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Verticle asymptote at 4, zeroes at 0 and 3, hole when x=5, no horizontal asymptotes

User NullNoname
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Final answer:

The question asks for the creation of a rational function with given features including vertical asymptote, zeroes, and a hole. One possible function that satisfies this is f(x) = x(x-3)(x-5)/(x-4)(x-5), with the appropriate characteristics.

Step-by-step explanation:

The student is asking about creating a rational function based on given characteristics: a vertical asymptote at x=4, zeroes at x=0 and x=3, a hole when x=5, and no horizontal asymptotes. To construct such a function, we can utilize these attributes as follows:

  • A vertical asymptote at x=4 suggests that there is a factor in the denominator that becomes zero at x=4. Therefore, the denominator could be (x-4).
  • Zeroes at x=0 and x=3 indicate that x and (x-3) are factors in the numerator, since setting the numerator equal to zero should give us these x-values.
  • A hole at x=5 is created when there is a factor in both the numerator and the denominator that cancel out. Hence, (x-5) should be present in both the numerator and the denominator.

Combining this information, one possible function that satisfies all these conditions is:

f(x) = \frac{x(x-3)(x-5)}{(x-4)(x-5)}

This function has the required zeroes and vertical asymptote, a removable discontinuity (hole) at x=5, and since the degrees of the numerator and denominator are equal, there is no horizontal asymptote.

User Linh
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