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Thread: turning point, coordinates

  1. #1
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    turning point, coordinates

    Given that $\displaystyle 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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    $\displaystyle y'=10x+a=0$
    $\displaystyle y'(b)=10b+a=0$ (1)

    $\displaystyle y=a=5b^2+ab+b$ (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 $\displaystyle 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:

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

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

    So now you have two equations:

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

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

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

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