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Math Help - Parabola

  1. #1
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    Parabola

    Hello, could someone show me how I would answer these problems.


    Sketch the graph of the parabola


    y
    = 3x^2 12x


    Circle the vertex, and state both the domain and the range (using
    interval notation).

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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Troy S View Post
    Hello, could someone show me how I would answer these problems.


    Sketch the graph of the parabola


    y
    = 3x^2 12x


    Circle the vertex, and state both the domain and the range (using
    interval notation).

    We see that the coefficient of x^2 is positive, thus we have an upward opening parabola. now when graphing any function, we want to know the x- and y-intercepts, if any, so let's start there.

    y = 3x^2 - 12x

    for x-intercepts, y = 0
    => 3x^2 - 12x = 0
    => 3x(x - 4) = 0
    so 3x = 0 or x - 4 = 0
    => x = 0, x = 4 are the y-intercepts

    for y-intercepts, x = 0
    => y = 0 is the y-intercept

    for a parabola y = ax^2 + bx + c, the x-value of the vertex is given by:
    x = -b/2a
    for our parabola, a = 3, b = -12, c = 0
    so for vertex,
    x = 12/6 = 2
    when x = 2, y = 3(2)^2 - 12(2) = 12 - 24 = -12
    so our vertex is at (2, -12)

    so we have all we need to graph the parabola. remember a parabola has the shape of a cup. ours will open up, cutting the x-axis at x = 0 and x = 4, and cutting the y-axis at y = 0. it's vertex, which in this case will be it's minimum point, will be at (2,-12)

    another way to do this is to complete the square:
    if a parabola is written in the form:
    y = a(x - h)^2 + k
    then the vertex is given by (h,k), the y-intercept is given by y = a(-h)^2 + k, and the x-intercepts are given by x = h +/- sqrt(-k/a)

    so for this problem, if we complete the square we end up with:

    y = 3x^2 - 12x
    => y = 3(x^2 - 4x)
    => y = 3(x^2 - 4x + (-2)^2 - (-2)^2)
    => y = 3[(x - 2)^2 - 4]
    => y = 3(x - 2)^2 - 12

    so the vertex is (2,-12), the y-intercept is, y = 3(-2)^2 - 12 = 0, and the x-intercepts are, x = 2 +/- sqrt(12/3) = 2 +/- 2 = 0 or 4

    exactly what we got before. if you have problems completing the square, say so. i do not know which method you use.

    now for the domain and range.

    the domain is the set of all possible inputs. that is, the set of all x-values for which the function is defined. examining the function y, you will realize there is no restriction on the values of x we can use. and so, as with all polynomials, this function is defined everywhere as far as the domain is concerned. so the domain is given by:
    dom(y) = (-infinity, infinity)

    the range is the set of all possible outputs, that is all the y-values for which the function is defined. of course you see, that no matter what x we plug in, we will always have an output greater than or equal to -12 for y, as that is our minimum point. in other words, for all y-values below -12, our graph does not exist. so the range is given by:
    ran(y) = [-12, infinity). Note that i have a square bracket at -12, since we can be equal to -12

    the graph is below, the vertex has a red dot over it:
    Attached Thumbnails Attached Thumbnails Parabola-parabola.gif  
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