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Math Help - turning point, coordinates

  1. #1
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    turning point, coordinates

    Given that  y = 5x^2 +ax + b has a turning point at (b,a). Where a does not equal 0, find a and b.



    I know that I need to get two equations to solve simultaneously, but not sure what the equations should look like.
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  2. #2
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    y'=10x+a=0
    y'(b)=10b+a=0 (1)

    y=a=5b^2+ab+b (2)
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Tweety View Post
    Given that  y = 5x^2 +ax + b has a turning point at (b,a). Where a does not equal 0, find a and b.



    I know that I need to get two equations to solve simultaneously, but not sure what the equations should look like.
    If you don't like calculus, start by completing the square:

    5\left(x^2+\frac{ax}{5}+~~~\right)+b = 5\left(x^2+\frac{ax}{5}+\frac{a^2}{100}\right)+b-\frac{a^2}{20} = 5\left(x+\frac{a}{10}\right)+\left(b-\frac{a^2}{20}\right)

    So it has a turning point at \left(-\frac{a}{10}, b-\frac{a^2}{20}\right). However, you are given that its turning point is at (b,a)

    So now you have two equations:

    b=-\frac{a}{10}~~~~~(1)
    a=b-\frac{a^2}{20}~~~~~(2)

    Spoiler:
    Subbing into the second equation, we have a=-\frac{a}{10}-\frac{a^2}{20} \implies a = \frac{-2a-a^2}{20} \implies 20a=-2a-a^2 \implies a^2+22a = 0

    Solving this equation gives us a=-22 (it can't be 0 from you initial condition) and it follows from (1) that b=-\frac{11}{5}.

    So your equation is \boxed{5x^2-22x-\frac{11}{5}}
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