Question

(x^2)/(x^(4)-81) find the horizontal, vertical, and oblique asymptotes

Answer 1

ok, first lets find the vertical asymptotes - do you know how to find these?

Answer 2

no i dont

Answer 3

sorry - ignore that - a vertical asymptote is found by finding a value for x that makes y infinite

Answer 4

that generally means finding a value for x that will make the denominator of a rational polynomial (i.e. a polynomial in terms of a fraction) is to zero

Answer 5

can you see what value of x would make the denominator zero?

Answer 6

i.e. we want to find values for x such that:\[x^4-81=0\]

Answer 7

9?

Answer 8

\(9^2=81\) so this doesn't work

Answer 9

you have a 4'th power here

Answer 10

oh so 3?

Answer 11

almost: \(3^4=81\) yes, but there is also another value - can see what that would be?

Answer 12

hint: \(2^2=4\) and \((-2)^2=4\)

Answer 13

-3

Answer 14

perfect! so now we know the two vertical asymptotes, they are x=3 and x=-3

Answer 15

the horizontal and/or oblique asymptotes are found by looking at what the given polynomial will "tend" to as x tends to plu/minus infinity

Answer 16

so, in this case we have:\[y=\frac{x^2}{x^4-81}\]now think about what the denominator would tend to as x tends to infinity - can you see that the "- 81" term will become more and more insignificant as x tends to infinity?

Answer 17

yeah

Answer 18

good, so we can express this as:\[\lim_{x\rightarrow\infty}\frac{x^2}{x^4-81}=\frac{x^2}{x^4}=\frac{1}{x^2}\]does that make sense?

Answer 19

ya kinda

Answer 20

ok - let me know if you need more clarification.

now also notice that since we have even powers in the numerator AND the denominator, the the limit will be the same for x tending to minus infinity - agreed?

Answer 21

*then the

Answer 22

ok

Answer 23

good: we can express this as follows:\[\lim_{x\rightarrow\pm\infty}\frac{x^2}{x^4-81}=\frac{x^2}{x^4}=\frac{1}{x^2}\]

Answer 24

now, what do you think the value of \(\displaystyle\frac{1}{x^2}\) will be as x tends to plus/minus infinity?

Answer 25

imagine x getting bigger and bigger - what will happen to the value of \(\displaystyle\frac{1}{x^2}\) in this case?

Answer 26

i have no idea.

Answer 27

I'm sry i am so horrible at math

Answer 28

np - let me try and explain it....

Answer 29

lets take different values of x:

x=10 ---> 1/x^2 = 1/100 = 0.01
x=100 ---> 1/x^2 = 1/10000 = 0.0001
...

can you see the value of 1/x^2 gets smaller and smaller as x gets bigger and bigger?

Answer 30

yes

Answer 31

good, so eventually, when x is plus/minus infinity, 1/x^2 will become zero

Answer 32

so we have shown that as x tends to plus/minus infinity, the value of \(\displaystyle\frac{x^2}{x^4-81}\) tends to zero.

this means we have a horizontal asymptote of y=0.

Answer 33

you can see these asymptotes here: http://www.wolframalpha.com/input/?i=asymptotes+%28x^2%29%2F%28x^%284%29-81%29

Answer 34

let me know if you are still unsure of anything here.

Answer 35

cool thank you so much!

Answer 36

yw :)