algebra ...

    3x^2 + 6x - 9 = 0

    3(x^2 + 2x - 3) = 0

    3(x + 3)(x - 1) = 0
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    After 3x^2 + 6x - 9 = 0, you wrote 3x(x - 1) = 0, which is false. Expand the left-hand side of the false equation to see why.

    Instead, divide the equation 3x^2 + 6x - 9 = 0 by 3:

    x^2 + 2x - 3= 0

    Then, factor the right-hand side of the equation:

    (x - 1)(x + 3) = 0

    Set each factor on the right-hand side of the equation equal to 0.

    x - 1 = 0 or x + 3 = 0

    Solve each equation in terms of x. The solutions are the critical values.
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    Quote Originally Posted by NOX Andrew View Post
    After 3x^2 + 6x - 9 = 0, you wrote 3x(x - 1) = 0, which is false. Expand the left-hand side of the false equation to see why.

    Instead, divide the equation 3x^2 + 6x - 9 = 0 by 3:

    x^2 + 2x - 3= 0

    Then, factor the right-hand side of the equation:

    (x - 1)(x + 3) = 0

    Set each factor on the right-hand side of the equation equal to 0.

    x - 1 = 0 or x + 3 = 0

    Solve each equation in terms of x. The solutions are the critical values.

    so x=1 or x =-3

    what do i do with the 3 outside the brackets? 3x = 0 ?
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  5. #5
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    No, only set factors with an x equal to 0. In other words, don't multiply a constant by an x and then set the product equal to 0. If you really wanted to, you could set just 3 equal to 0. However, 3 = 0 is false, so there are no solutions to that specific equation.
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