Question 202900: Which one of the following is true? (1) The function, f(x)=2x^2+1/x^2-1, does not have a horizontal asymptote. (2) The graph of a rational function cannot have x-intercepts. (3) The horizontal asymptote for the graph of y=4x-1/x+3 is x=-3. (4) The graph of f(x)=2x+4/x^2-4 has only one vertical asymptote. THIS IS A TOUGH ONE....HELP.
Answer by solver91311(24713)   (Show Source):
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1) Has a horizontal asymptote at  because the numerator and denominator are of the same degree, hence there is a horizontal asymptote at the value of the lead coefficient of the numerator divided by the lead coefficient of the denominator. This one is false.
2) The graph of a rational function can have  intercepts. The graph will intercept the -axis at any zero of the numerator polynomial that is not also a zero of the denominator polynomial. Consider  . This one is false.
3) As in #1, the horizontal asymptote is at  where  is the value of the lead coefficient of the numerator divided by the lead coefficient of the denominator. Hence, the horizontal asymptote is at  not  . This one is false.
4) This one looks false on the face of it, because you would expect a vertical asymptote at any zero of the denominator polynomial. However, note that -2 is a zero of the denominator, but it is also a zero of the numerator. Hence, while there is a discontinuity at -2, there is no asymptote. In order to prove it, you need a little trick from the Calculus called L'Hôpital's Rule. It says that if:
 = 0\text{ or }\pm\infty)
and
= 0\text{ or }\pm\infty)
and if
}{g'(x)}) exists, then
}{g(x)} = \lim_{x\rightarrow c}\ \frac{f'(x)}{g'(x)})
So for the function given in #4,
 = 2) and
 = 2x) and
}{g'(x)} = \frac{2}{2(-2)}=-\frac{1}{2})
not  as would be the case if there were an asymptote at that point.
Therefore #4 is the true statement.
John
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