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jhonyy9 (jhonyy9):
3rd
OpenStudy (wasiqss):
linear
OpenStudy (hoblos):
a straight line corresponds to a linear function
OpenStudy (anonymous):
whoops fail posted wrong file -____-
OpenStudy (anonymous):
i ment to post this
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jhonyy9 (jhonyy9):
i think just the last one can being
OpenStudy (anonymous):
thats what i was thinking but im kinda effy
OpenStudy (anonymous):
It's definitely the 3rd one.
OpenStudy (anonymous):
how puddle?
jhonyy9 (jhonyy9):
why ?
- so if you check these digrammes so you can see that from this functions just the last one can result negativ roots
- so hence this guarantee that this is the right answer
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jhonyy9 (jhonyy9):
hope that you can understanding those my words
jhonyy9 (jhonyy9):
ok ?
OpenStudy (anonymous):
i can jhony no worrys thanks for the help
jhonyy9 (jhonyy9):
ok
good luck
bye
OpenStudy (anonymous):
Functions in the form1/x+a form asymptotes which is what you have on your graph.
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OpenStudy (anonymous):
If you would like visual proof. Type y=-1/(x+2) into google. They have a new graphing system.
OpenStudy (anonymous):
alright
OpenStudy (anonymous):
If you subbed in -2 into the function to form the graph you'd end up with -1/0, which is undefined. This is what creates the two lines. For example. Take the simple function 1/x. Sub in values for x between -2 and 2. This gives us the points (-2,-1/2), (-1,-1), (1,1) and (2,1/2). Plotting these points gives us: |dw:1333301666652:dw|