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Math Help - y=a(x-p)^2+q

  1. #1
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    y=a(x-p)^2+q

    Not quite sure how to factorise this one...

    I'm looking for the turning point of y=-4x^2+12x-5 so I want y in the form y=a(x-p)^2+q.

    Any input greatly appreciated.

    So far, I tried y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13
    Last edited by davidman; January 21st 2010 at 06:38 AM.
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  2. #2
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    Quote Originally Posted by davidman View Post
    Not quite sure how to factorise this one...

    I'm looking for the turning point of y=-4x^2+12x-5 so I want y in the form y=a(x-p)^2+q.

    Any input greatly appreciated.

    So far, I tried y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13
    hi

    y=-4[x^2-3x]-5

    =-4[(x-\frac{3}{2})^2-(\frac{3}{2})^2]-5

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

    since the leading coefficient is negative , this graph would be a parabola opening downwards , hence the maximum value is 4 , and it occurs when x=3/2
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  3. #3
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    Quote Originally Posted by davidman View Post
    Not quite sure how to factorise this one...

    I'm looking for the turning point of y=-4x^2+12x-5 so I want y in the form y=a(x-p)^2+q.

    Any input greatly appreciated.

    So far, I tried y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13
    hi
    y=-(4x^2-12x+5)=-[(2x-3)^2-9+5]=-(2x-3)^2+4.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by davidman View Post
    Not quite sure how to factorise this one...

    I'm looking for the turning point of y=-4x^2+12x-5 so I want y in the form y=a(x-p)^2+q.

    Any input greatly appreciated.

    So far, I tried y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9-\frac{5}{2}
    Factor out the leading coefficient from the first two terms

    y=-4(x^2-3x)-5

    Now complete the square. This involve adding and subtracting by the same quantity.

    y=-4(x^2-3x+\frac{9}{4}-\frac{9}{4})-5

    Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"

    y=-4(x^2-3x+\frac{9}{4})+9-5

    Now factor the trinomial square, and combine like terms

    y=-4(x-\frac{3}{2})^2+4
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  5. #5
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    Quote Originally Posted by Raoh View Post
    hi
    y=-(4x^2-12x+5)=-[(2x-3)^2-9+5]=-(2x-3)^2+4.
    -(2x-3)^2+4=-(2(x-\frac{3}{2}))^2+4=-4(x-\frac{3}{2})^2+4.
    i didn't read your question well.
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  6. #6
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    [quote=VonNemo19;442494]

    Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"
    y=-4(x^2-3x+\frac{9}{4})+9-5

    [\quote]


    Complete the square and you get \frac {9}{4} so where did the \frac {-9}{4} come from? Any the 9?
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  7. #7
    No one in Particular VonNemo19's Avatar
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    [QUOTE=Masterthief1324;442892]
    Quote Originally Posted by VonNemo19 View Post

    Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"
    y=-4(x^2-3x+\frac{9}{4})+9-5

    [\quote]


    Complete the square and you get \frac {9}{4} so where did the \frac {-9}{4} come from? Any the 9?
    Note that when add a quantity to an expression, you must also subtract the same quantity as to maintain equality. I distributed the -4 to the last term in the parentheses and "kicked it out" of the parentheses.

    Example:

    3(x^2+2x)=3(x^2+2x+1-1)=3(x^2+2x+1)+3(-1)=3(x^2+2x+1)-3=3(x+1)^2-3
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  8. #8
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    [quote=VonNemo19;448426]
    Quote Originally Posted by Masterthief1324 View Post
    Note that when add a quantity to an expression, you must also subtract the same quantity as to maintain equality. I distributed the -4 to the last term in the parentheses and "kicked it out" of the parentheses.

    Example:

    3(x^2+2x)=3(x^2+2x+1-1)=3(x^2+2x+1)+3(-1)=3(x^2+2x+1)-3=3(x+1)^2-3
    I don't quite understand. If it isn't a burden, can you refer me to an article that covers this?
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  9. #9
    No one in Particular VonNemo19's Avatar
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    [QUOTE=Masterthief1324;448941]
    Quote Originally Posted by VonNemo19 View Post

    I don't quite understand. If it isn't a burden, can you refer me to an article that covers this?
    Completing the Square: Solving Quadratic Equations

    Standard and vertex form of the equation of parabola and how it relates to a parabola's graph.
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  10. #10
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    Quote Originally Posted by davidman View Post
    I'm looking for the turning point of y=-4x^2+12x-5 so I want y in the form y=a(x-p)^2+q.
    You can also find the turning-point (of a quadratic equation) without completing the square. It's at \left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right). So for your equation, for example, with a = -4, b = 12, and c = -5, we have:

    \left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right) = \left(\frac{-12}{2(-4)}, -\frac{\left[(12)^2-4(-4)(-5)\right]}{4(-4)}\right) = \left(\frac{6}{4}, -\frac{(144-80)}{-16}\right) = \left(\frac{3}{2}, \frac{64}{16}\right) = \left(\frac{3}{2}, 4\right).
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