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Math Help - quadratic function problem

  1. #1
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    quadratic function problem

    Diagram shows the curve of a quadratic function f(x) = ax^2 + bx + c, where a, b and c are constant.

    a)find the value of a, b and c

    b)Determine the range value of x if f(x)  \geq 1
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  2. #2
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    Quote Originally Posted by mastermin346 View Post
    Diagram shows the curve of a quadratic function f(x) = ax^2 + bx + c, where a, b and c are constant.

    a)find the value of a, b and c

    b)Determine the range value of x if f(x) \geq 1
    From the point (0, -2): -2 = c.

    Update the model: y = ax^2 + bx - 2.

    From the point (-1, -3): -3 = a - b - 2.

    From the turning point (-1, -3): -1 = -b/(2a).

    Solve simultaneously for a and b.

    (b) cannot be done until you have correctly answered (a).
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  3. #3
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    The point at which the curve crosses the 'y' axis is the value of c. This being -2.
    by using the complete the square form, we can work out the rest of the equation.
    To do this, we use the vertex, or lowest point of the graph.
    y=(x+1)^2-3
    this gives us x^2+2x+1-3 or x^2+2x-2
    Thus. The values are ....

    I cannot recall an answer to part b. But I hope this has helped
    Last edited by mr fantastic; October 14th 2010 at 04:20 AM. Reason: Let the OP try and get the answers.
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  4. #4
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    hi!

    check my answer:

    f(x) = ax^2 + bx + c

    =ax^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b^2}{4}\right) +c

    =a\left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4} + c

    =a\left(x + \frac{b}{2}\right)^2 + \left(-\frac{b^2}{4} + c \right)

    x + \frac{b}{2} = 0

    -1 + \frac{b}{2} = 0

    b= 2


    minimum value = -3

    -\frac{b^2}{4} + c = -3

    c=-2

    the value of a is

    f(x) = ax^2 + bx + c

    at point (-1,-3)

    -3 = a + 2(-1) - 2

    -3 = a - 4

    a = 1

    is correct sir?
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  5. #5
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    Seems correct to me.

    You can always double check by putting in x=-1 and seeing if the y co-ordinate is correct.
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  6. #6
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    how about the range?
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  7. #7
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    x^2 + 2x - 2 \geq 1

    x^2 + 2x - 3 \geq 0

    Can you carry on?
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  8. #8
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    so,the range is f(x) \geq 1 and f(x) \leq -3?
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  9. #9
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    Yep.
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