Answer:
a) f(x) = (x-5)/((x-3)(x-10))
b) f(x) = (x-4)/((x+4)(x^2+1))
c) f(x) = 2(x-1)(x+1)/((x+3)(x-4))
d) f(x) = -2(x+5)(x-3)/((x+2)(x-5))
e) f(x) = -3(x^2-1)(x-2)/(x(x^2-9))
Step-by-step explanation:
Ordinarily, we think of a horizontal (or slant) asymptote as a line that the function nears, but does not reach. Some of these questions ask for the horizontal asymptote to be zero and for a function zero at a specific place. That is, the actual value of the function must be the same as the asymptotic value, at least at one location.
There are several ways this can happen:
- add a vertical asymptote on the same side of the zero as the required vertical asymptote. The function will cross the horizontal asymptote and then approach from the new direction.
- add a vertical asymptote on the other side of the zero from the required asymptote. The function zero will then be between the asymptotes, and the function will approach the asymptote in the expected way. (See the attachment)
- add complex zeros in the denominator. The function will cross the horizontal asymptote and approach it from the new direction. This does not add any asymptotes to the function.
To make the horizontal asymptote be zero, the degree of the denominator must be greater than the degree of the numerator. That is, there must be additional real or complex zeros in the denominator beyond those for the required vertical asymptotes.
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a) f(x) = (x-5)/((x-3)(x-10)) . . . . vertical asymptote added at x=10 to make the horizontal asymptote be zero
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b) f(x) = (x-4)/((x+4)(x^2+1)) . . . . complex zero added to the denominator to make the horizontal asymptote be zero
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c) f(x) = 2(x-1)(x+1)/((x+3)(x-4)) . . . . factor of 2 added to the numerator to make the horizontal asymptote be 2. Numerator and denominator degrees are the same. (See the second attachment.)
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d) f(x) = -2(x+5)(x-3)/((x+2)(x-5)) . . . . similar to problem (c)
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e) f(x) = -3(x^2-1)(x-2)/(x(x^2-9)) . . . . similar to the previous two problems (See the third attachment.)
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You remember that the difference of squares factors as ...
a² -b² = (a-b)(a+b)
so the factor that gives zeros at x=±3 can be written (x²-9).
