Question

is there an oblique asymptote here?

f(x)=x^2-6x+4 divided by x+5

Answer 1

asymptotes are not my friends....:(

Answer 2

I would explain this to you but with limits so i don't think is a goog idea lol

Answer 3

understood rock_mit182, math is quite complex

Answer 4

\[f(x)=\frac{ x^2-6x+4 }{ x+5 }\]

Answer 5

yes

Answer 6

write the general form equation first which is y=mx+n

Answer 7

I can do algebra with ease, the precalc is well not wanting to be friends! ;)

Answer 8

got it

Answer 9

then first find m for -infinity

Answer 10

ok

Answer 11

m is by definition:
\[\lim_{x \rightarrow \infty}\frac{ f(x) }{ x }\]

Answer 12

ok

Answer 13

so you've got\[\lim_{x \rightarrow \infty}\frac{ x^2-6x+4 }{ x^2+5x }\]

Answer 14

the coefficients of x^2 are equal so this limits tends to 1
and it's the same for +infinity

Answer 15

so m=1

Answer 16

\[n=\lim_{x \rightarrow \infty}f(x)-mx\]

Answer 17

but in the denominator it is only x+5

Answer 18

pay attention i've multiplied directly with x....
[f(x)/x]

Answer 19

oh

Answer 20

as I've said m=1 now we need to find only n

Answer 21

\[n=\lim_{x \rightarrow \infty}\frac{ x^2-6x+4 }{ x+5 }-x\]

Answer 22

\[n=\lim_{x \rightarrow \infty}\frac{ x^2-6x+4-x^2-5x }{ x+5 }\]

Answer 23

\[n=\lim_{x \rightarrow \infty}\frac{ -11x+4 }{ x+5 }=-11\]

Answer 24

i was reading some notes the teacher gave us on oblique asymptotes and see what you were talking about the m and n

Answer 25

so y=mx+n where m=1 and n=-11 so..
can you tell me the final answer and if you understood my explanation

Answer 26

y=x-11

Answer 27

you are a strict teacher, I like that! ;)

Answer 28

I understood what you were saying...thanks a bunch and a well deserved medal is yours!

Answer 29

Recognize that if the numerator is of one degree higher than that of the denominator, you're going to have a slant asymptote. Using long division, divide the denom. into the num. and you'll immediately have your slant asymptote. Label it: y = x - 11.