Consider the following verbal description of a rational function. Where are the asymptotes and point of discontinuity located for the described rational function?

f of x equals the quotient of x plus four and x squared minus sixteen

(1 point)
Responses

vertical asymptote: x=4

horizontal asymptote: y=0

point of discontinuity: when x=−4

vertical asymptote: x is equal to 4 horizontal asymptote: y is equal to 0 point of discontinuity: when x is equal to negative 4

vertical asymptote: x=4

horizontal asymptote: y=12

point of discontinuity: none

vertical asymptote: x is equal to 4 horizontal asymptote: y is equal to 1 half point of discontinuity: none

vertical asymptote: x=4

horizontal asymptote: none

point of discontinuity: none

vertical asymptote: x is equal to 4 horizontal asymptote: none point of discontinuity: none

vertical asymptote: y=0

horizontal asymptote: x=4

point of discontinuity: when x=−4

vertical asymptote: y is equal to 0 horizontal asymptote: x is equal to 4 point of discontinuity: when x is equal to negative 4
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1 answer

To analyze the function \( f(x) = \frac{x + 4}{x^2 - 16} \), we need to determine the asymptotes and points of discontinuity.

  1. Factor the denominator: \[ x^2 - 16 = (x - 4)(x + 4) \] This means that the function can be simplified as follows: \[ f(x) = \frac{x + 4}{(x - 4)(x + 4)} \]

  2. Identify the point of discontinuity: The function has a factor of \( x + 4 \) in both the numerator and denominator. This indicates a removable discontinuity at \( x = -4 \). Therefore, there is a point of discontinuity at \( x = -4 \).

  3. Identify vertical asymptotes: Vertical asymptotes occur where the denominator is zero but the numerator is not zero. Thus:

    • The denominator \( (x - 4)(x + 4) = 0 \) at \( x = 4 \) and \( x = -4 \).
    • Since \( x = -4 \) is a removable discontinuity (as discussed), the vertical asymptote will only occur at \( x = 4 \).
  4. Identify horizontal asymptotes: To find horizontal asymptotes, we look at the degrees of the numerator and denominator. In this case, the degree of the numerator is 1 and the degree of the denominator is 2. For rational functions:

    • If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).

Now, summarizing this information:

  • Vertical asymptote: \( x = 4 \)
  • Horizontal asymptote: \( y = 0 \)
  • Point of discontinuity: \( x = -4 \)

Given this analysis, the correct response is:

vertical asymptote: \( x = 4 \); horizontal asymptote: \( y = 0 \); point of discontinuity: when \( x = -4 \).

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