College Algebra Quiz: End Behavior Via Degree Comparison

What this quiz covers

This quiz focuses on End Behavior Via Degree Comparison, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The function approaches $y = \frac{5}{2}$ as $x \to \pm\infty$
2. As $x \to +\infty$, $r(x) \to +\infty$ and as $x \to -\infty$, $r(x) \to +\infty$
3. As $x \to +\infty$, $r(x) \to +\infty$ and as $x \to -\infty$, $r(x) \to -\infty$ (correct answer)
4. The function approaches $y = 0$ as $x \to \pm\infty$

Explanation: The numerator has degree 3 and the denominator has degree 2. When the numerator's degree exceeds the denominator's degree, there is no horizontal asymptote and the function grows without bound. Since the leading term of the rational function behaves like $\frac{5x^3}{2x^2} = \frac{5x}{2}$, as $x \to +\infty$, $r(x) \to +\infty$, and as $x \to -\infty$, $r(x) \to -\infty$. Choice A incorrectly applies the equal-degree rule. Choice B ignores the sign change for negative x-values. Choice D incorrectly applies the denominator-higher-degree rule.

Question 2

1. The function approaches a horizontal asymptote at $y = \frac{2}{5}$
2. The function has no horizontal asymptote and grows without bound in both directions (correct answer)
3. The function approaches a horizontal asymptote at $y = 0$
4. The function approaches different finite values as $x$ approaches $+\infty$ and $-\infty$

Explanation: The numerator has degree 3 and the denominator has degree 2. When the numerator's degree is greater than the denominator's degree, there is no horizontal asymptote and the function grows without bound as $x \to \pm\infty$. Choice A incorrectly applies the equal-degree rule. Choice C incorrectly applies the numerator-lower-degree rule. Choice D is impossible for rational functions with polynomial numerator and denominator.

Question 3

1. As $x \to \pm\infty$, $h(x) \to \frac{4}{3}$
2. As $x \to \pm\infty$, $h(x) \to 0$ (correct answer)
3. The function has vertical asymptotes that determine the end behavior
4. As $x \to \pm\infty$, $h(x) \to \frac{2}{1} = 2$

Explanation: The numerator has degree 2 and the denominator has degree 4. When the denominator's degree is greater than the numerator's degree, the horizontal asymptote is $y = 0$, so $h(x) \to 0$ as $x \to \pm\infty$. Choice A incorrectly applies the equal-degree rule using leading coefficients. Choice C confuses vertical asymptotes (which affect behavior near specific x-values) with end behavior. Choice D incorrectly uses constant terms instead of considering degrees.

Question 4

1. As $x \to \pm\infty$, $k(x) \to \frac{7}{6}$
2. As $x \to \pm\infty$, $k(x) \to \frac{7}{2}$
3. The function grows without bound as $x \to \pm\infty$
4. As $x \to \pm\infty$, $k(x) \to 0$ (correct answer)

Explanation: When analyzing the end behavior of rational functions, you need to compare the degrees of the numerator and denominator polynomials. This comparison tells you what happens as $x$ approaches positive or negative infinity. First, let's identify the degrees. The numerator $7x^2 + 3x - 4$ has degree 2. For the denominator, expand $(2x - 1)(x + 3)(x - 5)$ to see it's a degree 3 polynomial (the highest power term will be $2x \cdot x \cdot x = 2x^3$). Since the denominator's degree (3) is greater than the numerator's degree (2), the function approaches 0 as $x \to \pm\infty$. Think of it this way: as $x$ gets very large, the denominator grows much faster than the numerator, making the entire fraction approach zero. Choice A is incorrect because $\frac{7}{6}$ would be the horizontal asymptote if both polynomials had degree 2, with leading coefficients 7 and 6. Choice B suggests $\frac{7}{2}$, which might result from incorrectly comparing just the leading coefficients of the numerator (7) and the first factor of the denominator (2). Choice C is wrong because the function doesn't grow without bound—that would happen if the numerator's degree exceeded the denominator's degree. Remember this pattern: when the denominator's degree exceeds the numerator's degree in a rational function, the horizontal asymptote is always $y = 0$. This is one of the most reliable rules for determining end behavior quickly.

Question 5

1. $y = \frac{1}{2}$
2. $y = -\frac{1}{2}$ (correct answer)
3. $y = \frac{2}{5}$
4. $y = -\frac{2}{5}$

Explanation: Both polynomials have degree 4, so the horizontal asymptote is the ratio of leading coefficients: $\frac{-3}{6} = -\frac{1}{2}$. Choice A has the wrong sign. Choices C and D incorrectly use the coefficients of the $x^2$ and $x^3$ terms rather than the leading coefficients.

Question 6

1. The student is incorrect; the horizontal asymptote should be $y = \frac{2}{7}$ using different coefficients
2. The student is incorrect; the horizontal asymptote should be $y = \frac{-1}{2}$ based on the constant terms
3. The student is incorrect; when degrees are equal, there is no horizontal asymptote
4. The student is correct; both polynomials have degree 4 and the ratio of leading coefficients is $\frac{1}{3}$ (correct answer)

Explanation: When analyzing rational functions for horizontal asymptotes, you need to compare the degrees of the numerator and denominator polynomials. The rule depends on this comparison: if the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. Let's examine this function carefully. The numerator is $x^4 + 2x^3 - 5x^2 + 1$, which has degree 4 with leading coefficient 1. The denominator is $3x^4 - x^3 + 7x - 2$, which also has degree 4 with leading coefficient 3. Since both polynomials have the same degree, the horizontal asymptote is $y = \frac{1}{3}$, confirming the student's reasoning. Choice A incorrectly suggests using different coefficients (possibly the coefficients of $x^3$ terms), but horizontal asymptotes depend only on the highest-degree terms when degrees are equal. Choice B mistakenly uses the constant terms (1 and -2), but these don't determine horizontal asymptotes. Choice C reflects a fundamental misunderstanding—when degrees are equal, there is indeed a horizontal asymptote, and it equals the ratio of leading coefficients. Only when the numerator's degree exceeds the denominator's degree is there no horizontal asymptote. The student's analysis is completely correct, making D the right answer. Study tip: For rational functions, memorize the horizontal asymptote rules: same degrees → ratio of leading coefficients; numerator degree less → y = 0; numerator degree greater → no horizontal asymptote. Always identify the leading coefficient as the coefficient of the highest-degree term, regardless of term order.

Question 7

1. $y = \frac{3}{7}$
2. $y = \frac{1}{2}$
3. $y = -\frac{1}{2}$ (correct answer)
4. $y = -2$

Explanation: When analyzing the end behavior of rational functions, you need to focus on the highest-degree terms in both the numerator and denominator, since these dominate the function's behavior as x approaches positive or negative infinity. To find the horizontal asymptote, compare the degrees and leading coefficients. Both the numerator and denominator have degree 5, so the horizontal asymptote equals the ratio of the leading coefficients. The leading term in the numerator is $-2x^5$ (coefficient: -2), and the leading term in the denominator is $4x^5$ (coefficient: 4). Therefore, the horizontal asymptote is $y = \frac{-2}{4} = -\frac{1}{2}$, which is answer choice C. Let's examine why the other answers are incorrect. Answer A ($y = \frac{3}{7}$) incorrectly uses the coefficients of the $x^3$ term (3) and $x^2$ term (7), but these lower-degree terms don't determine end behavior. Answer B ($y = \frac{1}{2}$) takes the correct leading coefficients but forgets the negative sign in the numerator. Answer D ($y = -2$) uses only the leading coefficient of the numerator while ignoring the denominator entirely. Remember this key pattern: when the degrees of numerator and denominator are equal, the horizontal asymptote is always the ratio of the leading coefficients. Focus only on the highest-degree terms and ignore everything else when determining end behavior—this will save you time and prevent errors from getting distracted by coefficients that don't matter.

Question 8

1. As $x \to \pm\infty$, $f(x) \to 3$ (correct answer)
2. As $x \to \pm\infty$, $f(x) \to 0$
3. As $x \to +\infty$, $f(x) \to +\infty$ and as $x \to -\infty$, $f(x) \to -\infty$
4. As $x \to \pm\infty$, $f(x) \to \frac{3}{5}$

Explanation: For rational functions, end behavior is determined by comparing the degrees of the numerator and denominator. Both the numerator ($3x^4 - 2x^2 + 1$) and denominator ($x^4 + 5x - 7$) have degree 4. When degrees are equal, the horizontal asymptote is the ratio of leading coefficients: $\frac{3}{1} = 3$. Therefore, as $x \to \pm\infty$, $f(x) \to 3$. Choice B incorrectly assumes the numerator has lower degree. Choice C incorrectly assumes the numerator has higher degree. Choice D incorrectly uses coefficients from non-leading terms.

Question 9

1. $y = 1$ (correct answer)
2. $y = 0$
3. $y = \frac{1}{4}$
4. There is no horizontal asymptote

Explanation: When expanded, the numerator becomes $x^3 - 2x^2 - 5x + 6$ (leading coefficient 1) and the denominator becomes $x^3 + 3x^2 - 14x - 80$ (leading coefficient 1). Since both polynomials have degree 3, the horizontal asymptote is $y = \frac{1}{1} = 1$. Choice B applies the wrong degree comparison rule. Choice C might result from incorrectly identifying leading coefficients. Choice D incorrectly assumes that factored forms somehow eliminate horizontal asymptotes.