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  1. #1
    RRH
    RRH is offline
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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 = ( \frac{3}{4} , \frac{49}{8})

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

    y-int = 5

    domain is (-inf,+inf)

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

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

    Completing the Square

    <br />
f(x) = -2x^2+3x+5<br />

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

    <br />
-2(x^2-\frac{3}{2}x)+5<br />

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

    <br />
-2(x^2-\frac{3}{2}x+\frac{9}{16}-\frac{9}{16})+5<br />

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

    <br />
-2(x^2-\frac{3}{2}x+\frac{9}{16})+5+\frac{9}{8}<br />

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

    <br />
-2(x-\frac{3}{4})^2+\frac{49}{8}<br />

    To determine the y-int take the original function 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 -2(x-\frac{3}{4})^2+\frac{49}{8} and substitute 0 for y

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

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

    Divide both sides by -2

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

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

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

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

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

    <br />
   -2(x^2-\frac{3}{2}x)+5<br />
    well done
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  3. #3
    Super Member
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    Quote Originally Posted by RRH View Post
    ...

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

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

    Vertex = \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 \left( \frac{3}{4} , \frac{49}{8} \right)
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  5. #5
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    While \frac{1}{2} produces: . \frac{1}{2}

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



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

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

    Also use \infty for: . \infty

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