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In the xy-plane, the graph of a rational function f has a vertical asymptote at x=-5. Which of the following could be an expression for f(x) ? (A) ((x-5)(x+5))/(2(x-5)) (B) ((x-4)(x+5))/((x-1)(x+5)) (C) ((x+1)(x+5))/((x-5)(x+2)) (D) ((x-5)(x-3))/((x-3)(x+5))

In the xy x y -plane, the graph of a rational function f f has a vertical asymptote at x=−5 x=-5 . Which of the following could be an expression for f(x) f(x) ?\newline(A) (x−5)(x+5)2(x−5) \frac{(x-5)(x+5)}{2(x-5)} \newline(B) (x−4)(x+5)(x−1)(x+5) \frac{(x-4)(x+5)}{(x-1)(x+5)} \newline(C) (x+1)(x+5)(x−5)(x+2) \frac{(x+1)(x+5)}{(x-5)(x+2)} \newline(D) (x−5)(x−3)(x−3)(x+5) \frac{(x-5)(x-3)}{(x-3)(x+5)}

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Q. In the xy x y -plane, the graph of a rational function f f has a vertical asymptote at x=−5 x=-5 . Which of the following could be an expression for f(x) f(x) ?\newline(A) (x−5)(x+5)2(x−5) \frac{(x-5)(x+5)}{2(x-5)} \newline(B) (x−4)(x+5)(x−1)(x+5) \frac{(x-4)(x+5)}{(x-1)(x+5)} \newline(C) (x+1)(x+5)(x−5)(x+2) \frac{(x+1)(x+5)}{(x-5)(x+2)} \newline(D) (x−5)(x−3)(x−3)(x+5) \frac{(x-5)(x-3)}{(x-3)(x+5)}
  1. Identify Vertical Asymptote: Identify the characteristic of a vertical asymptote in a rational function. A vertical asymptote occurs where the denominator of a rational function is zero, but the numerator is not zero at the same xx-value.
  2. Examine Option (A): Examine option (A) to see if it has a vertical asymptote at x=−5x=-5.\newlineThe denominator of option (A) is 2(x−5)2(x-5), which is zero when x=5x=5, not x=−5x=-5. Therefore, it cannot have a vertical asymptote at x=−5x=-5.
  3. Examine Option (B): Examine option (B) to see if it has a vertical asymptote at x=−5x=-5. The denominator of option (B) is (x−1)(x+5)(x-1)(x+5), which is zero when x=−5x=-5. However, the numerator also has a factor of (x+5)(x+5), which would cancel out the (x+5)(x+5) in the denominator, eliminating the vertical asymptote at x=−5x=-5.
  4. Examine Option (C): Examine option (C) to see if it has a vertical asymptote at x=−5x=-5. The denominator of option (C) is (x−5)(x+2)(x-5)(x+2), which is zero when x=5x=5, creating a vertical asymptote at x=5x=5. Since we are looking for a vertical asymptote at x=−5x=-5, this option is not correct.
  5. Examine Option (D): Examine option (D) to see if it has a vertical asymptote at x=−5x=-5. The denominator of option (D) is (x−3)(x+5)(x-3)(x+5), which is zero when x=−5x=-5. The numerator is (x−5)(x−3)(x-5)(x-3), which does not cancel out the (x+5)(x+5) in the denominator. Therefore, this function has a vertical asymptote at x=−5x=-5.

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