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Math Help - help on using the mid value to determine the vertex of a quadratic function?

  1. #1
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    help on using the mid value to determine the vertex of a quadratic function?

    I was giving an example that completely lost me.

    The quadratic function is: 2x^2 - 5x - 12

    Let y = 0, factor, and solve.

    0= (2x + 3)(x - 4)

    x = -3/2 or x = 4

    The mid value is:

    x = -3/2 + 4
    ________
    2

    = -3/2 + 8/2
    _________
    2

    = 5/2/2

    = 5/2 x 1/2

    = 5/4

    Im completely confused at the step where the 4 became 8/2, and the steps beyond, any help?
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  2. #2
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    Quote Originally Posted by danzig View Post
    Im completely confused at the step where the 4 became 8/2, and the steps beyond, any help?
    4 = 4/1 = 8/2 this is a very basic idea. This was done so the fractions could be added together. When adding fractions you need a common demoninator.

    Any of what I have said make sense?

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  3. #3
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    Well yes I understand how 4 = 8/2.. I'm just bad with fractions .. The part that confuses me is after the fractions are added together to get 5/2/2.. How did that become 5/2 x 1/2? Any explanation?
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    \displaystyle \frac{a}{b}\div \frac{c}{d} = \frac{a}{b}\times  \frac{d}{c}

    You follow?
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  5. #5
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    Quote Originally Posted by danzig View Post
    Well yes I understand how 4 = 8/2.. I'm just bad with fractions .. The part that confuses me is after the fractions are added together to get 5/2/2.. How did that become 5/2 x 1/2? Any explanation?
    You have -3/2 + 8/2 = 5/2 and that gets divided by 2.

    I suggest you extensively review fractions in order to get much better with them - if you are to have any hope of understanding what you will be required to do in your mathematical studies. It is simply not good enough to say "I'm just bad with fractions".
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  6. #6
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    I suppose, I'll take that in mind. Was just looking for a quicker explanation as I am in grade 11 functions.. and exams are coming up. I'll review fractions.. :/
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