AP Precalculus -1.9 Rational Functions and Vertical Asymptotes- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -Link- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -Link- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
We need \( f \) to have a vertical asymptote at \( x = 4 \), not a hole.
For that, \( x = 4 \) must be a zero of the denominator with higher multiplicity than in the numerator.
Note \( 4 – x = -(x – 4) \), so \( (4-x)^3 = [-(x-4)]^3 = -(x-4)^3 \) up to a constant factor.
Check each option:
(A) Numerator: \( (x-4)^2 \) (multiplicity 2). Denominator: \( (4-x)^3 = -(x-4)^3 \) (multiplicity 3 in \( x-4 \) factor). → Denominator exponent 3 > numerator exponent 2 → vertical asymptote at \( x = 4 \). ✓
(B) Same as (A) in listing, probably mislabeled duplicate in original.
(C) Numerator: \( (x-4)^3 \) (multiplicity 3). Denominator: same \( (4-x)^3 \) → same multiplicity → no asymptote (hole if factors match exactly after simplification).
(D) Denominator does not have \( x-4 \) factor, so \( x = 4 \) not a vertical asymptote.
✅ Answer: (A)
▶️ Answer/Explanation
Vertical asymptotes of \( h(x) = \frac{f(x)}{g(x)} \) occur at real zeros of \( g(x) \) that are not also zeros of \( f(x) \) with at least the same multiplicity.
From the given answer reasoning: zeros of \( g \) are \( x = -2 \) and \( x = 3 \).
Check \( f(x) \): If \( f \) does not have zeros at \( x = -2 \) or \( x = 3 \), or has them with lower multiplicity, then vertical asymptotes exist there.
Given answer: \( x = -2 \) and \( x = 3 \) only ⇒ not \( x = 2 \), meaning \( g \) does not have a zero at \( x = 2 \) or \( f \) cancels it.
✅ Answer: (C) \( x = -2 \) and \( x = 3 \) only
▶️ Answer/Explanation
Vertical asymptote at \( x = -5 \) means denominator has zero at \( x=-5 \) that is not canceled by numerator.
Check each:
(A) Denominator \( 2(x-5) \) → zero at \( x=5 \) only ⇒ no asymptote at \( x=-5 \).
(B) Factor \( (x+5) \) cancels ⇒ hole, not asymptote.
(C) Denominator zero at \( x=5 \) and \( x=-2 \), not \( x=-5 \) ⇒ no asymptote at \( x=-5 \).
(D) Denominator zero at \( x=3 \) and \( x=-5 \); \( x=3 \) cancels, \( x=-5 \) remains ⇒ vertical asymptote at \( x=-5 \).
✅ Answer: (D)
▶️ Answer/Explanation
1. Analyze the limits at -5:
As \(x \to -5^-\) (approaching from the left), \(f(x)\) decreases without bound (\(-10 \to -100 \to -1000\)).
As \(x \to -5^+\) (approaching from the right), \(f(x)\) increases without bound (\(10 \to 100 \to 1000\)).
2. Interpret the behavior:
When the limit of the absolute value of a function approaches infinity as \(x\) approaches a constant \(c\), the graph has a vertical asymptote at \(x=c\).
✅ Answer: (C)
▶️ Answer/Explanation
A vertical asymptote occurs where the denominator is zero but the numerator is not zero (after canceling any common factors).
We want \( x = -5 \) to be a vertical asymptote, so \( (x+5) \) must be a factor of the denominator and not cancel completely with the numerator.
Check each option:
(A) Factor \( (x+5) \) in numerator, \( (x-5) \) in denominator — no factor \( (x+5) \) in denominator. So no asymptote at \( x=-5 \).
(B) Factor \( (x+5) \) appears in numerator and denominator — cancels, giving a hole at \( x=-5 \), not an asymptote.
(C) Factor \( (x+5) \) in numerator, denominator factors are \( (x-5) \) and \( (x+2) \) — no \( (x+5) \) in denominator, so no asymptote at \( x=-5 \).
(D) Factor \( (x+5) \) appears only in denominator (numerator has \( (x-5)(x-3) \)), so \( x=-5 \) makes denominator zero but numerator nonzero. This gives a vertical asymptote at \( x=-5 \).
✅ Answer: (D)
▶️ Answer/Explanation
The variable $x$ represents the percent of pure water expressed as a decimal ($1 = 100\%$).
The limit $\lim_{x \to 1^-}$ indicates $x$ is approaching $1$ ($100\%$ pure water) from the left.
As $x$ approaches $1$, the denominator $(x – 1)$ approaches $0$, causing the cost $C(x)$ to approach $\infty$.
The result $\infty$ means the cost increases rapidly without bound as purity nears $100\%$.
An infinite limit implies a vertical asymptote, meaning $x$ can never actually reach $1$.
Therefore, the cost to achieve absolute $100\%$ purity is physically and economically impossible.
This matches Option b, which correctly identifies $x$ as pure water and notes the impossibility of $100\%$.
▶️ Answer/Explanation
To find the domain, we must determine for which values of \( x \) the denominator is zero.
Set the denominator equal to zero: \( x^3 + 10x^2 + 16x = 0 \).
Factor out the common term \( x \): \( x(x^2 + 10x + 16) = 0 \).
Factor the quadratic expression: \( x(x + 8)(x + 2) = 0 \).
Solve for \( x \): \( x = 0 \), \( x + 8 = 0 \Rightarrow x = -8 \), or \( x + 2 = 0 \Rightarrow x = -2 \).
The function is undefined at these values, so the domain is all real numbers except \( 0, -8, \) and \( -2 \).
Correct Option: (C)
▶️ Answer/Explanation
▶️ Answer/Explanation
We are analyzing the limit of \( f(x) \) as \( x \to 1 \) from the right (since \( x > 1 \)).
First, evaluate the numerator at \( x = 1 \): \( -2(1+3) = -2(4) = -8 \).
Next, consider the denominator \( (x-1) \). As \( x \) approaches 1 from the right, \( (x-1) \) is a very small positive number.
We are dividing a negative constant (\( -8 \)) by a positive number approaching zero (\( 0^+ \)).
The result will be a negative number with an infinitely large magnitude.
Mathematically, \( \lim_{x \to 1^+} \frac{-8}{x-1} = -\infty \).
Therefore, the output values decrease without bound.
Correct Option: (A)
▶️ Answer/Explanation
A vertical asymptote occurs when the denominator equals zero and the factor does not completely cancel with the numerator. Since \(4 – x = -(x – 4)\), we rewrite: \[ (4-x)^3 = -(x-4)^3 \] For option (A), simplifying gives: \[ f(x) = -\frac{x+1}{x-4} \] Because a factor of \((x-4)\) remains in the denominator, there is a vertical asymptote at \(x = 4\).
1. A vertical asymptote occurs where the denominator is zero after simplification.
2. Note that \(4-x = -(x-4)\).
3. In (A), cancel \((x+1)^2\) and simplify powers of \((x-4)\).
4. This gives \(f(x) = -\frac{x+1}{x-4}\).
5. Since \((x-4)\) remains in the denominator, it does not fully cancel.
6. Therefore, \(x=4\) is a vertical asymptote.
7. In (B) and (C), \((x-4)\) cancels completely → hole, not asymptote.
8. In (D), denominator is zero at \(x=-4\), not \(x=4\).
▶️ Answer/Explanation
The correct answer is (C).
A vertical asymptote for a rational function \(h(x) = \frac{f(x)}{g(x)}\) occurs where the denominator \(g(x)\) is zero and the numerator \(f(x)\) is non-zero.
From the graph of \(g\), the curve intersects the x-axis at \(x = -2\) and \(x = 3\), so \(g(-2) = 0\) and \(g(3) = 0\).
From the graph of \(f\), at these x-values, the function is non-zero: \(f(-2) \neq 0\) (negative) and \(f(3) \neq 0\) (positive).
Since the denominator is zero and the numerator is non-zero at \(x = -2\) and \(x = 3\), both lines are vertical asymptotes.
Note that at \(x = 2\), \(f(2) = 0\) but \(g(2) \neq 0\), which results in an x-intercept for \(h(x)\), not an asymptote.
▶️ Answer/Explanation
2. The numerator is already factored: \((x+10)(x^2 – 10x + 100)\).
3. The common factor \((x+10)\) cancels, so \(g(x) = \frac{x^2 – 10x + 100}{x-10}\), where \(x \neq -10\).
4. Since the factor cancels, \(x = -10\) creates a removable discontinuity (hole).
5. The remaining denominator \(x-10 = 0\) gives a vertical asymptote at \(x = 10\).
6. Therefore, the graph has a hole at \(x = -10\), so option (B) is correct.
▶️ Answer/Explanation
Explanation:
A vertical asymptote occurs where the denominator equals zero and the factor does not cancel with the numerator.
1. A vertical asymptote at \(x=-5\) requires a factor \((x+5)\) in the denominator.
2. In (A), \((x-5)\) cancels, leaving \( \frac{x+5}{2} \); no asymptote at \(x=-5\).
3. In (B), \((x+5)\) cancels completely, creating a hole at \(x=-5\), not an asymptote.
4. In (C), the denominator does not contain \((x+5)\); no asymptote at \(x=-5\).
5. In (D), \((x-3)\) cancels, leaving \( \frac{x-5}{x+5} \).
6. Since \((x+5)\) remains in the denominator, \(x=-5\) gives a vertical asymptote.
▶️ Answer/Explanation
(A)
i. The graph of \( f(x) \) shows that \( f(2) \approx 1 \), so \( f^{-1}(1) \approx 2 \). From the table, \( g(2) = 1 \). Thus, \( g(f^{-1}(1)) \approx 1 \).
Since \( f(x) \) is continuous and increasing, the inverse is defined for values in its range.
The estimate is a single value, \( 1 \), as the graph aligns precisely with the point. No other values exist due to the strictly increasing nature of \( f(x) \).
ii. The graph of \( f(x) \) crosses zero at \( x=1 \), so \( f(1)=0 \), and from the table, \( g(1)=0 \).
Thus, \( h(1) \) is undefined (\( 0/0 \) form).
The limit as \( x \to 1 \) of \( h(x) \) exists and equals \( 1 \), since both \( g(x) \) and \( f(x) \) resemble \( \log_2(x) \), making \( h(x)=1 \) elsewhere.
This indicates a removable discontinuity at \( x=1 \). No other discontinuities for \( x>0 \), as \( f(x) \neq 0 \) elsewhere in the domain.
The location is \( x=1 \), type: removable.
(B)
i. As \( x \to \infty \), \( \ln(x) \to \infty \), so \( 4.99 – \ln(x) \to -\infty \), and \( \frac{1}{4.99 – \ln(x)} \to 0^- \).
The term \( e^{2 \sin(\sqrt{x})} \) oscillates between \( e^{-2} \) and \( e^{2} \), remaining bounded.
By the squeeze theorem, since the amplitude approaches \( 0 \), \( \lim_{x \to \infty} j(x) = 0 \).
The end behavior is that \( j(x) \) approaches \( 0 \). The limit exists.
ii. As \( x \to 0^+ \), \( \ln(x) \to -\infty \), so \( 4.99 – \ln(x) \to +\infty \), and \( \frac{1}{4.99 – \ln(x)} \to 0^+ \).
Also, \( \sqrt{x} \to 0^+ \), \( \sin(\sqrt{x}) \to 0^+ \), so \( e^{2 \sin(\sqrt{x})} \to e^0 = 1 \).
Thus, \( j(x) \to 0 \cdot 1 = 0 \).
The limit is \( \lim_{x \to 0^+} j(x) = 0 \).
No vertical asymptote at \( x=0 \), as the function approaches a finite value, not \( \pm \infty \).
(C)
The table shows \( g(x) \) values: at \( x=0.5, 1, 2, 4, 7, 8 \), \( g(x)=-1, 0, 1, 2, 2.807, 3 \).
When \( x \) doubles (\( 0.5 \) to \( 1 \), \( 1 \) to \( 2 \), \( 2 \) to \( 4 \), \( 4 \) to \( 8 \)), \( \Delta g = +1 \) consistently.
This pattern matches logarithmic functions, where outputs increase by a constant for multiplicative input changes.
Linear would require constant \( \Delta g \) for constant \( \Delta x \), but \( \Delta x \) varies while \( \Delta g =1 \) for doublings.
Quadratic would show constant second differences, but here first differences are not constant.
Exponential would show constant ratios in outputs for additive input changes, which does not fit.
Thus, best modeled by a logarithmic function like \( g(x) = \log_2(x) \).