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Math Help - Finding X intercepts in a parabola

  1. #1
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    Finding X intercepts in a parabola

    Hi all! I found myself needing some desperate help on a homework problem of mine, so I found this site on google, I can already say i can see myself on here frequently lol.

    I have homework that wants me to find the vertex, and x and y intercepts' of a parabola. I can very easily find the vertex and the y intercepts with no problem, but the X intercept is giving me some trouble.

    The problem is f(x)=-3X^2-4X+5

    My professor has tought me to factor the problem and it has worked on a previous problem, but it doesn't work on this one. The back of the book says the X intercepts are -2.12 and .79 but no clue how to get that. Hope you guys can help

    Edit: sorry, I just realized I was probably supposed to post this in the Math Challenging Problems forum, my apologies
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  2. #2
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    The x-intercepts of y = ax^2+bx+c can be found via the formula

     x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}

    In your case a=-3, b=-4 and c= 5

    Have a go!
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  3. #3
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    Ty so much for the reply I got it now.
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  4. #4
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    you could also write it in the form:

    ax^2 + bx + c = p(x+q)^2+r

    this will give the vertex (-q,r), and p will tell you whether its a max or min.

    setting it equal to 0 will solve for x

    the book answers are approximations only
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