1. ## Absolute Value Graph

Can someone please show me the steps for graphing $\displaystyle f(x)=-\lvert x+3 \rvert +2$

Preferably by splitting it into two equations, one for $\displaystyle x\geq-3$ and one for $\displaystyle x<3$

Though any other method that I can use consistently would also help.

Thanks

2. Would you be able to graph $\displaystyle g(x)=-|x|$?

This graph should be translated along both the axes appropriately to obtain the graph of f(x).

3. Originally Posted by Stroodle
Can someone please show me the steps for graphing $\displaystyle f(x)=-\lvert x+3 \rvert +2$

Preferably by splitting it into two equations, one for $\displaystyle x\geq-3$ and one for $\displaystyle x<3$

Though any other method that I can use consistently would also help.

Thanks

(edit) Sorry. I just realised I put this in the wrong thread...
first rewrite $\displaystyle -\lvert x+3 \rvert$ without using the absolute value symbol you will have

$\displaystyle -\lvert x+3 \rvert=\left\{ \begin{array}{rcl} -x-3 & \mbox{if} & x>-3 \\ x+3 & \mbox{if} & x\leq-3 \end{array}\right.$

according to this

the graph of it

now $\displaystyle f(x)=-\lvert x+3 \rvert+2$

$\displaystyle f(x)=-\lvert x+3 \rvert+2=\left\{ \begin{array}{rcl} -x-3+2 & \mbox{if} & x>-3 \\ x+3+2 & \mbox{if} & x\leq-3 \end{array}\right.$

$\displaystyle \Rightarrow f(x)=-\lvert x+3 \rvert+2=\left\{ \begin{array}{rcl} -x-1 & \mbox{if} & x>-3 \\ x+5 & \mbox{if} & x\leq-3 \end{array}\right.$

the graph of it like the graph above just rise it 2 unit up like this

4. Hello
i think i have what you need
Graphing Absolute-Value Functions

if you want to see whether your graphing is correct or not,use this software :

Or this one :
SourceForge.net: Maxima -- GPL CAS based on DOE-MACSYMA: Files

5. Hello, Stroodle!

Graph: .$\displaystyle f(x)\:=\:-|x+3| +2$
We know what: $\displaystyle f(x) \:=\:|x|$ looks like.
Code:
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\   |   /
\  |  /
\ | /
\|/
- - - - - - + - - - - - -
|
|

Then: .$\displaystyle f(x) \:=\:|x\:{\color{red}+\:3}\:|$ moves the graph 3 units to the left.
Code:
\       / |
\     /  |
\   /   |
\ /    |
- - - * - - + - - - - - -
-3     |
|

Then: .$\displaystyle f(x)\:=\:{\color{red}-}|x+3|$ inverts the graph vertically.
Code:
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-3     |
- - - * - - + - - - - - -
/ \    |
/   \   |
/     \  |
/       \ |
|

Finally, $\displaystyle f(x) \:=\:-|x+3| \:{\color{red}+ 2}$ moves the graph up 2 units.
Code:
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(-3,2)   |
*     |
/ \    |
- - - / - \ - + - - - - - -
/     \  |
/       \ |
|

There!

6. Thanks heaps for your help guys, really appreciate it. Didn't expect such detailed answers.
I've got it now