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Math Help - Quadratics (again)

  1. #1
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    Quadratics (again)

    In the equation y=3t^2 - 7x + 4

    I got the x-intercept points as -8 and -8.33

    But then got the turning point co-ordinates as 1.17, -0.08
    using the x=-b/2a formula then substituting into the quadratic equation.

    This doesn't seem like the correct answer, could someone please let me know what I have done wrong?
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  2. #2
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    Wait.. I think I know where I did a wrong calculation... are the x-intercept points 1.33 and 1?
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  3. #3
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    Quote Originally Posted by J4553 View Post
    In the equation y=3t^2 - 7x + 4

    I got the x-intercept points as -8 and -8.33

    I must assume you didn't mean for there to be a variable other than x in your equation.

    y=3x^2-7x+4

    (3x-4)(x-1)=0

    x=\frac{4}{3} \ \ or \ \ x=1

    Quote Originally Posted by J4553 View Post
    But then got the turning point co-ordinates as 1.17, -0.08
    using the x=-b/2a formula then substituting into the quadratic equation.

    This doesn't seem like the correct answer, could someone please let me know what I have done wrong?
    Unless you are told to do so, I believe I would leave the vertex as fractions rather than rounding them.

    V(7/6, -1/12)
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