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Thread: Graphing helpppp...

  1. #1
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    Graphing helpppp...

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

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

    range: $\displaystyle [0,+ /infty ) $


    2.) $\displaystyle F(x) = |3x + 2|$


    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.)$\displaystyle F(x) = \frac{x^3 - 2x^2}{x-2}$

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

    range: $\displaystyle [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 $\displaystyle x=2$
    as $\displaystyle x-2$ is zero.

    When $\displaystyle x \ne 2$ we may write:

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

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

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

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

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

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

    To sketch this we can rewrite the function as:

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

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

    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:
    $\displaystyle 1)\;f(x) \:= \:\frac{x^3 - 2x^2}{x-2}$

    To repeat what Captain Black said . . .

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

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

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

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



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

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

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

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

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

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

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