1. ## 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

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

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

4. Here is graph for F(x)=|3x+2|.

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