SOLUTION: How do i graph y-2=[x+3] and y=[2x-1]? I don't really get the how to graph with the brackets.

Algebra ->  Graphs -> SOLUTION: How do i graph y-2=[x+3] and y=[2x-1]? I don't really get the how to graph with the brackets.      Log On


   

Question 160415This question is from textbook Algebra 2
: How do i graph y-2=[x+3] and y=[2x-1]? I don't really get the how to graph with the brackets. This question is from textbook Algebra 2

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Brackets are just another form of parentheses, but I wonder if you really meant to write "absolute-value" bars?.
If they are just brackets, then...
Graph:
y-2+=+%28x%2B3%29 and
y+=+%282x-1%29
You don't really need the brackets/parentheses here at all!
y-2+=+x%2B3 and
y+=+2x-1 (Green line)
Rewrite the first equation as: y+=+x%2B5 (Red line)
graph%28400%2C400%2C-10%2C10%2C-5%2C15%2Cx%2B5%2C2x-1%29
------------------------------------
If you really meant "absolute-value" bars rather than "brackets", then...
Graph:
1) y-2+=+abs%28x%2B3%29 and
2) y+=+abs%282x-1%29
Remove the absolute-value bars to get:
1) y-2+=+%28x%2B3%29 or y-2+=+-%28x%2B3%29 Write these equation in the slope-intercept form for easier graphing:
1) y+=+x%2B5(Red line) or y+=+-x-1(Green line)
Now we'll graph these first:
graph%28400%2C400%2C-5%2C5%2C-5%2C5%2Cx%2B5%2C-x-1%29
Now equation 2)
2) y+=+abs%282x-1%29 remove the absolute-value bars to get:
y+=+2x-1 (Red line) or y+=+-%282x-1%29 which is y+=+-2x%2B1 (Green line) So let's graph these two:
graph%28400%2C400%2C-5%2C5%2C-5%2C5%2C2x-1%2C-2x%2B1%29