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osanseviero Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 2x1 }{ x+2 }\]
 9 months ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.0
Horizontal: x+2= 0, x=2...how for the vertcial one?
 9 months ago

rach_ell Group TitleBest ResponseYou've already chosen the best response.0
The vertical asymptote would be x=2 (because adding 2 to 2 would give you a zero in the denominator), and the horizontal asymptote would be 2 (as calculated by dividing 2x/x)
 9 months ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.0
oh...sorry, I did the vertical before...and what would the horizontal be?
 9 months ago

ranga Group TitleBest ResponseYou've already chosen the best response.1
Vertical asymptote is when y goes to infinity. Horizontal asymptote is when x goes to infinity. To find the horizontal asymptote, find the limit of (2x  1) / (x + 2) as goes to infinity.
 9 months ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.0
y=2 :)
 9 months ago

osanseviero Group TitleBest ResponseYou've already chosen the best response.0
So the range is (inf,2)U(2,INF) ?
 9 months ago

ranga Group TitleBest ResponseYou've already chosen the best response.1
Yes the range is correct. Domain is (inf, 2) U (2, inf)
 9 months ago
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