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Mathematics 20 Online
OpenStudy (anonymous):

file

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

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

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.

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|

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