Graphing please

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Graphing please

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1439231312962:dw| Will medal!
|dw:1439231406656:dw|

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

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