That was legit the first thing i did but didnt get the right answer… unless i suck at algebrauh do u just use discriminant
confused tho on why we let it <0 bc one asymptote is like the denominator has one real solution right so idkthis 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
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?confused tho on why we let it <0 bc one asymptote is like the denominator has one real solution right so idk
wait yes but how do u know theres a horizontal im confused loldiscriminant < 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?
i did get the -4a<b<4a inequality but thanks for clearing my doubtthis 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
1 over anything has a horizontal asymptote, but yeah you have to visualise the graphwait 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
pretty much yeah... think of any curve that is completely above the axis lets say x^2+1. when u graph it .... you will see what i meanwait 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
More precisely, has a horizontal asymptote iff at least one of or is non-zero.1 over anything has a horizontal asymptote, but yeah you have to visualise the graph
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.More precisely, has a horizontal asymptote iff at least one of or is non-zero.
approaches positive infinity on both sides (since a > 1 > 0), thus both of the limits are non-zero and has a horizontal asymptote.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.
I see what has happened, I misread your post as saying that the limit of had to be non-zero. Whoops!approaches positive infinity on both sides (since a > 1 > 0), thus both of the limits are non-zero and has a horizontal asymptote.