Dayv O. answered • 03/24/21
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Hi, anyway began thinking and after considering some sort of -e-x+K
I thought what about 1/(!+x2) and what could be done
does this fit the answer, f(x)=g(x)*x-3)/x-3)
where g(x)=(5x2)/(1+x2)
Dayv O.
same as f(x)=(5x^3-15x^2)/((1+x^2)*(x-3) asymptote=5- ;;;;; for asymptote 5+, then f(x)= -(5x^3-15x^2)/((1+x^2)*(x-3)
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03/25/21
Dayv O.
well did graph and no, for asymptote 5+ must use f(x)=10-(5x^3-15x^2)/((1+x^2)*(x-3)
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03/25/21
Dayv O.
well, analyzing more, only f(x)=(5x^3-15x^2)/((1+x^2)*(x-3) can fit question. I think, let me know if I am wrong.
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03/25/21
Dayv O.
should read f(x)=g(x)*(x-3)/x-3), this is my hole at x=3, that is (x-3)/x-3)03/25/21