Asked by j
The vertical asymptote is x =
.
The horizontal asymptote is y =
.
All Answers 1
Answered by
GPT-5 mini
AI
I don't see the function or graph — could you paste the rational function or describe the graph? Without that I can't give the specific asymptotes.
If you want to find them yourself, here are the quick rules:
- Vertical asymptotes: factor numerator and denominator. Any x-value that makes the denominator zero and does not cancel with a factor in the numerator gives a vertical asymptote x = that value. If a factor cancels, you get a hole, not a vertical asymptote.
- Horizontal asymptotes for rational functions f(x) = p(x)/q(x) (degrees n = deg p, m = deg q):
- If n < m, horizontal asymptote y = 0.
- If n = m, horizontal asymptote y = (leading coeff of p)/(leading coeff of q).
- If n > m, there is no horizontal asymptote (if n = m + 1 there is a slant/oblique asymptote found by polynomial division; if n > m + 1 the end behavior is like a polynomial of degree n − m).
Examples:
- f(x) = (2x^2+3x+1)/(x^2−1): vertical asymptotes x = 1 and x = −1 (denominator zeros), horizontal asymptote y = 2 (leading coef ratio 2/1).
- f(x) = x/(x^2+1): no vertical asymptote (denominator never zero), horizontal asymptote y = 0.
- f(x) = (x^2−1)/(x−1) = x+1 with a hole at x = 1 (canceled factor), no vertical asymptote; behaves like the line y = x+1.
Send the function or graph and I'll compute the specific asymptotes.
If you want to find them yourself, here are the quick rules:
- Vertical asymptotes: factor numerator and denominator. Any x-value that makes the denominator zero and does not cancel with a factor in the numerator gives a vertical asymptote x = that value. If a factor cancels, you get a hole, not a vertical asymptote.
- Horizontal asymptotes for rational functions f(x) = p(x)/q(x) (degrees n = deg p, m = deg q):
- If n < m, horizontal asymptote y = 0.
- If n = m, horizontal asymptote y = (leading coeff of p)/(leading coeff of q).
- If n > m, there is no horizontal asymptote (if n = m + 1 there is a slant/oblique asymptote found by polynomial division; if n > m + 1 the end behavior is like a polynomial of degree n − m).
Examples:
- f(x) = (2x^2+3x+1)/(x^2−1): vertical asymptotes x = 1 and x = −1 (denominator zeros), horizontal asymptote y = 2 (leading coef ratio 2/1).
- f(x) = x/(x^2+1): no vertical asymptote (denominator never zero), horizontal asymptote y = 0.
- f(x) = (x^2−1)/(x−1) = x+1 with a hole at x = 1 (canceled factor), no vertical asymptote; behaves like the line y = x+1.
Send the function or graph and I'll compute the specific asymptotes.
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