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Math Help - completing the square

  1. #1
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    completing the square

    If the function ax^2+bx+c has a minimum value -5 when x=-1 and 0 when x=-2 , find the values of a , b and c .

    This is what i did :

    Let f(x)=ax^2+bx+c .. completing the square

    =a[(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}]

    =a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}

    x=-1 when y=-5

    x=-\frac{b}{2a} ... b=2a -- 1

    \frac{4ac-b^2}{4a}=-5 ... 4ac-b^2=-20a -- 2

    Combining gives me ac-a^2=5a --3

    x=-2 when y=0

    -2=-\frac{b}{2a} --- 4a=b ... 4

    4ac-b^2=0 ... 5

    combining , ac-4a^2=0 --- 6

    Now , combining 3 and 6 , 3a^2-5a=0 .. a=0 or a=\frac{5}{3}

    i just couldnt find my mistake . thanks .
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  2. #2
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    line of symmetry is x = -b / 2a = -1, so b = 2a.
    x = -2 is a zero and just 1 left of line of symmetry, so other zero must be 1 right of line of symmetry at x = 0.
    finally, (-2,0) is 1 left but 5 up from min (-1,-5), so vertical stretch is 5 = a.

    so 5x(x + 2) = 5x + 10x, a = 5, b = 10, c = 0
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  3. #3
    MHF Contributor red_dog's Avatar
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    f(-2)=0\Rightarrow 4a-2b+c=0

    (When x=-2 the value of is not minimum)
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  4. #4
    Senior Member DeMath's Avatar
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    f\left( { - 2} \right) = 4a - 2b + c = 0.

    f{\left( { - 1} \right)_{\min }} = a - b + c =  - 5 \Rightarrow f'\left( { - 1} \right) =  - 2a + b = 0.

    Then we have a system of linear equations:

    \left\{ \begin{gathered}<br />
  4a - 2b + c = 0, \hfill \\<br />
  a - b + c =  - 5, \hfill \\<br />
   - 2a + b = 0. \hfill \\ <br />
\end{gathered}  \right.
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  5. #5
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    Good gosh, were you required to solve the problem using complete the square? You can see that the above solutions are much easier. Anyway, I found one error in your steps:

    Quote Originally Posted by thereddevils View Post
    If the function ax^2+bx+c has a minimum value -5 when x=-1 and 0 when x=-2 , find the values of a , b and c .

    This is what i did :

    Let f(x)=ax^2+bx+c .. completing the square

    =a[(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}]

    =a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}

    x=-1 when y=-5

    x=-\frac{b}{2a} ... b=2a -- 1

    \frac{4ac-b^2}{4a}=-5 ... 4ac-b^2=-20a -- 2

    Combining gives me ac-a^2=5a --3
    That should be ac-a^2={\color{red}-}5a. I didn't look further I got tired.


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    Last edited by mr fantastic; July 18th 2009 at 04:02 AM. Reason: Replaced a word
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by yeongil View Post
    Good gosh, were you required to solve the problem using complete the square?
    I offer my solution for testing
    Last edited by mr fantastic; July 18th 2009 at 04:02 AM.
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  7. #7
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    Quote Originally Posted by DeMath View Post
    Quote Originally Posted by yeongil View Post
    Good gosh, were you required to solve the problem using complete the square?
    I offer my solution for testing
    Oops, my comment wasn't directed towards you, DeMath, but to thereddevils.


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    Last edited by mr fantastic; July 18th 2009 at 04:03 AM.
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