Question

Find the oblique asymptote for the graph of y=x^3/x^(2)-x, i got x+1 am i right ??

Answer 1

what was your work to get that answer?

Answer 2

i used long divison

Answer 3

so you got... x + 1 + r(x)/(x^2 - x) from long division?
and then selected y = x+1 as your oblique asymptote

Answer 4

yes and my resulatant was x+0

Answer 5

right. so by definition, as x approaches larger and larger numbers, our remainder is x/(x^2 - x) will simply disappear (denominator >> numerator, 1/big number --> 0). I think your answer looks good. :)

Answer 6

it helps to imagine the behavior for very very large numbers:
y = x + 1 + 1/x-1
anything with larger degree in denominator than numerator approaches 0 for the same reasoning as above
then y = x + 1 is the end behavior.

Answer 7

what does this mean ?? What combinations of vertical, horizontal, and oblique asymptotes are not possible for a
rational function?

Answer 8

what are the conditions for each type of asymptote?

Answer 9

what do u mean ? like ?

Answer 10

For a vertical asymptote, you need a factor in the denominator of x that allows you to get 0
like f(x) = (x + 1)/(x - 1) x=1 is a vertical asymptote

Answer 11

what would tell you that you have a horizontal asympttote? or oblique asymptote?

Answer 12

and most relevant to your question.. can you have both based on what is required?

Answer 13

confused :?

Answer 14

wait i looked at my notes here what i says "the degree of the numerator P(x) must be one more than the degree of the denominator "

Answer 15

that sounds good for oblique asymptotes. what about horizontal? :)

Answer 16

not sure about that

Answer 17

my best hint is that it also has to do with degrees of the numerator / denominator.

Answer 18

like, would y = 1/(x - 1) have a horizontal asymptote? y = x/(x - 1)? y = x^2 / (x - 1) ?

Answer 19

not really

Answer 20

again, for big values of x: y = 1/(x-1) goes to y = 1/biiiig number, so it goes to 0. thus, horizontal asymptote is at y=0.

for y = x/ (x-1), you have an equally large number for numerator and denominator (the 1 is unimportant because it is so comparatively small). horizontal asymptote is at y = 1/1 = 1

but y = x^2 / (x-1) is the same problem we just had (removing the x in numerator/denominator). it had an oblique asymptote, right?

Answer 21

yea

Answer 22

the first two are degree(numerator) <= degree(denominator), and they had horizontal asymptotes.
the last was degree(numerator) > degree(denominator), so that we had an oblique asymptote.

Answer 23

so back to this question:
What combinations of vertical, horizontal, and oblique asymptotes are not possible for a
rational function?
can the degree of a numerator be both greater than - and less than/equal to the degree of the denominator to give us both horizontal and oblique asymptotes?

Answer 24

its can't be less than but more than for sure

Answer 25

In a slightly more math-y perspective:
\( \deg (\text{numerator} ) \le \deg ( \text{denominator} ) \) => Horizontal Asymptote
\( \deg ( \text{numerator) } > \deg ( \text{denominator} ) \) ==> Oblique Asymptote
If we know both of these values, can both statements be true simultaneously? (a yes/no question)
it is more or less like asking: can you look left /and/ right at the same time?
Hope this makes more sense like this.

Answer 26

or we just call the degree of numerator N, and degree of denominator D:
\( D \le N \) and \( D > N \)
these simply have nothing in common, and i invite you to try any two numbers (N,D) to see if you can make both true.