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Does anyone know how to do this? (1 Viewer)

member 6003

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this is basically picking the answer that fits the best, which is why you might be confused.



since then we can say that
satisfies the condition
 

synthesisFR

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this is basically picking the answer that fits the best, which is why you might be confused.



since then we can say that
satisfies the condition
confused tho on why we let it <0 bc one asymptote is like the denominator has one real solution right so idk
 

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confused tho on why we let it <0 bc one asymptote is like the denominator has one real solution right so idk
discriminant < 0 means no x intercepts so that when you do 1 over, there is no vertical asymptotes but there is still one horizontal asymptote. does that clear the confusion?
 

synthesisFR

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discriminant < 0 means no x intercepts so that when you do 1 over, there is no vertical asymptotes but there is still one horizontal asymptote. does that clear the confusion?
wait yes but how do u know theres a horizontal im confused lol
are we meant to just quickly visualise a funciton like that using ext 1 skills or what
 

Luukas.2

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More precisely, has a horizontal asymptote iff at least one of or is non-zero.
But here, both of those limits are zero, and the horizontal asymptote is - as it must be for for where is a concave up parabola.
 

yanujw

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But here, both of those limits are zero, and the horizontal asymptote is - as it must be for for where is a concave up parabola.
approaches positive infinity on both sides (since a > 1 > 0), thus both of the limits are non-zero and has a horizontal asymptote.
 

Luukas.2

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approaches positive infinity on both sides (since a > 1 > 0), thus both of the limits are non-zero and has a horizontal asymptote.
I see what has happened, I misread your post as saying that the limit of had to be non-zero. Whoops!
 

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