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Find the slant asymptote of f(x)=18x^2/9x-8?
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divide it all by x methinks or the sensible option would be to divide it thru
I know you divide 18x^2 by 9x-8 and I get 2x+1 with a remainder of 7x-8 and it just doesn't look right nor does the computer think it is right.
2x +16/9 ---------- 9x-8/18x^2 -18x^2+16x ------------ 16x -16x - (something else that goes away)
ah I see. By doing this all of my x's are out of the bottom, which leaves me with a semi clean remainder. Thank you for helping me figure out where I was going wrong!
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