## sara1234 Group Title State the horizontal asymptote of the rational function. one year ago one year ago

1. satellite73

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

2. satellite73

for the vertical asymptotes, set the denominator equal to zero and solve for $$x$$

3. sara1234

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

4. satellite73

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?

5. sara1234

yeah i know what degree is

6. sara1234

what if the degree of the numnater is bigger

7. satellite73

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)

8. satellite73

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

9. satellite73

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

10. sara1234

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

11. satellite73

yes, always

12. satellite73

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

13. sara1234

why 5?

14. satellite73

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

15. satellite73

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 $$y=\frac{2}{5}$$

16. sara1234

State the horizontal asymptote of the rational function. 9x^2-3x-8 --------- 4x^2-5x+3 so this would be y=9/4?

17. satellite73

sure would

18. sara1234

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

19. satellite73

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

20. satellite73

k

21. sara1234

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

22. satellite73

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