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Math Help - Graphing helpppp...

  1. #1
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    Graphing helpppp...

    Sketch the graph of the function and determine its domain and range:
    1.)  F(x) = \frac{x^3 - 2x^2}{x-2}

    the answer from the back of our book is
    domain: {x|x is not equal to 2}; - i dont know why

    range: [0,+ /infty )


    2.) F(x) = |3x + 2|


    thank you very much
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Sketch the graph of the function and determine its domain and range:
    1.)  F(x) = \frac{x^3 - 2x^2}{x-2}

    the answer from the back of our book is
    domain: {x|x \mbox{ is not equal to }2}; - i dont know why

    range: [0,+ \infty )
    The domain is the set on which the function is defined, as division by
    zero is not allowed the function is not defined when x=2
    as x-2 is zero.

    When x \ne 2 we may write:

     F(x) = \frac{x^3 - 2x^2}{x-2}=\frac{x^2(x-1)}{x-2}=x^2

    which may take any value from 0 and upwards, so the range is: [0, \infty)

    So the graph of the function is identical to they of y=x^2 except that there
    is a hole at x=2

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post

    2.) F(x) = |3x + 2|
    First the domain of this function is all of \mathbb{R} as it is defined for all real numbers.

    As F(x) \ge 0 the smallest value that F can take is 0 and there is clearly no largest value that it can take. So its range is [0,\infty).

    To sketch this we can rewrite the function as:

    <br />
F(x) = \left\{ {3x + 2,\ x \ge -2/3 \atop<br />
-3x-2,\ x < -2/3} \right.<br />

    So the graph is these two line segments with a join at x=-2/3

    RonL
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  4. #4
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    Here is graph for F(x)=|3x+2|.
    Attached Thumbnails Attached Thumbnails Graphing helpppp...-image1.gif  
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  5. #5
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    Hello, ^_^Engineer_Adam^_^!

    Sketch the graph of the function and determine its domain and range:
    1)\;f(x) \:= \:\frac{x^3 - 2x^2}{x-2}

    To repeat what Captain Black said . . .

    We see that x \neq 2.
    . . \boxed{\text{Domain: all } x \neq 2} .or: \boxed{(-\infty,\,2) \cup (2,\,\infty)}

    Since x \neq 2, we have: . f(x) \:=\:\frac{x^2(x - 2)}{x-2} \:=\:x^2

    We have a parabola, y = x^2, with a "hole" at (2,4).
    Code:
                      |
              *       |       *
                      |
               *      |      *
                *     |     o(2,4)
                  *   |   *
            ---------***--------
                      |

    Hence: . \boxed{\text{Range: all }y \geq 0} .or: [0,\,\infty)



    2)\;f(x) \:= \:|3x + 2|

    We know that the graph of y \:=\:|x| is a V with its vertex at the origin.

    The graph of  y \:=\:|3x| is the same V but it "rises faster".

    The graph of y \:=\:|3x + 2| \:=\:\left|3\left(x + \frac{2}{3}\right)\right|
    . .is the same graph moved \frac{2}{3} units to the left.

    The graph looks like this:
    Code:
            \         | /
             \        |/
              \       /
               \     /|
                \   / |
                 \ /  |
          --------*---+------
                -2/3  |

    Therefore: . \begin{array}{cc}\text{Domain} & (-\infty,\,\infty) \\ \text{Range} & [0,\,\infty) \end{array}

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