## babydoll332 2 years ago State the horizontal asymptote of the rational function. f(x) = x^2+6x-8/x-8

1. L.T.

Can you find the derivative of this function first?

2. L.T.

Set the derivative equal to zero

3. babydoll332

i normally pluged it into the graph

4. timo86m
5. L.T.

All right, I'll assume you can't do that now. Is this the equation:$x ^{2}+\frac{ 6x-8 }{ x-8 }$

6. babydoll332

i see what u posted @timo86m but idk what the horrizontal is

7. babydoll332

@L.T. ok what would i do

8. Outkast3r09

there is a slant retricemptote.. not a horizontal =/

9. Outkast3r09

lol asymptote... lame chat =]

10. satellite73

the degree of the top is greater than the degree of the bottom therefore no horizontal asymptote no calc needed, that is all

11. L.T.

If what I wrote was the equation, then you find the derivative of each term in the equation, using the quotient rule on the second term.

12. babydoll332

reall no horizontal ?

13. satellite73

if the degree of the bottom was greater, then the horizontal asymptote would be $$y=0$$ if the degree of the top is greater no horizontal asymptote

14. Outkast3r09

yes there is a slant use long division

15. Outkast3r09

if you ned to find the slant asymptote

16. satellite73

if the degrees are the same, it is the ration of the leading coefficients that is it

17. babydoll332

so the anserw is no horizontal asymptote

18. satellite73

that is correct. there is none

19. timo86m

it has a slanted asymptote with a slope of 1045/1058

20. babydoll332

ok thank you so much

21. Outkast3r09

how'd you find that @timo86m

22. timo86m

take derivitive of (x^2+6x-8)/(x-8) ${\frac {2\,x+6}{x-8}}-{\frac {{x}^{2}+6\,x-8}{ \left( x-8 \right) ^{2} }}$ then just plug in a very high or very low value for x It has roughly a V asymptote of 8.5

23. L.T.

My bad; must have misread the question :(