Question

f(x)=(x^2)/(x-9)^2
Find the vertical and horizontal asymptotes.

My problem is I don't know the way to find the vertical and horizontal asymptotes. Is there any algebraic way to find them?

Answer 1

find that x for which y tends to infinty --> vertical asymp.
and that y for which x tends to infinty -->horizontal "

Answer 2

infinity*

Answer 3

What you say is like definition of "vertical asymp" and "horizontal asymp" right.
I am looking for whether there is specific formula to solve it?

Answer 4

\[\frac{x^{2}}{(x-9)^{2}} \]
Vertical asymptote: \((x-9)^2=0\)

Answer 5

so if I want to get vertical asymptote, I just make the denominator equals to zero?

Answer 6

Yes, for the vertical asymptote.

Answer 7

when x = 9 denominator = 0
vertical asymptote at x = 9

Answer 8

how about horizontal one?

Answer 9

find x in terms of y
and see where does x tends to infinty

Answer 10

you mean ,use y to express x?

y=x^2 over (x-9)^2
then y(x-9)^2=x^2
then... x=sth *y ,right?

Answer 11

you can take root of both sides..that'll give you sqrt(y) and -sqrt(y) and it'll help you get x in a much simplified form

Answer 12

sqrt i mean by root**

Answer 13

you mean dy/dx*y ' = [x^2 over (x-9)^2] ' ?

derivative both sides, right???


haha... English is my second language.. thanks for your patience!

Answer 14

You don't differentiate to find the horizontal asympotes :/

Answer 15

i misunderstood ?

Answer 16

i mean taking square root of both sides..
like for eg.
if x^2 = 4
then x=2 or -2
you take sqrt here..was it clearer mam ?

Answer 17

ohhh ! yep

Answer 18

x / x-9 :
as x approaches negative infinity y approaches 0

Answer 19

I found a way to find horizontal asymptotes!!!!

1) Put equation or function in standard form.
2) Remove everything except the biggest exponents of x found in the numerator and denominator.

Answer 20

thanksssssssss you guyyyyyssss <3 <3 <3

Answer 21

hmm..

Answer 22

haha