anonymous
  • anonymous
does anyone know how to do piecewise functions ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Go ahead and post a problem. Eventually someone with the talent will probably be able to walk you through the solution. You may have to be patient though, many of the East Coast/UK/Indian posters will have logged off by now.
anonymous
  • anonymous
What I'm saying is - keep checking back.
anonymous
  • anonymous
ok thank much for the help

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anonymous
  • anonymous
f(X)=(2/3)x-3 if x less than or equal to 3 3x-10 if x is greater than 3
anonymous
  • anonymous
Do you need help graphing this?
anonymous
  • anonymous
yes
anonymous
  • anonymous
OK, it appears that you'll be graphing 2 rays. I will follow the slope-intercept equation y=(2/3)x - 3. But you'll only graph the part of the line that starts (above or below) 3 on the x-axis. This ray will proceed to the left, it needs a SOLID end-point. Do you follow?
anonymous
  • anonymous
no i dont so u want me to graph y=(2/3)x - 3?
anonymous
  • anonymous
The other ray (not a full line) will follow the line y=3x-10, with a OPEN endpoint above or below the 3 on the x-axis - and proceeds to the right.
anonymous
  • anonymous
Sorry, finishing a comment. Yes can you graph y=(2/3)x - 3 and y=3x-10 if you had too. (otherwise I'll go over point - intercept graphing with you.)
anonymous
  • anonymous
i can individually
anonymous
  • anonymous
Great. Look at the second graph. http://www.wolframalpha.com/input/?i=y%3D3x-10+and+y%3D%282%2F3%29x-3 Now your going to graph both lines but erase 1/2 of each line where the two lines cross. KEEP the blue line going to the right (for y=3x-10 if x>3) and only KEEP the part of the purple line that goes to the left ( for the x<= 3 condition.
anonymous
  • anonymous
so i am suppose to erase the bottom part of the blue where they meet and going back
anonymous
  • anonymous
ahhh
anonymous
  • anonymous
Correct, DON'T show the blue line left of where the two lines intercect. (Again where using the second graph, not the first).
anonymous
  • anonymous
You'll want to use two different colors or solid and dashed lines.
anonymous
  • anonymous
then what exactly am i doing with the purple line?
anonymous
  • anonymous
The purple line will be the rest of your graph left of the (2,-1) intersection point. The net effect is the line looks "bent" like a shallow "v".
anonymous
  • anonymous
ahhhhhhh why lol jk
anonymous
  • anonymous
???
anonymous
  • anonymous
so i am not supose to do anything to the purple line?
anonymous
  • anonymous
no
anonymous
  • anonymous
ok ok i thank you guys for all your help!!!!!!!!!!!!......:)
anonymous
  • anonymous
Right - if you started out by graphing both lines - you would _erase_ the purple line to the RIGHT of (2,-1) --> but keep the purple line LEFT of (2,-1). Do the same but in reverse for the blue line. So looking at the graph - coming from the right. ONLY blue ray, SOLID purple dot at (2,-1) then ONLY purple ray to negative infinity.
anonymous
  • anonymous
np that is way i am here i am in 8 so if you need help
anonymous
  • anonymous
Sorry - lag in the reply posting.
anonymous
  • anonymous
lol its ok
anonymous
  • anonymous
I found a white board. Go to this URL in another tab. http://www.twiddla.com/491709
anonymous
  • anonymous
i am there
anonymous
  • anonymous
I didn't put in the x or y-axis.
anonymous
  • anonymous
ok i did that if u can see it on the board
anonymous
  • anonymous
Do you think you got it now the function follows the blue line for all x inputs above 3 and the purple line for inputs 3 or less. The out-put "steps" over to another rule (function) at x=3.
anonymous
  • anonymous
So you set?
anonymous
  • anonymous
yes i do...and i thank u alot for ur help...:)
anonymous
  • anonymous
OK - good luck.

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