Rational Function Features - Math 3

A rational function has the form $r(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials. If $r(x)$ has vertical asymptotes at $x = -2$ and $x = 5$ , and a hole at $x = 3$ , which expression could represent $Q(x)$ ?
Answer: $(x+2)(x-5)(x-3)$ For vertical asymptotes at $x = -2$ and $x = 5$ , the denominator $Q(x)$ must have factors $(x+2)$ and $(x-5)$ that do not completely cancel with the numerator. For a hole at $x = 3$ , both the numerator and denominator must have the factor $(x-3)$ , and this factor must cancel completely. Therefore, $Q(x)$ must contain $(x+2)$ , $(x-5)$ , and $(x-3)$ . Choice A satisfies this. Choice B is missing the $(x-3)$ factor needed for the hole. Choice C would work but has an extra factor. Choice D has wrong signs for the asymptote factors.
For the rational function $k(x) = \frac{3x^3 - 12x^2 + 9x}{x^3 - 6x^2 + 11x - 6}$ , what happens as $x$ approaches positive infinity?
Answer: The function approaches $y = 3$ because the ratio of leading coefficients determines the horizontal asymptote behavior. For large values of $x$ , the behavior is determined by the leading terms. Both numerator and denominator have degree 3, so the horizontal asymptote is the ratio of leading coefficients: $\frac{3x^3}{x^3} = 3$ . Even though this function can be factored (numerator: $3x(x^2-4x+3) = 3x(x-1)(x-3)$ ; denominator factors as $(x-1)(x-2)(x-3)$ ), and factors cancel to give $\frac{3x}{x-2}$ after removing holes, the end behavior is still determined by $y = 3$ . Choice B incorrectly assumes the degrees change after factoring. Choice C is wrong about the direction. Choice D gives an incorrect asymptote value.
A rational function $p(x) = \frac{(x-a)^2(x-b)}{(x-a)(x-c)(x-d)}$ where $a$ , $b$ , $c$ , and $d$ are distinct real numbers. Which statement best describes the key features of this function?
Answer: The function has a hole at $x = a$ , vertical asymptotes at $x = c$ and $x = d$ , and crosses the x-axis at $x = b$ . After canceling the common factor $(x-a)$ , we get $p(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}$ . The canceled factor creates a hole at $x = a$ . The remaining denominator factors $(x-c)$ and $(x-d)$ create vertical asymptotes at $x = c$ and $x = d$ . The remaining numerator factor $(x-b)$ has multiplicity 1, so the function crosses (not touches) the x-axis at $x = b$ . Choice A incorrectly states the function touches the x-axis at $x = b$ (which would require even multiplicity). Choice B ignores the cancellation and incorrectly claims a vertical asymptote at $x = a$ . Choice D incorrectly claims $x = a$ is a zero rather than a hole.
Consider the rational function $r(x) = \frac{x^4 - 5x^3 + 6x^2}{x^3 - 6x^2 + 9x}$ . What is the most complete description of this function's key features?
Answer: After factoring, the function simplifies to $r(x) = \frac{x-2}{x-3}$ with a hole at $x = 0$ and vertical asymptote at $x = 3$ . Factor the numerator: $x^4 - 5x^3 + 6x^2 = x^2(x^2 - 5x + 6) = x^2(x-2)(x-3)$ . Factor the denominator: $x^3 - 6x^2 + 9x = x(x^2 - 6x + 9) = x(x-3)^2$ . So $r(x) = \frac{x^2(x-2)(x-3)}{x(x-3)^2} = \frac{x(x-2)}{x-3}$ after canceling $x$ and one factor of $(x-3)$ . The canceled factor $x$ creates a hole at $x = 0$ . The remaining $(x-3)$ in the denominator creates a vertical asymptote at $x = 3$ . Choice A keeps the extra $x$ factor incorrectly. Choice B incorrectly claims $x = 3$ is both a hole and asymptote. Choice C keeps the wrong factors after cancellation.
A rational function $f(x) = \frac{ax^2 + bx + c}{dx^2 + ex + f}$ has a horizontal asymptote at $y = 3$ . If $d = 2$ , what must be true about the value of $a$ ?
Answer: $a = 6$ For a rational function where the numerator and denominator have the same degree, the horizontal asymptote is $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$ . Here, the horizontal asymptote is $y = \frac{a}{d} = \frac{a}{2} = 3$ . Solving for $a$ : $a = 6$ . Choice B gives $y = \frac{3}{2}$ . Choice C gives $y = \frac{3/2}{2} = \frac{3}{4}$ . Choice D is incorrect because the horizontal asymptote depends specifically on the ratio of leading coefficients.
A rational function has the form $g(x) = \frac{P(x)}{(x+2)^3(x-1)^2}$ where $P(x)$ is a polynomial. If $g(x)$ has exactly one vertical asymptote, what must be true about $P(x)$ ?
Answer: $P(x)$ must have $(x+2)^2(x-1)^2$ as a factor, leaving only $(x+2)$ in the denominator for the asymptote. The denominator has potential vertical asymptotes at $x = -2$ and $x = 1$ . For exactly one vertical asymptote, we must eliminate one of these. If $P(x)$ has $(x+2)^2(x-1)^2$ as a factor, then $g(x) = \frac{(x+2)^2(x-1)^2 \cdot Q(x)}{(x+2)^3(x-1)^2} = \frac{Q(x)}{x+2}$ for some polynomial $Q(x)$ . This leaves only one vertical asymptote at $x = -2$ . Choice A leaves $(x+2)$ in the denominator but doesn't address $x = 1$ , giving two asymptotes. Choice B leaves $(x+2)^3$ in the denominator, still giving two asymptotes. Choice C is too vague and doesn't specify the correct multiplicities needed.
A rational function has the form $h(x) = \frac{ax^2 + bx + c}{(x-3)^2(x+1)}$ where $a$ , $b$ , and $c$ are constants. If the function has a horizontal asymptote at $y = 2$ , which additional condition would create a hole in the graph?
Answer: The numerator must have $(x-3)$ as a factor, requiring $9a + 3b + c = 0$ to be satisfied. Since the horizontal asymptote is $y = 2$ , we need $a = 2$ . For a hole to exist, a factor in the denominator must cancel with the numerator. The denominator has factors $(x-3)^2$ and $(x+1)$ . If the numerator has factor $(x-3)$ , then one factor of $(x-3)$ cancels, leaving $(x-3)$ in the denominator (still a vertical asymptote) and creating a hole. For $(x-3)$ to be a factor of the numerator, we need the numerator to equal zero when $x = 3$ : $a(3)^2 + b(3) + c = 9a + 3b + c = 0$ . Choice B would create a hole but at $x = -1$ . Choice C would completely cancel $(x-3)^2$ , removing the vertical asymptote entirely. Choice D is incorrect about the result.
The function $m(x) = \frac{x^3 + 2x^2 - 9x - 18}{x^2 + x - 6}$ has a specific type of discontinuity. After performing polynomial long division, what can be concluded about the function's behavior?
Answer: The function has a slant asymptote $y = x + 1$ , a hole at $x = -3$ , and a vertical asymptote at $x = 2$ . First, factor the denominator: $x^2 + x - 6 = (x+3)(x-2)$ . Factor the numerator by grouping: $x^3 + 2x^2 - 9x - 18 = x^2(x+2) - 9(x+2) = (x^2-9)(x+2) = (x-3)(x+3)(x+2)$ . So $m(x) = \frac{(x-3)(x+3)(x+2)}{(x+3)(x-2)}$ . The factor $(x+3)$ cancels, creating a hole at $x = -3$ . This gives $m(x) = \frac{(x-3)(x+2)}{x-2} = \frac{x^2-x-6}{x-2}$ . Performing polynomial long division: $x^2-x-6 = (x-2)(x+1) - 4$ , so $m(x) = x+1 - \frac{4}{x-2}$ . This shows a slant asymptote $y = x+1$ and vertical asymptote at $x = 2$ . Choice A misses the hole. Choice B misses the vertical asymptote. Choice D incorrectly includes $x = -3$ as a vertical asymptote.
The rational function $h(x) = \frac{x^3 - 8}{x^2 + 2x + 4}$ appears to have a denominator that doesn't factor over the reals. However, there's a hidden relationship. What is the correct analysis of this function?
Answer: The function simplifies to $h(x) = x - 2$ because the numerator and denominator share a common factor. The key insight is recognizing that $x^3 - 8 = (x-2)(x^2+2x+4)$ using the difference of cubes formula: $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ . So $h(x) = \frac{(x-2)(x^2+2x+4)}{x^2+2x+4} = x-2$ (for $x \neq$ the complex roots of $x^2+2x+4=0$ ). Since the denominator has no real zeros, there are no holes or asymptotes in the real plane - the function is simply the line $y = x-2$ . Choice A incorrectly assumes the original form determines the behavior. Choice C correctly identifies no real asymptotes but wrong about the simplified form. Choice D mentions irrelevant complex asymptotes and incorrect end behavior.
Consider the rational function $f(x) = \frac{x^2 - 9}{x^2 - 2x - 3}$ . After simplifying this function, what is the behavior at $x = 3$ ?
Answer: There is a removable discontinuity (hole) at $x = 3$ First, factor the numerator and denominator: $f(x) = \frac{(x-3)(x+3)}{(x-3)(x+1)}$ . Since $(x-3)$ appears in both numerator and denominator, it cancels out, leaving $f(x) = \frac{x+3}{x+1}$ for $x \neq 3$ . This creates a removable discontinuity (hole) at $x = 3$ . Choice A is wrong because the factor cancels rather than creating an asymptote. Choice C is wrong because the original function is undefined at $x = 3$ . Choice D is wrong because horizontal asymptotes are not located at specific x-values.
For the rational function $k(x) = \frac{x^2 - 1}{x^3 - x^2 - x + 1}$ , determine the total number of vertical asymptotes and holes combined.
Answer: One vertical asymptote and one hole Factor the numerator: $x^2 - 1 = (x-1)(x+1)$ . Factor the denominator by grouping: $x^3 - x^2 - x + 1 = x^2(x-1) - 1(x-1) = (x^2-1)(x-1) = (x-1)(x+1)(x-1) = (x-1)^2(x+1)$ . The function becomes $k(x) = \frac{(x-1)(x+1)}{(x-1)^2(x+1)}$ . The factor $(x+1)$ cancels completely, creating a hole at $x = -1$ . One factor of $(x-1)$ cancels, leaving $(x-1)$ in the denominator, creating a vertical asymptote at $x = 1$ . So there is one vertical asymptote and one hole.
Consider the piecewise-defined function where $f(x) = \frac{x^2 - 9}{x + 3}$ for $x \neq -3$ and $f(-3) = 5$ . How does this compare to the rational function $g(x) = \frac{x^2 - 9}{x + 3}$ that is undefined at $x = -3$ ?
Answer: Function $f$ has a jump discontinuity at $x = -3$ while $g$ has a removable discontinuity at the same point. For both functions, $\frac{x^2-9}{x+3} = \frac{(x-3)(x+3)}{x+3} = x-3$ when $x \neq -3$ . So $\lim_{x \to -3} f(x) = \lim_{x \to -3} g(x) = -3-3 = -6$ . Function $g$ has a hole (removable discontinuity) at $x = -3$ since the limit exists but the function is undefined there. Function $f$ has $f(-3) = 5 \neq -6$ , so while the limit exists, the function value differs from the limit, creating a jump discontinuity (the function 'jumps' from the limit value to the defined value). Choice A ignores that $f(-3) = 5$ . Choice C incorrectly categorizes both as the same type of discontinuity. Choice D correctly identifies the discontinuity types but wrongly claims different limits.
The rational function $g(x) = \frac{2x^3 + x^2 - 8x - 4}{x^2 + 3x + 2}$ can be simplified by factoring. What is the horizontal asymptote of the simplified function?
Answer: The function has no horizontal asymptote because the degree of the numerator exceeds the degree of the denominator. First, factor the denominator: $x^2 + 3x + 2 = (x+1)(x+2)$ . For the numerator, factor by grouping: $2x^3 + x^2 - 8x - 4 = x^2(2x+1) - 4(2x+1) = (x^2-4)(2x+1) = (x-2)(x+2)(2x+1)$ . So $g(x) = \frac{(x-2)(x+2)(2x+1)}{(x+1)(x+2)}$ . After canceling $(x+2)$ , we get $g(x) = \frac{(x-2)(2x+1)}{x+1} = \frac{2x^2-3x-2}{x+1}$ . The simplified function has degree 2 in the numerator and degree 1 in the denominator, so there is no horizontal asymptote. Choices B, C, and D all incorrectly assume a horizontal asymptote exists.
Consider the rational function $f(x) = \frac{x^2 - 4x - 5}{x^2 - 6x + 5}$ . After factoring both numerator and denominator, which statement correctly describes the behavior of this function?
Answer: The function has a vertical asymptote at $x = 1$ and a hole at $x = 5$ where the function is undefined. Factoring the numerator: $x^2 - 4x - 5 = (x-5)(x+1)$ . Factoring the denominator: $x^2 - 6x + 5 = (x-5)(x-1)$ . So $f(x) = \frac{(x-5)(x+1)}{(x-5)(x-1)}$ . The factor $(x-5)$ cancels, creating a hole at $x = 5$ . The remaining denominator factor $(x-1)$ creates a vertical asymptote at $x = 1$ . Choice A incorrectly identifies both zeros of the original denominator as vertical asymptotes. Choice C reverses which discontinuity is which. Choice D incorrectly claims both are holes.

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