Question

Asymptote question, writing below :)

Answer 1

I said that x=4 was the vertical asymptote,
& that there was no horizontal asymptote, is that correct?
Also how would I found the oblique (slant) asymptote?

Answer 2

Vertical Asymptote is indeed 4.
No Horizontal Asymptote because the limit is not a real number.
The slant asymptote can be calculated as following (think y-y1=m(x-x1)):

y - f(a) = f'(a)(x-a)

Answer 3

what's a supposed to be?

Answer 4

"a"*

Answer 5

@Mr.Mosterd

Answer 6

did you learn derivatives already?

Answer 7

i don't think so no

Answer 8

Sorry,
If I recall correctly the difference quotient is (f(x)-f(a))/(x-a)
So a would be 4.

I'm trying to solve it but I seem to be doing something wrong :p
I'll recheck

Answer 9

oh ok np :) on my notes it says:
An oblique asymptote, which is a slanted asymptote, is foudn by dividing the numerator of the rational function by its denominatn the quotient may be disregarded

Answer 10

found*

Answer 11

So x+4. Seems right :p
I'm guessing my method is about sth else :p I'll have to recheck

Answer 12

if the degree of the numerator is greater than the degree denominator by 1, then it will have an oblique asymptote, as you said, it is found by dividing numerator and denominator..

vertical asymptote is the zero of the denominator..

there will be a horizontal asymptote if the degree of the denominator is greater than or equal to the degree of the numerator..

Answer 13

yeah i found the vertical & horizontal, so x^2-9/x+4 woudl be what i'd have 2 do 2 get the asymptote of the oblique?

Answer 14

if there's an oblique asymptote, there will be no horizontal asymptote..

Answer 15

for the equation of oblique asymptote i got x-4..

Answer 16

I know there isn't a horizontal asymptote, and oh ok i see

Answer 17

@satellite73