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Math Help - help on precalc. quadratic functions, ratios,

  1. #1
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    help on precalc. quadratic functions, ratios,

    any help on any of these is appreciated. i took precalc so long ago, i'm finding it really difficult to recall how to do certain things, and i didnt want to clutter the board with all these different questions.




    Put the quadratic function in standard form (i.e. .


    Correct Answer:


    --------------------------------------------------------

    Given that , find the ratio .
    Correct Answer:

    ---------------------------------------------------------

    Use the properties of exponents to simplify the expression

    Correct Answer:


    ------------------------------------------------------
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  2. #2
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    Quote Originally Posted by rafaeli View Post
    any help on any of these is appreciated. i took precalc so long ago, i'm finding it really difficult to recall how to do certain things, and i didnt want to clutter the board with all these different questions.




    Put the quadratic function in standard form (i.e. .


    Correct Answer:
    This is completing the square.

    When in the form ax^2 + bx + c where a, \ b, \ c are constants; you can complete the square by \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c.

    x^2 + 5x+1
    a=1, \ b=5, \ c=1
    \left(x + \frac{5}{2}\right)^2 -  \left(\frac{5}{2}\right)^2 + 1
    \left(x + \frac{5}{2}\right)^2 - \frac{25}{4} + 1
    \left(x + \frac{5}{2}\right)^2 - \frac{21}{4}

    Quote Originally Posted by rafaeli View Post
    --------------------------------------------------------

    Given that , find the ratio .
    Correct Answer:
    This question just involves algebra. Apply the rules provided and simplify.

    f(x+t) = \frac{x+t}{x+t+1}

    f(x) = \frac{x}{x+1}

    \implies \frac{f(x+t) - f(x)}{t} = \frac{\frac{x+t}{x+t+1} - \frac{x}{x+1}}{t}

     \frac{\frac{(x+t)(x+1) - (x)(x+t+1)}{(x+t+1)(x+1)}}{t}

     \frac{(x+t)(x+1) - (x)(x+t+1)}{(x+t+1)(x+1)(t)}

    \frac{t}{(x+t+1)(x+1)(t)}

    \frac{1}{(x+t+1)(x+1)}

    Quote Originally Posted by rafaeli View Post
    ---------------------------------------------------------

    Use the properties of exponents to simplify the expression

    Correct Answer:


    ------------------------------------------------------
    This question used the rules of indices.

    \left(\frac{9^{\frac14}}{2^{\frac12} 4^{\frac14}}\right)^{-2}

    \left(\frac{9^{-\frac12}}{2^{-1} 4^{-\frac12}}\right)

    \left(\frac{\frac13}{\left(\frac12\right)\left(\fr  ac12\right)}\right)

    \left(\frac{\left(\frac13\right)}{\left(\frac14\ri  ght)}\right)

    \left(\frac43\right)
    Last edited by Simplicity; June 29th 2008 at 01:12 PM. Reason: Typo Corrected
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  3. #3
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    Quote Originally Posted by rafaeli View Post
    any help on any of these is appreciated. i took precalc so long ago, i'm finding it really difficult to recall how to do certain things, and i didnt want to clutter the board with all these different questions.




    Put the quadratic function in standard form (i.e. .


    Correct Answer:


    ------------------------------------------------------
    Edit: Air, check your first step in completing the square. Should've added 1 instead of subtract. Just a typo, since the answer checked. I'm still too slow today,


    Just complete the square.

    f(x)=(x^2+5x+ \ \ ?)+1 - \ \ ?

    Take half the coefficient of your linear term, square it and add it to make a perfect square trinomial, then subtract it to equalize the equation.

    f(x)=\left(x^2+5x+\frac{25}{4}\right)+1-\frac{25}{4}

    f(x)=\left(x+\frac{5}{2}\right)^2+1-\frac{25}{4}

    f(x)=\left(x+\frac{5}{2}\right)^2-\frac{21}{4}
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  4. #4
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    how do you get this?



    Quote Originally Posted by Air View Post
    This question just involves algebra. Apply the rules provided and simplify.

    f(x+t) = \frac{x+t}{x+t+1}

    f(x) = \frac{x}{x+1}

    \implies \frac{f(x+t) - f(x)}{t} = \frac{\frac{x+t}{x+t+1} - \frac{x}{x+1}}{t}

     \frac{\frac{(x+t)(x+1) - (x)(x+t+1)}{(x+t+1)(x+1)}}{t}

     \frac{(x+t)(x+1) - (x)(x+t+1)}{(x+t+1)(x+1)(t)}

    \frac{t}{(x+t+1)(x+1)(t)}

    \frac{1}{(x+t+1)(x+1)}

    [/tex]
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  5. #5
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    Quote Originally Posted by rafaeli View Post
    how do you get this?
    You are given that f(x) = \frac{x}{x+1} which is a given function and works for all value of x. If you are asked to work out the value of the function when x=2 (for example) when you would insert 2 into the x places hence:

    f(x) = \frac{x}{x+1}
    f(2) = \frac{2}{2+1} = \frac{2}{3}

    Similarly, when you are asked to work out f(x+t), this means you are working out the value when x = (x+t) hence inserting x+t into x places.

    f(x) = \frac{x}{x+1}
    f((x+t)) = \frac{(x+t)}{(x+t)+1} =  \frac{x+t}{x+t+1}
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  6. #6
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    thank you Air for clearing that up for me. you've been a great help.
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