OpenStudy (rayford):

Identify the horizontal asymptote of f(x)= 4x/7

5 years ago
OpenStudy (cwrw238):

are you sure you have question right?

5 years ago
OpenStudy (rayford):

one moment i'll draw it :)

5 years ago
OpenStudy (rayford):

|dw:1345224680095:dw|

5 years ago
OpenStudy (cwrw238):

this equation represents a straight line through the origin

5 years ago
OpenStudy (rayford):

well here are the choices..

5 years ago
OpenStudy (rayford):

y=4/7 y=0 y=7/4 no horizontal asytope

5 years ago
OpenStudy (rayford):

soo...it's no horizontal?

5 years ago
OpenStudy (cwrw238):

yes

5 years ago
OpenStudy (rayford):

just making sure thank you very much.

5 years ago
OpenStudy (cwrw238):

yw

5 years ago
OpenStudy (rayford):

and thank you cuzzin for last time :)

5 years ago
OpenStudy (anonymous):

Yes, no horizontal. For a horizontal asymptote to occur, either (1) the degree of the numerator is less than that of the denominator (in which case, HA is y=0), or (2) the degrees in the numerator and denominator are equal, in which case you would divinde the leading coefficient in the numerator by the leading coefficient in the denominator. This will give you your HA.

5 years ago
OpenStudy (anonymous):

No prob

5 years ago
OpenStudy (rayford):

you guys are life savers.. thank you so much

5 years ago
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