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Math Help - plz help!! test tomorrow!

  1. #1
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    plz help!! test tomorrow!

    here is what i need to do_

    ''Rewrite the following functions in the form y=a(x-h)^2+k.Find the vertex.''

    here is my problem_

    y = x^2 + 4x - 9

    ----------
    I tried to learn this but i just can't...my teacher is 2 lazy and not willing to spend any extra time with me. Also not a single stundent in my class knows this stuff. so please HELP! This is my last test and if i dont pass it i will have 2 go to summer school or even worse.

    here some more problems_( i know these will be on a test)

    y = x^2 - 6x + 16

    y = x^2 - 8x - 1

    y = x^2 + 5x + 2

    thanks
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  2. #2
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    Quote Originally Posted by alientype
    here is what i need to do_

    ''Rewrite the following functions in the form y=a(x-h)^2+k.Find the vertex.''

    here is my problem_
    y = x^2 + 4x - 9
    ----------
    ...
    here some more problems_( i know these will be on a test)
    y = x^2 - 6x + 16
    y = x^2 - 8x - 1
    y = x^2 + 5x + 2
    thanks
    Hello,

    with all your problems you have to do the following steps:

    If the factor before x^2 is 1 (as it is with your problems) you can calculate a complete square and an additional summand:
    y = x^2 + 4x - 9 = x^2 + 4x +\left({4\over 2}\right)^2-\left({4\over 2}\right)^2-9=(x+2)^2-13
    Coordinates of the Vertex are V(-2,-13).

    1. step: Divide the coefficient of x (not x^2!!) by 2.
    2. step:Square this quotient.
    3. step: Add this square and subtract it at once
    4. step: The first three summands are a complte square.
    5. step: What you've subtracted and the constant together give the y-coordinate of the vertex.

    y = x^2 - 6x + 16 = x^2 - 6x +\left({6\over 2}\right)^2-\left({6\over 2}\right)^2+16=(x-3)^2+7
    Coordinates of the Vertex are V(3,7).

    y = x^2 - 8x - 1 = x^2 - 8x +\left({8\over 2}\right)^2-\left({8\over 2}\right)^2-1=(x-4)^2-17
    Coordinates of the Vertex are V(4,-17).

    y = x^2 + 5x + 2 = x^2 + 5x +\left({5\over 2}\right)^2-\left({5\over 2}\right)^2+2=(x+\frac{5}{2})^2-\frac{15}{4}
    Coordinates of the Vertex are V\left(-\frac{5}{2},-\frac{15}{4}\right).

    Greetings

    EB
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  3. #3
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    Hello, alientype!

    Rewrite the following functions in the form y \:=\:a(x-h)^2+k.
    Find the vertex.

    1)\;y \,= \,x^2 + 4x - 9
    This requires "Completing the Square".
    I hope you're famiiar with it . . .

    We have: . y\;=\;x^2 + 4x - 9

    Take one-half of the x-coefficient and square it:
    . . \frac{1}{2}(4) = 2\quad\Rightarrow\quad 2^2 = 4

    On the right side, add 4 and subtract 4:
    . . . y \;= \;x^2 + 4x + 4  +\, 9 - 4


    And we have: . x + 2)^2 \,+ \,5" alt="y \;\:= \;\x + 2)^2 \,+ \,5" />
    . . . . . . . . . . . . . . . . . . . \searrow\;\;\;\swarrow
    Therefore, the vertex is: . (-2,\,5)


    2)\;y \:= \:x^2 - 6x + 16
    We have: . \frac{1}{2}(-6) = -3\quad\Rightarrow\quad(-3)^2 = 9

    Add and subtract 9: . y \;= \;x^2 - 6x + 9  + 16 - 9

    And we have: . y \;=  \;(x - 3)^2 \,+ \,7
    . . . . . . . . . . . . . . . . . . . \searrow\;\;\,\swarrow
    Therefore, the vertex is: . . . (3,7)


    Think you can do the rest of them now?
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  4. #4
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    thks alot guys i think i can do them now! 20 more probs. 2 go lol!
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  5. #5
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    Quote Originally Posted by Soroban
    ...

    We have: . y\;=\;x^2 + 4x - 9

    Take one-half of the x-coefficient and square it:
    . . \frac{1}{2}(4) = 2\quad\Rightarrow\quad 2^2 = 4

    On the right side, add 4 and subtract 4:
    . . . y \;= \;x^2 + 4x + 4  +\, 9 - 4


    And we have: . x + 2)^2 \,+ \,5" alt="y \;\:= \;\x + 2)^2 \,+ \,5" />
    . . . . . . . . . . . . . . . . . . . \searrow\;\;\;\swarrow
    Therefore, the vertex is: . (-2,\,5)
    ...
    Hello, Soroban,

    I don't want to be a know-it-all (I found this expression in my dictionary and I don't know if it is appropriate here), but in comparison with the problem and your solution I believe that you changed the sign before the 9.

    Greetings

    EB
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