# Absolute Value Graph

• Jun 16th 2009, 01:29 AM
Stroodle
Absolute Value Graph
Can someone please show me the steps for graphing $f(x)=-\lvert x+3 \rvert +2$

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

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

Thanks
• Jun 16th 2009, 01:48 AM
alexmahone
Would you be able to graph $g(x)=-|x|$?

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

• Jun 16th 2009, 04:56 AM
Amer
Quote:

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

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

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 $-\lvert x+3 \rvert$ without using the absolute value symbol you will have

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

according to this
Attachment 11910

the graph of it
Attachment 11911

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

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

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

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

Attachment 11912
• Jun 16th 2009, 04:57 AM
Raoh
Hello(Hi)
i think i have what you need(Nod)
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
• Jun 16th 2009, 05:10 AM
Soroban
Hello, Stroodle!

Quote:

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

                  |               \  |  /               \  |  /                 \ | /                 \|/       - - - - - - + - - - - - -                   |                   |

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

        \      / |         \    /  |           \  /  |           \ /    |       - - - * - - + - - - - - -           -3    |                   |

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

                  |                   |           -3    |       - - - * - - + - - - - - -           / \    |           /  \  |         /    \  |         /      \ |                   |

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

                  |         (-3,2)  |             *    |           / \    |     - - - / - \ - + - - - - - -         /    \  |         /      \ |                   |

There!

• Jun 16th 2009, 08:17 AM
Stroodle
Thanks heaps for your help guys, really appreciate it. Didn't expect such detailed answers.
I've got it now :)