Question

horizontal and vertical asymptotes

Answer 1

Give an example of a rational function that has a horizontal asymptote at y = 0 and a vertical asymptote at,
x = 2 and x = 1.

Answer 2

how would i even go about doing this?

Answer 3

hint : if numerator degree is less than denominator, then it will have a horizontal asymptote y = 0

Answer 4

hint2 : vertical asymptotes occur when a factor in denominator equals 0

Answer 5

wait what? o.o ive tried (x+1)/(x^2+x+1)
(x+2)/(x^2+2x+2) and ive keep going on with that but i just keep getting a really screwed up like

Answer 6

*line

Answer 7

try this :

\(\large \frac{1}{(x-2)(x-1)}\)

Answer 8

clearly, the denominator equals 0 when x = 2 or 1

Answer 9

so it will have vertical asymptotes at x =2 and x = 1

Answer 10

okay that was magic, but how did you get that equation?

Answer 11

ive just cooked it up from the given info

Answer 12

we need vertical asymptotes at,
x = 2 and x = 1.

Answer 13

.-. how? is there like a formula for it?

Answer 14

so, x = 2 and x = 1 should make the denominator 0

Answer 15

that means we need to have factors (x-2) and (x-1) in the denominator

Answer 16

the simplest denominator is (x-2)(x-1)

Answer 17

so if i needed x=8 then i would do (x-8) in the denominator?

Answer 18

thats right, you got it !!

Answer 19

and 1 goes on top to make it simple? that way its less then the denominator.

Answer 20

correct !

Answer 21

okay thank you c:

Answer 22

below all will have vertical asymptotes at x = 2 and 1, adn a horizontal asumptote y = 0 :

\(\large \frac{1}{(x-2)(x-1)} \)

\(\large \frac{x}{(x-2)(x-1)} \)


\(\large \frac{x^{100}}{(x-2)(x-1)(x^{800 } - 2x^{300} + \pi x^3 - ex^2)} \)

Answer 23

the simplest ofcourse is the first one

Answer 24

yeah i like the first one best XD people should avoid the last one at all costs.... and thank you ^.^

Answer 25

Very clever :)