Question

can you explain how to find the end behavior asymptote of 1/x^2+1

Answer 1

the wh0?

Answer 2

what do you mean?

Answer 3

the worksheet says to find the end B.A. how?

Answer 4

B.A?

Answer 5

ohh the ahemm, heheh

Answer 6

behavior asymptote

Answer 7

I guess you can always just graph it

Answer 8

so im assuming its y=0?

Answer 9

well, it has a horizontal asymptote and vertical

Answer 10

the numerator's degree is less than the denominator's, thus the horizontal will be at y=0, yes

Answer 11

hmm well.... wait. a sec... x^2+1 has no real zeros, so,I guess no vertical asympotes

Answer 12

ok...ive lost all this info since precal...now im taking cal and im forgetting simple things lol

Answer 13

whats the rule for the exponents being equal

Answer 14

If degree of num'r = degree of den'r, then the horizontal asymptote is the line

y = p/q

where p is the leading coefficient of the num'r, q is the leading coefficient of the den'r.

Answer 15

\(\bf \cfrac{an^\square}{bd^\square}\implies \textit{horizontal asymptote} = \cfrac{a}{b}\)

Answer 16

E.g., in a function like \(y=\dfrac{3x-1}{2x+7}\) the HA is y=3/2. the leading terms "take over" everything else for large values of x, and the x^n cancel out.

Answer 17

ok..thank you

Answer 18

You're welcome, and welcome to Open Study! :)

Answer 19

seems like a great site...so bigger exp for numerator means no ha?

Answer 20

right, if the numerator's degree is bigger, then no HA
if it's bigger by "1", then it'd be an "oblique" asymptote

Answer 21

ok...so can you explain how to find vert asymptotes?

Answer 22

ohh, at the zeros of the denominator

Answer 23

the zeros of the denominator so long they don't make the numerator zero

Answer 24

ok