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

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

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

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

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

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

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

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

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

    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.

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

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

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

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

    $\displaystyle \left(\frac43\right)$
    Last edited by Simplicity; Jun 29th 2008 at 12:12 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.

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

    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.

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

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

    $\displaystyle f(x)=\left(x+\frac{5}{2}\right)^2-\frac{21}{4}$
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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.

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

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

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

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

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

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

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

    [/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 $\displaystyle f(x) = \frac{x}{x+1}$ which is a given function and works for all value of $\displaystyle x$. If you are asked to work out the value of the function when $\displaystyle x=2$ (for example) when you would insert $\displaystyle 2$ into the $\displaystyle x$ places hence:

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

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

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