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Math Help - Questions about y = ax˛+bx+c.

  1. #1
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    Questions about y = ax˛+bx+c.

    f(x) : ax˛+bx+c, a≠ 0

    [have to find the a(x-h)˛+k form first]

    Find an expression in terms of a, b, c for: (without using graph or calculator)
    (i) An equation of the axis of symmetry
    (ii) The maximum or minimum value
    (iii) The coordinated of the vertex
    (iv) The domain and the range
    (v) The y-intercept of the graph of the function
    (vi) The zeroes of the function

    Discuss how you can predict the number of zeroes for a given quadratic function of the form y= ax˛+bx+c, a≠ 0. Support the validity of your prediction with some examples.



    My attempt
    (ax˛ + bx + b/2 - b/2) + c
    (ax˛ + bx + b/2) + c - b/2


    (v) y intercept is (0,c)
    Last edited by mr fantastic; May 11th 2010 at 06:55 PM.
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  2. #2
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    Quote Originally Posted by Divina View Post
    f(x) : ax˛+bx+c, a≠ 0

    [have to find the a(x-h)˛+k form first]
    find a(x-h)^2+kform first....

    ax^2+b x+c=0 (divided by a)
    x^2+\frac{b}{a} x+ \frac {c}{a}=0(substract by \frac {c}{a})

    x^2+\frac{b}{a} x =-\frac {c}{a} (add by (\frac {b}{2a})^2)

    (x+\frac{b}{2a})^2=-\frac {c}{a}+(\frac {b}{2a})^2

    (x+\frac{b}{2a})^2+(\frac {c}{a}-(\frac {b}{2a})^2)=0

    so,
    h=-\frac{b}{2a}

    k=(\frac {c}{a}-(\frac {b}{2a})^2)
    or k=-\frac{b^2-4ac}{4a^2}

    -----I hope it'll help U-----
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  3. #3
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    Quote Originally Posted by pencil09 View Post
    find a(x-h)^2+kform first....

    ax^2+b x+c=0 (divided by a)
    x^2+\frac{b}{a} x+ \frac {c}{a}=0(substract by \frac {c}{a})

    x^2+\frac{b}{a} x =-\frac {c}{a} (add by (\frac {b}{2a})^2)

    (x+\frac{b}{2a})^2=-\frac {c}{a}+(\frac {b}{2a})^2

    (x+\frac{b}{2a})^2+(\frac {c}{a}-(\frac {b}{2a})^2)=0

    so,
    h=-\frac{b}{2a}

    k=(\frac {c}{a}-(\frac {b}{2a})^2)
    or k=-\frac{b^2-4ac}{4a^2}

    -----I hope it'll help U-----
    Thank you very much, that helped a lot
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