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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.
What I'm saying is - keep checking back.
ok thank much for the help
f(X)=(2/3)x-3 if x less than or equal to 3 3x-10 if x is greater than 3
Do you need help graphing this?
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?
no i dont so u want me to graph y=(2/3)x - 3?
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.
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.)
i can individually
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.
so i am suppose to erase the bottom part of the blue where they meet and going back
Correct, DON'T show the blue line left of where the two lines intercect. (Again where using the second graph, not the first).
You'll want to use two different colors or solid and dashed lines.
then what exactly am i doing with the purple line?
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".
ahhhhhhh why lol jk
so i am not supose to do anything to the purple line?
ok ok i thank you guys for all your help!!!!!!!!!!!!......:)
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.
np that is way i am here i am in 8 so if you need help
Sorry - lag in the reply posting.
lol its ok
I found a white board. Go to this URL in another tab. http://www.twiddla.com/491709
i am there
I didn't put in the x or y-axis.
ok i did that if u can see it on the board
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.
So you set?
yes i do...and i thank u alot for ur help...:)
OK - good luck.