I am working through a packet as I prepare for a test.
y= (2x^3-8x^2-x+4)/(x^2-6x+9)
I found one vertical asymptote at x=3 from the factors (x-3)^2
I found no horizontal asymptote - my rationale is the highest degree of the numerator is larger than the highest degree of the denominator. Where did I go wrong? My answers say there should be two asymptotes? Am I doing this wrong, overthinking, etc.?
2007-10-12 02:17:59 · 2 answers · asked by CHILL 1 in Science & Mathematics ➔ Mathematics
there is only one vertical asymptote... correct!
there is no horizontal asymptote.... correct again... your analysis is still correct...
but there is an oblique asymptote... you can have that if the degree of the numerator is one more than the degree of the denominator...
simply divide... the quotient becomes the oblique asymptote..
quotient: 2x + 4
thus oblique asymptote: y = 2x+4
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2007-10-12 02:32:12 · answer #1 · answered by Alam Ko Iyan 7 · 4⤊ 0⤋
There is another asymptote for x==> infinity (it is not vertical nor horizontal)
lim y/x= 2 = m (possible slope
lim y-mx = lim (2x^3-8x^2-x+2-2x^3+12x^2-18x)/(x^2-6x+9)=4
so the other asymptote is
y=2x+4
2007-10-12 02:26:28 · answer #2 · answered by santmann2002 7 · 1⤊ 0⤋