Question

State the horizontal asymptote of the rational function.

Answer 1

numerator degree is 1, denominator has degree 2 and since 2 is larger than 1 the horizontal asymptote is \(y=0\)

Answer 2

for the vertical asymptotes, set the denominator equal to zero and solve for \(x\)

Answer 3

i dont get how you got y=o pplz help with horizontal

Answer 4

if the degree of the numerator is smaller than the degree of the denominator, you will have a horizontal asymptote at \(y=0\) aka the \(x\)- axis
do you know what i mean by degree?

Answer 5

yeah i know what degree is

Answer 6

what if the degree of the numnater is bigger

Answer 7

ok then that is all you need to consider
think about what would happen if you put in say \(x=10^6\)
you would have a million in the numerator but you would have \(10^{12}\) in the denominator, and get a very small number (close to zero)

Answer 8

if the degree of the numerator is larger, no horizontal asymptote

Answer 9

and if the degrees are the same, it is the ratio of the leading coefficients

Answer 10

so its allways y=0 when the top is smaller and none when the top is larger

Answer 11

yes, always

Answer 12

and if they are the same, for example
\[\frac{2x^2+3x-3}{5x^2+10}\] i would be \(y=\frac{2}{5}\)

Answer 13

why 5?

Answer 14

you got that? three cases
1) top is larger: none
2) bottom is larger :\(y=0\)
3) the degrees are equal: \(y=\text{ratio of leading coefficients}\)

Answer 15

in my example
\[\frac{2x^2+3x-3}{5x^2+10}\] the degrees are both two
the leading coefficient of the numerator is 2
the leading coefficent of the denominator is 5
so horizontal asymptote is \(y=\frac{2}{5}\)

Answer 16

State the horizontal asymptote of the rational function.
9x^2-3x-8
---------
4x^2-5x+3

so this would be y=9/4?

Answer 17

sure would

Answer 18

thanks your the best!! can you fan me so i can message you if i need help plz?

Answer 19

you have to admit this is rather easy, right? i mean once you know what you are doing

Answer 20

k

Answer 21

yeah i posted this because i am home schooled and i have no one to teach me the concepts

Answer 22

that must be tough
guess you can always post here. there are lots of on line resources, but here you may get a direct answer