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

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

Mathematics
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Can you find the derivative of this function first?
Set the derivative equal to zero
i normally pluged it into the graph

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https://www.google.com/search?q=x%5E2%2B6x-8%2Fx-8&rlz=1C1CHFX_enUS470US470&oq=x%5E2%2B6x-8%2Fx-8&sugexp=chrome,mod=11&sourceid=chrome&ie=UTF-8#hl=en&safe=off&rlz=1C1CHFX_enUS470US470&sclient=psy-ab&q=(x%5E2%2B6x-8)%2F(x-8)&oq=(x%5E2%2B6x-8)%2F(x-8)&gs_l=serp.3...2141.18850.1.19323.8.8.0.0.0.0.120.792.5j3.8.0.les%3B..0.0...1c.1.Q4G1o6RnZBs&psj=1&bav=on.2,or.r_gc.r_pw.r_cp.r_qf.&fp=35197abecd6f379b&biw=1422&bih=776
All right, I'll assume you can't do that now. Is this the equation:\[x ^{2}+\frac{ 6x-8 }{ x-8 }\]
i see what u posted @timo86m but idk what the horrizontal is
@L.T. ok what would i do
there is a slant retricemptote.. not a horizontal =/
lol asymptote... lame chat =]
the degree of the top is greater than the degree of the bottom therefore no horizontal asymptote no calc needed, that is all
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.
reall no horizontal ?
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
yes there is a slant use long division
if you ned to find the slant asymptote
if the degrees are the same, it is the ration of the leading coefficients that is it
so the anserw is no horizontal asymptote
that is correct. there is none
it has a slanted asymptote with a slope of 1045/1058
ok thank you so much
how'd you find that @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
My bad; must have misread the question :(

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