## anonymous one year ago Graphing please

1. anonymous

|dw:1439231312962:dw| Will medal!

2. anonymous

|dw:1439231406656:dw|

3. anonymous

@Michele_Laino

4. Michele_Laino

first graph: we have to consider these 2 cases: 1) $x - 2 \geqslant 0$ and: 2)$x - 2 < 0$

5. anonymous

Alright. But for this hw assignmnet, I dont have to show work, i just have to graph it.

6. anonymous

So, is there a website where I can go on to type it in?

7. Michele_Laino

I understand, nevertheless I can not give you the graph, since I have to guide you to your solution

8. anonymous

Ok:)

9. Michele_Laino

in the first case your function is: $\Large \begin{gathered} y = - \frac{1}{3}\left( {x - 2} \right) + 2 = - \frac{x}{3} + \frac{2}{3} + 2 \hfill \\ \hfill \\ y = - \frac{x}{3} + \frac{8}{3} \hfill \\ \end{gathered}$

10. anonymous

Yes:)

11. anonymous

So, then i would just plug in some points to get the answer

12. Michele_Laino

in thesecond case, the equation of your function is: $\Large \begin{gathered} y = - \frac{1}{3} \cdot - \left( {x - 2} \right) + 2 = \frac{x}{3} - \frac{2}{3} + 2 \hfill \\ \hfill \\ y = \frac{x}{3} + \frac{4}{3} \hfill \\ \end{gathered}$

13. 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)

14. anonymous

Oh! Okay, thank you:) I understand it now

15. Michele_Laino

|dw:1439232238778:dw|

16. 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)

17. anonymous

ok and then i can just graph it?

18. anonymous

thanks so much for all ur help!

19. Michele_Laino

|dw:1439232482291:dw|

20. Michele_Laino

similarly for second graph

21. anonymous

Makes sense :)

22. 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

23. anonymous

Kk:) I will do that

24. Michele_Laino

ok! :)