📝 Rational functions domain asymptotes (13 MCQs)

📖 Explanation: A rational function is precisely a quotient of two polynomial expressions, written as $P(x)/Q(x)$ where $Q(x)\neq0$. This definition matches option B, whereas the other options describe unrelated properties such as continuity or asymptotic behavior.

📖 Explanation: Factoring the numerator gives $(x-2)(x+2)$. Cancelling the common factor with the denominator leaves $f(x)=x+2$ for all $x\neq2$. The point $x=2$ is omitted, creating a removable hole rather than an infinite break, so option C is correct.

📖 Explanation: The denominator $x^{2}-9$ factors to $(x-3)(x+3)$. The function is undefined wherever this product equals zero, i.e., at $x=3$ and $x=-3$. All other real numbers are allowed, making option A the correct description of the domain.

📖 Explanation: Near the vertical asymptote $x=2$, evaluate the sign of the numerator and denominator. For $x$ just larger than 2, the numerator $x+1$ is positive, and the denominator $x-2$ is a small positive number, giving a positive quotient. Hence the function is positive to the right of the asymptote, matching option A.

📖 Explanation: The degrees of numerator and denominator are both 2, so the horizontal asymptote is the ratio of leading coefficients: $\frac{3}{2}$. Therefore the statement that the horizontal asymptote is $y=\frac{3}{2}$ is true, which corresponds to option A.

📖 Explanation: Horizontal asymptotes depend on the leading terms. The original function has leading coefficient ratio $2/1 = 2$. Among the choices, only $\frac{2x^{2}+1}{x^{2}+4}$ retains that ratio, giving the same horizontal asymptote $y=2$. Thus option A is correct.

📖 Explanation: The numerator $x^{2}-9$ factors as $(x-3)(x+3)$. Cancelling the factor $(x-3)$ with the denominator leaves $p(x)=x+3$ for all $x\neq3$. The point at $x=3$ is removed, producing a hole rather than a vertical asymptote, so option C is correct.

📖 Explanation: When $x$ grows without bound, the highest-degree terms dominate. The ratio of the leading coefficients $\frac{3}{2}$ determines the limit, because lower-degree terms become negligible. Hence $\displaystyle\lim_{x\to\infty}\frac{3x^{3}-x}{2x^{3}+5x^{2}}=\frac{3}{2}$, which matches option A.

📖 Explanation: A removable discontinuity occurs when both numerator and denominator vanish at the same point, allowing cancellation. Setting the numerator zero at $x=a$ gives $a^{2}-a=0$, factoring to $a(a-1)=0$. Thus $a$ must be 0 or 1. Option C lists exactly these values, making it the correct choice.

📖 Explanation: Differentiating $s(x)=\frac{x^{2}-4}{x^{2}-9}$ yields s&#039;(x)=\frac{-10x}{(x^{2}-9)^{2}}. The denominator is always positive except at the undefined points $x=\pm3$. Therefore the sign of s&#039;(x) is opposite to the sign of $x$. The function is increasing where $x<0$; the interval $(-\infty,-3)$ satisfies this, so option A is correct.

📖 Explanation: The end behavior of a rational function is governed by the degrees of its numerator and denominator. When these degrees are equal, the ratio of the leading coefficients determines a constant value approached as $x$ tends to $\pm\infty$. This explains why such functions have a horizontal asymptote, matching option B.

📖 Explanation: The original expression simplifies by factoring $x^{2}-1=(x-1)(x+1)$ and cancelling the common factor $(x-1)$. This reveals a hole at $x=1$ and reduces the function to $f(x)=x+1$ elsewhere. Recognizing and removing the factor is essential for accurate graphing, so option B is correct.

📖 Explanation: A vertical asymptote occurs where the denominator equals zero while the numerator remains non‑zero. For $f(x)=\frac{2}{x-5}$, the denominator vanishes at $x=5$, producing an infinite blow‑up there. Hence the vertical asymptote is the line $x=5$, which is option B.