Domain Restrictions, Holes, and Discontinuities - College Algebra

Consider the rational function $f(x) = \frac{x^2 - 9}{x^2 + 2x - 15}$ . After factoring both numerator and denominator completely, what can be concluded about the domain and discontinuities of this function?
Answer: The domain is all real numbers except $x = -5$ and $x = 3$ , with a hole at $x = 3$ and vertical asymptote at $x = -5$ First, factor completely: $f(x) = \frac{(x-3)(x+3)}{(x+5)(x-3)}$ . The factor $(x-3)$ appears in both numerator and denominator, creating a hole at $x = 3$ . The factor $(x+5)$ appears only in the denominator, creating a vertical asymptote at $x = -5$ . The domain excludes both $x = 3$ and $x = -5$ . Choice A incorrectly identifies both as vertical asymptotes. Choice C uses wrong values from incorrectly factoring the denominator. Choice D incorrectly excludes $x = 3$ from the domain restriction.
A student claims that the function $f(x) = \frac{x^2+x-6}{x^2-9}$ has a hole at $x = 3$ because both the numerator and denominator equal zero when $x = 3$ . What is wrong with this reasoning?
Answer: While both expressions equal zero at $x = 3$ , the factored forms show no common factor, so there is a vertical asymptote, not a hole Factor both expressions: numerator $x^2+x-6 = (x+3)(x-2)$ and denominator $x^2-9 = (x+3)(x-3)$ . While both equal zero at $x = 3$ , this alone doesn't determine the type of discontinuity. The factored forms show no common factor involving $(x-3)$ , so $x = 3$ creates a vertical asymptote, not a hole. A hole occurs only when the same linear factor appears in both numerator and denominator and can be canceled. Choice A gives an incorrect general rule. Choice B is wrong since $f(3)$ is undefined. Choice D correctly identifies the flaw in reasoning.
For which value of $k$ will the function $f(x) = \frac{x^2+kx-12}{x^2-16}$ have a removable discontinuity (hole) at $x = 4$ ?
Answer: $k = -1$ For a hole at $x = 4$ , the factor $(x-4)$ must divide both numerator and denominator. The denominator $x^2-16 = (x-4)(x+4)$ already contains $(x-4)$ . For the numerator to also contain $(x-4)$ , we need $x = 4$ to be a root: $4^2 + k(4) - 12 = 0$ , so $16 + 4k - 12 = 0$ , giving $4k = -4$ and $k = -1$ . We can verify: when $k = -1$ , the numerator becomes $x^2 - x - 12 = (x-4)(x+3)$ . Choice B gives $x^2 + x - 12 = (x+4)(x-3)$ , no $(x-4)$ factor. Choice C gives $x^2 + 3x - 12$ , which doesn't factor to include $(x-4)$ . Choice D gives $x^2 - 3x - 12$ , also without $(x-4)$ .
Consider the piecewise-defined function: $f(x) = \begin{cases} \frac{x^2-4}{x-2} & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases}$ . What type of discontinuity, if any, occurs at $x = 2$ ?
Answer: Removable discontinuity because the rational part has a hole at $x = 2$ For $x \neq 2$ , we have $f(x) = \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2$ . So $\lim_{x \to 2} f(x) = \lim_{x \to 2} (x+2) = 4$ . However, $f(2) = 5$ . Since the limit exists but doesn't equal the function value, this is a removable discontinuity. The function could be made continuous by redefining $f(2) = 4$ . Choice A is wrong because $f(2) \neq \lim_{x \to 2} f(x)$ . Choice B is incorrect because left and right limits are both 4. Choice D is wrong because the limit exists and is finite.
For the rational function $h(x) = \frac{x^3-8}{x^2-4}$ , what is the domain restriction and what type of discontinuity occurs at $x = 2$ ?
Answer: Domain: all real numbers except $x = \pm 2$ ; hole at $x = 2$ Factor the function: $h(x) = \frac{x^3-8}{x^2-4} = \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}$ . The factor $(x-2)$ appears in both numerator and denominator, creating a hole (removable discontinuity) at $x = 2$ . The factor $(x+2)$ appears only in the denominator, creating a vertical asymptote at $x = -2$ . The domain excludes both $x = 2$ and $x = -2$ . Choice A incorrectly identifies $x = 2$ as a vertical asymptote. Choice C incorrectly excludes $x = -2$ from domain restrictions. Choice D correctly identifies the removable discontinuity but uses less precise terminology.
Consider the function $k(x) = \frac{x^2-3x-10}{x^2-25}$ . If this function is graphed on a coordinate plane, which statement correctly describes its discontinuities?
Answer: There is a hole at $x = 5$ and a vertical asymptote at $x = -5$ Factor completely: $k(x) = \frac{x^2-3x-10}{x^2-25} = \frac{(x-5)(x+2)}{(x-5)(x+5)}$ . The factor $(x-5)$ appears in both numerator and denominator, creating a hole at $x = 5$ . The factor $(x+5)$ appears only in the denominator, creating a vertical asymptote at $x = -5$ . Choice B incorrectly places the hole at $x = -2$ (a zero of the reduced function, not a discontinuity). Choice C ignores the common factor cancellation. Choice D reverses the locations of the hole and asymptote.

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