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Math Help - find vertex of parabola

  1. #1
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    find vertex of parabola

    find the vertex of the parabola and give its direction of opening.

    x = 1/2y^2 + 3y-6
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    Two ways to find the vertex that I can think of are:

    Putting the eqn into turning pt form

    or

    Finding halfway between the x-intercepts if they exist

    Which way suits you?
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  3. #3
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    im not sure how to solve using those ways. i have to solve to go through by completing the square
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  4. #4
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    Quote Originally Posted by kenzie103109 View Post
    im not sure how to solve using those ways. i have to solve to go through by completing the square
    Completing the square is the same method as turning pt form.

    x = \frac{1}{2}y^2 + 3y-6

    First step is to tkae out half as a common factor

    x = \frac{1}{2}(y^2 + 6y-12)

    Now we need to change the constant term from inside the brackets so we can form a perfect square. To do this we take the co-efficient of the middle term, half it and square it. This gives +9 giving

    x = \frac{1}{2}((y^2 + 6y+9)-9-12)

    x = \frac{1}{2}((y^2 + 6y+9)-21)

    Then making the perfect square complete.

    x = \frac{1}{2}((y+3)^2 -21)

    Can you take it from here?
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