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Mathematics 16 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

Can you find the derivative of this function first?

OpenStudy (anonymous):

Set the derivative equal to zero

OpenStudy (anonymous):

i normally pluged it into the graph

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@L.T. ok what would i do

OpenStudy (anonymous):

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

OpenStudy (anonymous):

lol asymptote... lame chat =]

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

reall no horizontal ?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes there is a slant use long division

OpenStudy (anonymous):

if you ned to find the slant asymptote

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so the anserw is no horizontal asymptote

OpenStudy (anonymous):

that is correct. there is none

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok thank you so much

OpenStudy (anonymous):

how'd you find that @timo86m

OpenStudy (anonymous):

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

OpenStudy (anonymous):

My bad; must have misread the question :(

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