Question

For the given function, find the vertical and horizontal asymptote(s) (if there are any).
(2x^2+1)/(x^2-4)

Answer 1

None

x = 2, y = 2y = 0

x = 2, x = -2, y = 2

x = 2, y = 2, y = 1

Answer 2

Horizontal asymptote(s): evaluate\[\lim_{x \rightarrow \pm \infty}\frac{ 2x^2+1 }{ x^2-4 }\]First try guessing: what happens with both nominator and denominator when x becomes very large (neg or pos): they also become very large, so subtracting 4 or adding 1 doesn't make much difference anymore. This means that the nominator will be twice the denominator,so y = 2 is horizontal asymptote.

All this sounds a little sloppy, so there is a more sophisticated way of getting there:
It is a trick thet you can use when you have to calculate limits of fractions for x to infinity (as is the case here).

Divide everything by the highest power of x, which is x²:\[\lim_{x \rightarrow \pm \infty}\frac{ 2x^2+1 }{ x^2-4 }=\lim_{x \rightarrow \pm \infty}\frac{ 2+\frac{ 1 }{ x^2 } }{ 1-\frac{ 4 }{ x^2 } }=\]

Can you see now what this limit is?

Answer 3

oh i get it! thank you so much. i can tell you put a lot of work into this. i appreciate it:)

Answer 4

Don't forget the vertical asymptotes... ;)