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  1. #1
    RRH
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    test

    Only took me 45 minutes to type this up .... not setting any records here :P







    Please try using Stapel sources before using my results. I am trying to practice using latex and it took me approx 45 minutes to type this up (yes I am slow) and I do not want to lose my work; therefore, I am going to post it, but you should try to do the work yourself first.

    Vertex = ($\displaystyle \frac{3}{4} , \frac{49}{8}$)

    x-int = -1 and $\displaystyle \frac{5}{2}$

    y-int = 5

    domain is (-inf,+inf)

    range is (-inf, $\displaystyle \frac{49}{8}$]

    See explanation below to determine how I came to these results.

    Completing the Square

    $\displaystyle
    f(x) = -2x^2+3x+5
    $


    When completing the square the value of "a" should not be any number other than 1; therefore, factor out the -2

    $\displaystyle
    -2(x^2-\frac{3}{2}x)+5
    $

    Now divide $\displaystyle -\frac{3}{2}$ by 2 and square the result. The equation should now look like this.

    $\displaystyle
    -2(x^2-\frac{3}{2}x+\frac{9}{16}-\frac{9}{16})+5
    $

    To move the $\displaystyle -\frac {9}{16} $out of the parenthesis you have to multiply it by -2

    $\displaystyle
    -2(x^2-\frac{3}{2}x+\frac{9}{16})+5+\frac{9}{8}
    $

    Factor within the parenthesis and clean up outside the parenthesis. The vector format of your function should look like below

    $\displaystyle
    -2(x-\frac{3}{4})^2+\frac{49}{8}
    $

    To determine the y-int take the original function $\displaystyle f(x) = -2x^2+3x+5$ substitute 0 for x and you should get the y-int of 5

    To get the x-int take the vertex format of the function $\displaystyle -2(x-\frac{3}{4})^2+\frac{49}{8}$ and substitute 0 for y

    $\displaystyle -2(x-\frac{3}{4})^2+\frac{49}{8}=0$

    $\displaystyle -2(x-\frac{3}{4})^2=-\frac{49}{8}$

    Divide both sides by -2

    $\displaystyle (x-\frac{3}{4})^2=\frac{49}{16}$

    $\displaystyle \sqrt(x-\frac{3}{4})^2=\sqrt\frac{49}{16}$

    $\displaystyle x-\frac{3}{4}=\pm\frac{7}{4}$

    $\displaystyle x=\frac{3}{4}\pm\frac{7}{4}$

    the x-int = -1 , $\displaystyle \frac{5}{2}$
    Last edited by RRH; Oct 15th 2009 at 03:16 PM.
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  2. #2
    -1
    e^(i*pi)'s Avatar
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    Quote Originally Posted by RRH View Post
    $\displaystyle
    f(x) = -2x^2+3x+5
    $


    $\displaystyle
    -2(x^2-\frac{3}{2}x)+5
    $
    well done
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  3. #3
    Super Member
    earboth's Avatar
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    Quote Originally Posted by RRH View Post
    ...

    Vertex = ($\displaystyle \frac{3}{4} , \frac{49}{8} $)

    ...
    I've modified the quoted line a little bit:

    Vertex = $\displaystyle \left( \frac{3}{4} , \frac{49}{8} \right)$

    Click on the formula above to see the code.
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  4. #4
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    [tex]\left( \frac{3}{4} , \frac{49}{8} \right)[/tex] gives $\displaystyle \left( \frac{3}{4} , \frac{49}{8} \right)$
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  5. #5
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    While \frac{1}{2} produces: .$\displaystyle \frac{1}{2}$

    . . we can make smaller fractions with \tfrac{1}{2}: .$\displaystyle \tfrac{1}{2}$



    \frac{\frac{x}{2}}{\frac{y}{3}} produces: .$\displaystyle \frac{\frac{x}{2}}{\frac{y}{3}}$

    But we can make larger fractions with \dfrac:
    . . \frac{\dfrac{x}{2}}{\dfrac{y}{3}} produces: .$\displaystyle \frac{\dfrac{x}{2}}{\dfrac{y}{3}}$

    Also use \infty for: .$\displaystyle \infty$

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