Question

How do you find vertical asymptotes and horizontal asymptotes

Answer 1

the vertical asymptotes are the restrictions on x
basically set the denominator = to zero
the horizontal asymptotes is the coeffiecint of the highest degree variable in the numerator and denominator over each other.
\[\frac{ 2x-5 }{ x-2 }\]
the vert is 2 and the horizontal is 2/1
there are some stipulations on the horizontal though.
where \[\frac{ P(x) }{ Q(x) }\]
if degree of p > degree of q there is no horiz asymptote
if degree of p = degree of q the asymptote is as above.
if degree of p < degree of q the asymptote is 0

Answer 2

okay so the vertical asymptote is 2 because 2-2=0

Answer 3

yes. because that would make the function undefined if the denominator is =0

Answer 4

SO the vertical asymptote to this problem would be -1

Answer 5

yes, evaluate the degrees of both to determine the horizontal asymptote

Answer 6

if you need me to clarify the horizontal asymptotes I can type it out a little better.

Answer 7

3/1? I still don't understand that one very well

Answer 8

alright allow me to type it out better. it will take a second

Answer 9

I am supposed to use limits to be able verify my answers are correct

Answer 10

\[\frac{ P(x) }{ Q(x) }\]
evaluate the degrees of each where
\[\frac{ P(x) }{ Q(x) }>\frac{ P(x) }{ Q(x)}\]
there is no horizontal asymptote
and where
\[\frac{ P(x) }{ Q(x) }=\frac{ P(x) }{ Q(x) }\]
the asymptote is the highest coefficients of
\[\frac{ P(x) }{ Q(x) }\] is the horizontal asymptote
and where
\[\frac{ P(x) }{ Q(x) }<\frac{ P(x) }{ Q(x) }\]
the horizontal asymptote is 0

Answer 11

now evaluate the degree and think about the horizontal asymptote is again. crap I messed that up. lol

Answer 12

lol :)

Answer 13

\[P(x)=Q(x)\] assume all of those fractions with equations and inequalities say this

Answer 14

alright if your using limits. graph the function and find where y goes as x approaches from the left or right of your vertical asymptotes

Answer 15

does that make sense?

Answer 16

yes I think so.

Answer 17

so with an asymptote we will go to infinity either direction depending on. our case

Answer 18

Thank you for your help.

Answer 19

actually there is something special about this problem. you can factor the numerator and keep the conditions of the first statement. the graph will be a parabola with a discontinuity at -1

Answer 20

I may have misled you a bit. if you couldn't factor the numerator all of the above would be true factor the sum of 2 cubes on the numerator

Answer 21

you will get (x+1)(x^2-x+1) and the graph will follow the quadratic statement. keeping the restrictions on x from the first problem