anonymous
  • anonymous
Graphing please
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1439231312962:dw| Will medal!
anonymous
  • anonymous
|dw:1439231406656:dw|
anonymous
  • anonymous
@Michele_Laino

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Michele_Laino
  • Michele_Laino
first graph: we have to consider these 2 cases: 1) \[x - 2 \geqslant 0\] and: 2)\[x - 2 < 0\]
anonymous
  • anonymous
Alright. But for this hw assignmnet, I dont have to show work, i just have to graph it.
anonymous
  • anonymous
So, is there a website where I can go on to type it in?
Michele_Laino
  • Michele_Laino
I understand, nevertheless I can not give you the graph, since I have to guide you to your solution
anonymous
  • anonymous
Ok:)
Michele_Laino
  • Michele_Laino
in the first case your function is: \[\Large \begin{gathered} y = - \frac{1}{3}\left( {x - 2} \right) + 2 = - \frac{x}{3} + \frac{2}{3} + 2 \hfill \\ \hfill \\ y = - \frac{x}{3} + \frac{8}{3} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
Yes:)
anonymous
  • anonymous
So, then i would just plug in some points to get the answer
Michele_Laino
  • Michele_Laino
in thesecond case, the equation of your function is: \[\Large \begin{gathered} y = - \frac{1}{3} \cdot - \left( {x - 2} \right) + 2 = \frac{x}{3} - \frac{2}{3} + 2 \hfill \\ \hfill \\ y = \frac{x}{3} + \frac{4}{3} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
yes! for example, I consider the first function: \[\Large y = - \frac{x}{3} + \frac{8}{3}\] if x=0, then y=8/3, so that line passes at (0,8/3) whereas if y=0, then x=8, so that line passes at point (8,0)
anonymous
  • anonymous
Oh! Okay, thank you:) I understand it now
Michele_Laino
  • Michele_Laino
|dw:1439232238778:dw|
Michele_Laino
  • Michele_Laino
in the second case, I consider this function: \[\Large y = \frac{x}{3} + \frac{4}{3}\] now, if x=0, then y= 4/3, so that line passes at (0, 4/3) whereas if y=0, then x=-4, so that line passes at (4, 0)
anonymous
  • anonymous
ok and then i can just graph it?
anonymous
  • anonymous
thanks so much for all ur help!
Michele_Laino
  • Michele_Laino
|dw:1439232482291:dw|
Michele_Laino
  • Michele_Laino
similarly for second graph
anonymous
  • anonymous
Makes sense :)
Michele_Laino
  • Michele_Laino
for second graph, you have to consider these 2 subsequent cases: \[\Large \begin{gathered} x + 3 \geqslant 0 \hfill \\ x - 3 < 0 \hfill \\ \end{gathered} \] and repeat the procedure above
anonymous
  • anonymous
Kk:) I will do that
Michele_Laino
  • Michele_Laino
ok! :)

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