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Math Help - Need some major help!!! i'm about to go nuts~

  1. #1
    Junior Member mcdanielnc89's Avatar
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    Need some major help!!! i'm about to go nuts~

    I've understood algebraic functions mostly but the vertex of the parabola thing is about to drive me nuts.. here's the example problem...

    For the quadraic function f(x) = 5.71x^2 -3.135x -5.64, find the maximum or minimum value, the x-intercepts, and the y-intercept.
    Last edited by mcdanielnc89; August 13th 2008 at 03:21 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mcdanielnc89 View Post
    I've understood algebraic funtions mostly but the vertex of the parabola thing is about to drive me nuts.. here's the example problem...

    For the quadraic function f(x) = 5.71x^2 -3.135x -5.64, find the maximum or minimum value, the x-intercepts, and the y-intercept.
    First I will go through how we find the vertex of a parabolic equation y=ax^2+bx+c

    We should recall the vertex form of a parabola: y=a(x-h)^2+k

    where (h,k) is the vertex of the parabola.

    We need to rewrite ax^2+bx+c in vertex form.

    Let us first start off by grouping the first two terms together:

    (ax^2+bx)+c

    Now, pull out a common factor a:

    a\left(x^2+\frac{b}{a}x\right)+c

    Now complete the square:

    a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\right)+c-\frac{b^2}{4a}

    Thus, we see that we end up with:

    a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}

    So our vertex would then be \left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)

    Now use this to determine the vertex of f(x) = 5.71x^2 -3.135x -5.64, where a=5.71, b=-3.135, and c=-5.64.

    Does this make sense? If you still have questions, feel free to ask.

    --Chris
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  3. #3
    Junior Member mcdanielnc89's Avatar
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    ok i got this so please tell me where i went wrong if i did... also how do i figure out the minimum value?

    <br />
5.71(x^2 + -3.135/5.71 * x) + -5.64<br />

    <br />
5.71(x^2 + -3.135/5.71 * x + -3.135^2/4 * 5.71^2) + (-5.64 - (-3.135^2/4 * 5.71<br />

    <br />
5.71(x + -3.135/2 * 5.71)^2 + -5.64 - (-3.135^2/4 * 5.71<br />

    Which then I get:
    (-3.135/11.42 , 5.498/28.48)
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  4. #4
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    Hi McDaniel!
    You don't need to convert it to vertex form directly to your given equation. That would be too much trouble, and you can simply do it by plugging in the values for the variables, which is easier. Chris derived it for you. The vertex of a parabola is given by:

     V(x, y) = V\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)

    The function that you gave is in the form of:

    ax^2 + bx + c = 5.71x^2 -3.135x -5.64

    You have a, b, and c. Now simply plug and chug.
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  5. #5
    Junior Member mcdanielnc89's Avatar
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    (8.95 , -5.44) ??
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  6. #6
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    No that's incorrect

    a = 5.71
    b = -3.135
    c = -5.64
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  7. #7
    Junior Member mcdanielnc89's Avatar
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    ok i'm not understanding it then... this time i got..

    (8.95 , -5.43)
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  8. #8
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    -\frac{b}{2a} = -\frac{(-3.135)}{2(5.71)}

    c-\frac{b^2}{4a} = -5.64 - \frac{(-3.135)^2}{4(5.71)}
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  9. #9
    Junior Member mcdanielnc89's Avatar
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    Quote Originally Posted by Chop Suey View Post
    -\frac{b}{2a} = -\frac{(-3.135)}{2(5.71)}

    c-\frac{b^2}{4a} = -5.64 - \frac{(-3.135)^2}{4(5.71)}
    Ok, telkl me this much. tthis is how i'm entering it into the calculator..


    -(3.135)/2(5.71) then the other one isss

    -5.64 - (3.135)^2/4(5.71)



    am i even entering it right?
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  10. #10
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    I wrote it for you. And b is -3.135.
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  11. #11
    Junior Member mcdanielnc89's Avatar
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    Quote Originally Posted by Chop Suey View Post
    I wrote it for you. And b is -3.135.
    I cannot enter it into the calc. like u have showed it.. i have tie TI-84 Plus Silver Edition...


    Thisis how i have it

    -(-3.135)/2(5.71) then the other one isss

    -5.64 - (-3.135)^2/4(5.71)


    which ten I get...

    (8.95, -19.67)
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  12. #12
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mcdanielnc89 View Post
    I cannot enter it into the calc. like u have showed it.. i have tie TI-84 Plus Silver Edition...


    Thisis how i have it

    -(-3.135)/{\color{red}(}2(5.71){\color{red})} then the other one isss

    -5.64 - (-3.135)^2/{\color{red}(}4(5.71){\color{red})}


    which ten I get...

    (8.95, -19.67)
    The way you are inputting it is syntactically incorrect. This is how the calculator is reading what you're inputting:

    -(-3.135)/2(5.71)=-\frac{-3.135}{2}\cdot(5.71) which is not what we want

    -5.64 - (-3.135)^2/4(5.71)=-5.64-\frac{(-3.135)^2}{4}\cdot(5.71) which is not what you want.

    You need the extra set of parenthesis when you plug it into your calculator. This is how your calculator interprets it now...

    -(-3.135)/{\color{red}(}2(5.71){\color{red})}=-\frac{-3.135}{2(5.71)}

    -5.64 - (-3.135)^2/{\color{red}(}4(5.71){\color{red})}=-5.64-\frac{(-3.135)^2}{4(5.71)}

    Does this clarify things?

    --Chris
    Last edited by Chris L T521; August 13th 2008 at 02:50 PM.
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  13. #13
    Junior Member mcdanielnc89's Avatar
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    Quote Originally Posted by Chris L T521 View Post

    -(-3.135)/{\color{red}(}2(5.71){\color{red})}=-\frac{-3.135}{2(5.71)}

    -5.64 - (-3.135)^2/{\color{red}(}4(5.71){\color{red})}=-5.64-\frac{(-3.135)^2}{4(5.51)}

    Does this clarify things?

    --Chris
    I have to find the minimum value, the x-coordinates of the c-intercepts, and the y-intercept...

    I jsut still do not understand how to come up with the correct answers with me now inputting the calculations into the calculator right...
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  14. #14
    Junior Member mcdanielnc89's Avatar
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    Okay, I started over. Let's see if I can do it this time...


    How to find the vertex of a parabolic equation is "y = ax^2 + bx + c".

    As you have stated we should recall the vertex form of a parabola which is "y = a(x - h)^2 + k".

    (h, k) is the vertex of the parabola.

    We need to rewrite "ax^2 + bx + c" in vertex form.

    First we start by grouping the first two terms together: "5.71x^2 + (- 3.135)x^2 + c".

    Now we pull out a common factor a: "5.71 (x^2 + (-3.135 / 5.71x) + c".

    Complete the square: "5.71 (x^2 + (-3.135 / 5.71x + (-3.135^2 / 4(5.71^2) + -5.64 - (-3.135^2) / 4(5.71)".

    Then, we end up with "5.71 (x + (-3.135) / 4(5.71))^2 + -5.64 - (-3.135^2 / 4(5.71)".

    So, our vertex would be "(-(-3.135) / (2(5.71)), -5.64 - (-3.135^2 / (4(5.71))", which would be "(.27, -5.209692526 which is -5.21)".


    Have I gotten this much right?
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  15. #15
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mcdanielnc89 View Post
    Okay, I started over. Let's see if I can do it this time...


    How to find the vertex of a parabolic equation is "y = ax^2 + bx + c".

    As you have stated we should recall the vertex form of a parabola which is "y = a(x - h)^2 + k".

    (h, k) is the vertex of the parabola.

    We need to rewrite "ax^2 + bx + c" in vertex form.

    First we start by grouping the first two terms together: "5.71x^2 + (- 3.135)x^2 + c".

    Now we pull out a common factor a: "5.71 (x^2 + (-3.135 / 5.71x) + c".

    Complete the square: "5.71 (x^2 + (-3.135 / 5.71x + (-3.135^2 / 4(5.71^2) + -5.64 - (-3.135^2) / 4(5.71)".

    Then, we end up with "5.71 (x + (-3.135) / 4(5.71))^2 + -5.64 - (-3.135^2 / 4(5.71)".

    So, our vertex would be "(-(-3.135) / (2(5.71)), -5.64 - (-3.135)^2 / (4(5.71))", which would be "(.27, -5.209692526 which is -5.21)".


    Have I gotten this much right?
    I get (.27,-6.07).

    To get the proper y coordinate, treat it as if we are dealing with -5.64-\frac{3.135^2}{4(5.71)} since (-3.135)^2=3.135^2

    Otherwise, everything else is correct. (I just noticed a couple typos, but you came up with the correct setup at the end)

    Just keep in mind that the y coordinate of the vertex is the maximum or minimum value [depending on the orientation of the parabola].

    --Chris
    Last edited by Chris L T521; August 13th 2008 at 03:40 PM.
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