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Math Help - college pretest

  1. #1
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    college pretest

    for the graph of the quadratic function, identify the direction of the opening and the coordinates of the vertex. f(x)=(x+2)raised to the second -3

    A: downward 2,3
    B: upward -2,-3
    C: downward -2,-3
    D: upward 2,3


    identify the equation of the a quadractic function with the vertex (0,1/4) that passes through the point (-1,17/4) and opens upward

    A: f(x)= 4x raised to the second + 1/4
    B: f(x)= -4x raissed to the second - 1/4
    C: f(x)= 4x raised to the second - 1/4
    D: f(x)= -4x raised to the second - 1/2
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  2. #2
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    First question

    So you have f(x)=(x+2)^2-3

    Do you know about derivatives? Taking the derivative will help.
    First expand:

    f(x) = x^2 + 4x + 1

    Take the first derivative:

    f'(x) = 2x + 4

    This is a new function describing the original function's slope given an x. The vertex of the parabola will have a slope of zero, so set 0 = 2x + 4 and solve for x. This will yield the x-value of the vertex, and you can plug it back into the original function to find its corresponding y.

    Take the dervative one more time:

    f''(x) = 2

    This new function gives you the the rate with which the slope of the original function changes. It's useful to determine concavity in a situation like this. As a rule of thumb, if the function is a positive constant, then the parabola opens upward. If it's negative, then it opens downward. In this case, it's positive.
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  3. #3
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    Quote Originally Posted by Brooke
    for the graph of the quadratic function, identify the direction of the opening and the coordinates of the vertex. f(x)=(x+2)raised to the second -3

    A: downward 2,3
    B: upward -2,-3
    C: downward -2,-3
    D: upward 2,3
    f(x) = (x+2)^2 - 3

    This is of the (slightly more) general form: y = a(x-h)^2+k

    Since a = +1 is positive so we know that the parabola opens upward. The point (h, k) is the vertex, so upon comparing we get (h, k) = (-2, -3).

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Brooke
    identify the equation of the a quadractic function with the vertex (0,1/4) that passes through the point (-1,17/4) and opens upward

    A: f(x)= 4x raised to the second + 1/4
    B: f(x)= -4x raissed to the second - 1/4
    C: f(x)= 4x raised to the second - 1/4
    D: f(x)= -4x raised to the second - 1/2
    Frankly, I'd do this one by the old "plug 'n' chug" method. (You've got two points on the parabola and 4 equations. Only one of them can be the answer so see if those points fit the parabolas listed in the answer.)

    More analytically:
    f(x) = a(x-h)^2 + k

    You know that the vertex (h, k) = (0, 1/4) so
    f(x) = a(x-0)^2 + 1/4 = ax^2+1/4

    Since the other point (-1, 17/4) lies ABOVE this point (the vertex), the parabola must open upward. Thus a is positive.

    The only possibility open then, is answer A.

    -Dan
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  5. #5
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    Ahh right, I forgot about f(x)=a(x-h)^2+k!
    That is much easier than my method.
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