Asked by htht
What are the vertical and horizontal asymptotes for the function f (x) = StartFraction x squared + x minus 6 Over x cubed minus 1 EndFraction?
vertical asymptote: x = 1
horizontal asymptote: none
vertical asymptote: x = 1
horizontal asymptote: y = 0
vertical asymptote: x = –2, x = 3
horizontal asymptote: y = 0
vertical asymptote: x = –2, x = –3
horizontal asymptote: none
All Answers 1
Answered by
GPT-5 mini
AI
Factor and analyze:
- numerator x^2 + x − 6 = (x + 3)(x − 2)
- denominator x^3 − 1 = (x − 1)(x^2 + x + 1), and x^2 + x + 1 has no real roots.
There is a single real zero of the denominator at x = 1 (not canceled by the numerator), so a vertical asymptote x = 1. Since deg(numerator) = 2 < deg(denominator) = 3, the horizontal asymptote is y = 0.
Answer: vertical asymptote x = 1; horizontal asymptote y = 0.
- numerator x^2 + x − 6 = (x + 3)(x − 2)
- denominator x^3 − 1 = (x − 1)(x^2 + x + 1), and x^2 + x + 1 has no real roots.
There is a single real zero of the denominator at x = 1 (not canceled by the numerator), so a vertical asymptote x = 1. Since deg(numerator) = 2 < deg(denominator) = 3, the horizontal asymptote is y = 0.
Answer: vertical asymptote x = 1; horizontal asymptote y = 0.
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