▶️ Answer/Explanation
The average rate of change on $[a, b]$ is $\frac{W(b) – W(a)}{b – a}$.
For $0 \le t < 20$, $W(t) = 125 – 0.2t^2$, which is a downward-opening parabola.
The rate of change on $[5, 10]$ is $\frac{W(10)-W(5)}{5} = \frac{105-120}{5} = -3$.
The rate of change on $[10, 15]$ is $\frac{W(15)-W(10)}{5} = \frac{80-105}{5} = -5$.
Since $-5 < -3$, statement (B) is correct: the rate on $[10, 15]$ is less than on $[5, 10]$.
For $20 \le t \le 22$, $W(t)$ is linear with a constant slope of $-20$.
Thus, the rates on $[20, 21]$ and $[21, 22]$ are equal, making (D) incorrect.
▶️ Answer/Explanation
The general term of the expansion $(a + b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
For $p(x) = (x – 3)^5$, we have $n = 5$, $a = x$, and $b = -3$.
To find the $x^3$ term, we set the exponent of $x$ to $3$, so $n – k = 3$.
Solving for $k$ gives $5 – k = 3$, which means $k = 2$.
The term is $\binom{5}{2} (x)^3 (-3)^2$.
The binomial coefficient $\binom{5}{2} = \frac{5 \cdot 4}{2 \cdot 1} = 10$.
The coefficient of the $x^3$ term is therefore $10 \cdot (-3)^2$.
This matches option (B).
▶️ Answer/Explanation
The correct answer is (D).
The binomial theorem states $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k$.
For $(x + 3y)^5$, the power $n = 5$, so the expansion has $n+1 = 6$ terms.
The coefficients are found using the 5th row of Pascal’s Triangle: $1, 5, 10, 10, 5, 1$.
The powers of $x$ decrease from $5$ to $0$, while powers of $(3y)$ increase from $0$ to $5$.
Applying these coefficients yields $1x^5 + 5x^4(3y)^1 + 10x^3(3y)^2 + 10x^2(3y)^3 + 5x^1(3y)^4 + 1(3y)^5$.
This matches the expression provided in option (D).
▶️ Answer/Explanation
Concavity is determined by the behavior of the rate of change (the first derivative, $f’$).
A function is concave down on an interval if its rate of change is decreasing.
In the interval $0 < x < 1$, the rate of change is increasing, so the graph is concave up.
In the interval $1 < x < 2$, the rate of change is constant, so the graph is linear.
In the interval $2 < x < 3$, the rate of change is decreasing, which fits the definition of concave down.
In the interval $3 < x < 4$, the rate of change is constant, so the graph is linear.
Therefore, the graph of $f$ is concave down only on the interval $2 < x < 3$.
The correct option is (C).
▶️ Answer/Explanation
Explanation:
As the car moves with constant speed around a circular track, its horizontal position relative to the wall varies sinusoidally with time. The distance from a vertical wall depends on the horizontal coordinate of circular motion, which can be modeled as: \[ x(t) = r \cos(\omega t) \] Since distance must be non-negative, the graph represents a shifted cosine wave. Over three revolutions, the pattern repeats three times smoothly. Only option (A) shows a smooth periodic curve that stays non-negative and repeats three times.
1. Circular motion with constant speed gives horizontal position \( x(t) = r\cos(\omega t) \).
2. Distance from the wall depends on this horizontal coordinate.
3. The function is periodic with period \( T = \dfrac{2\pi}{\omega} \).
4. Three revolutions produce three identical cycles.
5. Distance cannot be negative, so the curve stays above the time axis.
6. Only graph (A) shows a smooth repeating periodic wave three times.
▶️ Answer/Explanation
The rate of change of a function is represented by the slope of the tangent line to the graph.
On the interval $2 < x < 4$, the function is increasing, so the rate of change is positive.
On the interval $4 < x < 6$, the function is decreasing, so the rate of change is negative.
Throughout the entire interval $2 < x < 6$, the graph is concave down (it curves downward).
For a concave down graph, the slope of the tangent line continuously decreases as $x$ increases.
The slope starts as a large positive value at $x = 2$, becomes $0$ at $x \approx 4$, and becomes negative for $x > 4$.
Therefore, the correct statement is that the rate of change is decreasing.
Correct Option: (D)
▶️ Answer/Explanation
The initial height at $t = 0$ is $h(0) = 15.24$ meters, but the positive linear term $+4.4t$ indicates the ball is thrown upward first.
The time to reach maximum height is $t = \frac{-b}{2a} = \frac{-4.4}{2(-4.9)} \approx 0.449$ seconds.
The maximum height is $h(0.449) = -4.9(0.449)^2 + 4.4(0.449) + 15.24 \approx 16.228$ meters.
To find when it hits the ground, set $h(t) = 0$ and solve $-4.9t^2 + 4.4t + 15.24 = 0$ using the quadratic formula.
The positive root is $t \approx 2.269$ seconds, which is the total time from the window to the ground.
The time from the maximum height to the ground is $2.269 – 0.449 = 1.820$ seconds.
Therefore, the height increases to $16.228$ m, then decreases for $1.820$ seconds until it hits the ground.
Correct Option: (C)
▶️ Answer/Explanation
The “rate of change of $f(x)$” corresponds to the first derivative, $f'(x)$.
On the interval $3 < x < 4$, the rate of change is positive, meaning $f'(x) > 0$.
A positive first derivative indicates that the function $f$ is increasing.
The table also specifies that the rate of change is decreasing.
This implies the second derivative, $f”(x)$, is negative because $f'(x)$ is falling.
A negative second derivative ($f”(x) < 0$) means the graph of $f$ is concave down.
Combining these facts, $f$ is increasing and concave down, which is option (A).
▶️ Answer/Explanation
The rate of change of depth $\frac{dh}{dt}$ is inversely proportional to the cross-sectional area of the vase.
A concave up graph indicates that $\frac{d^2h}{dt^2} > 0$, meaning the vase must be narrowing as height increases.
Vase (D) starts with a wide base that curves inward, causing the water level to rise faster and faster (concave up).
A steady and steep increase implies a constant, high rate of change, which occurs in a narrow, cylindrical neck.
Vase (D) features a long, uniform, and very narrow neck, which produces a linear and steep depth-time relationship.
Vase (B) lacks the uniform vertical neck required for a “steady” increase, as its width continues to change at the top.
Vase (A) and (C) do not match the concave up start because they begin by widening at the bottom.
Therefore, Vase (D) is the correct choice as it perfectly aligns with both geometric descriptions in the graph.
▶️ Answer/Explanation
The correct option is (B).
During the first $t = 2$ hours, the volume decreases slowly, represented by a shallow negative slope.
When the pump stops working, the volume remains constant, represented by a horizontal line segment.
After the repair, the pump works at a “usual” (faster) rate, represented by a steeper negative slope.
The graph must intersect the $x$-axis (volume $= 0$) exactly at $t = 6$ hours.
Graph (A) is incorrect because it ends at $t = 10$ hours.
Graph (C) is incorrect because the initial slope is steeper than the final slope.
Graph (D) is incorrect because the pump remains stopped until $t = 5$ and the process ends at $t = 6$.
▶️ Answer/Explanation
The function $N(t)$ represents the total number of customers, so $N(t) = \int_{0}^{t} R(x) dx$.
Since $R(t)$ is a rate of arrival, $R(t) \ge 0$, which means $N(t)$ must be a non-decreasing function.
The derivative of the total accumulation is the rate, so $N'(t) = R(t)$ and $N”(t) = R'(t)$.
When $R(t)$ is increasing ($0 < t < 4$ and $8 < t < 12$), $N(t)$ is concave up ($N”(t) > 0$).
When $R(t)$ is decreasing ($4 < t < 8$ and $12 < t < 16$), $N(t)$ is concave down ($N”(t) < 0$).
Graph (D) starts at $(0,0)$, always increases, and switches concavity at $t = 4, 8, 12$.
Therefore, the correct graph is (D).
▶️ Answer/Explanation
The volume of a sphere is given by $V = \frac{4}{3}\pi r^{3}$, so the radius is $r = \sqrt[3]{\frac{3V}{4\pi}}$.
Since the radius $r$ increases as volume $V$ increases, the graph must have a positive slope.
The problem states the radius increases at a decreasing rate with respect to volume.
This means the slope $\frac{dr}{dV}$ is positive but getting smaller, indicating a concave down shape.
Graph (A) shows a decreasing radius, which contradicts the expansion of the sphere.
Graph (B) shows an increasing radius with a decreasing rate (concave down), matching the description.
Graph (C) shows an increasing rate (concave up), and Graph (D) shows a decreasing radius.
Therefore, the correct graph depicting this relationship is (B).
▶️ Answer/Explanation
The graph shows an S-shaped (sigmoidal) curve.
Initially, the slope is increasing, meaning the depth rises faster as the container narrows.
This indicates the bottom of the container is wide and gets narrower.
In the middle, the slope is steepest where the container is narrowest.
Then, the slope decreases, meaning the depth rises slower as the container widens again.
This matches a container that is wide at the bottom, narrow in the middle, and wide at the top.
Therefore, the correct container is (B).
▶️ Answer/Explanation
(C) \( x^4 + 8x^3 + 24x^2 + 32x + 16 \)
Using the Binomial Theorem, \[ (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \] Substituting \( a = x \) and \( b = 2 \): \[ (x+2)^4 = x^4 + 4x^3(2) + 6x^2(2^2) + 4x(2^3) + 2^4 \] \[ = x^4 + 8x^3 + 24x^2 + 32x + 16 \] Thus, the correct answer is (C).
1. Apply the binomial formula: \( (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \).
2. Substitute \( a = x \), \( b = 2 \).
3. Compute each term: \( 4x^3(2) = 8x^3 \).
4. Compute \( 6x^2(2^2) = 6x^2(4) = 24x^2 \).
5. Compute \( 4x(2^3) = 4x(8) = 32x \).
6. Compute \( 2^4 = 16 \).
7. Combine all terms: \( x^4 + 8x^3 + 24x^2 + 32x + 16 \).
8. Therefore, the equivalent expression is option (C).
▶️ Answer/Explanation
▶️ Answer/Explanation
2. Compute each term: \( 4(3)=12 \), \( 6(9)=54 \), \( 4(27)=108 \), \( 3^4=81 \).
3. Therefore, \( (x+3)^4 = x^4 + 12x^3 + 54x^2 + 108x + 81 \).
4. Comparing with \( x^4 + ax^3 + bx^2 + cx + d \):
5. \( a=12 \), \( b=54 \), \( c=108 \), \( d=81 \).
6. The greatest value among \( 12, 54, 108, 81 \) is \( 108 \).
7. Hence, the greatest coefficient is \( c \).
▶️ Answer/Explanation
1. Factor (D): \(x^2-4=(x-2)(x+2)\), and \(x^2-x-6=(x-3)(x+2)\).
2. The common factor \((x+2)\) cancels, producing a hole at \(x=-2\).
3. The simplified function is \( \dfrac{x-2}{x-3} \), so \(x=3\) is a vertical asymptote.
4. The numerator gives an \(x\)-intercept at \(x=2\).
5. Since numerator and denominator have equal degree, the horizontal asymptote is \(y=1\).
6. Other choices have no common factor, so they have no hole.
7. Therefore, only choice (D) satisfies all four conditions.
▶️ Answer/Explanation
1. By the Remainder Theorem, compute \( f(-3) \).
\( f(-3) = 2(-27) – 3(9) – 23(-3) + 12 = -54 – 27 + 69 + 12 = 0. \)
2. Since \( f(-3) = 0 \), the remainder is \(0\).
3. Perform synthetic division by \(-3\): the quotient is \( 2x^2 – 9x + 4 \).
4. Compute the discriminant:
\( \Delta = (-9)^2 – 4(2)(4) = 81 – 32 = 49. \)
5. Since \( \Delta = 49 > 0 \), the quadratic factors over the real numbers.
6. Factor: \( 2x^2 – 9x + 4 = (2x – 1)(x – 4). \)
Therefore, the remainder is \(0\) and the quotient factors into real linear factors.
▶️ Answer/Explanation
(D)
1. Divide \(6x^4\) by \(3x^2\) to get \(2x^2\); subtract to obtain \(9x^3\).
2. Divide \(9x^3\) by \(3x^2\) to get \(3x\); subtract to obtain \(6x^2\).
3. Divide \(6x^2\) by \(3x^2\) to get \(2\); subtract to obtain remainder \(7x – 5\).
4. Since the remainder is \(7x – 5 \neq 0\), \(g(x)\) is not a factor of \(f(x)\).
5. Degree of \(f(x)\) is \(4\); degree of \(g(x)\) is \(2\).
6. The degree difference is \(2\), so the quotient is quadratic (not linear).
7. A slant asymptote occurs only when the degree difference is \(1\).
8. Therefore, the correct statement is (D).
▶️ Answer/Explanation
1. Compute first differences: \( -2 – (-3) = 1 \), \( 1 – (-2) = 3 \), \( 6 – 1 = 5 \), \( 13 – 6 = 7 \).
2. First differences are \( 1, 3, 5, 7 \) (not constant).
3. Compute second differences: \( 3 – 1 = 2 \), \( 5 – 3 = 2 \), \( 7 – 5 = 2 \).
4. Second differences are constant and equal to \( 2 \).
5. A constant second difference indicates a quadratic function.
6. Therefore, the correct statement is that \( g \) is best modeled by a quadratic function.
7. Hence, option (D) is correct.
▶️ Answer/Explanation
(A) The interval from \( A \) to \( B \)
1. A function is increasing where \( f'(x) > 0 \), meaning the graph rises from left to right.
2. From \( A \) to \( B \), the graph rises to a local maximum at \( B \), so \( f'(x) > 0 \).
3. The point \( C \) is labeled as the point of inflection, so concavity changes at \( C \).
4. Therefore, for all \( x < C \), the graph is concave down (\( f”(x) < 0 \)).
5. Since the interval \( A \) to \( B \) lies entirely before \( C \), it is concave down.
6. Hence, \( f \) is increasing and concave down on \( (A, B) \).
▶️ Answer/Explanation
(D) \(0 < t < 3\) and \(8 < t < 10\)
Velocity is decreasing where the graph of \(v(t)\) has a negative slope. From \(t=0\) to \(t=3\), the graph slopes downward continuously. From \(t=3\) to \(t=5\), the slope is positive (velocity increasing). From \(t=5\) to \(t=8\), the slope is zero (constant velocity). From \(t=8\) to \(t=10\), the slope is negative again. Therefore, velocity decreases only on \(0 < t < 3\) and \(8 < t < 10\).
1. Velocity decreases when the slope of \(v(t)\) is negative.
2. From \(0 < t < 3\), the line segment slopes downward ⇒ decreasing.
3. From \(3 < t < 5\), the slope is positive ⇒ increasing.
4. From \(5 < t < 8\), the slope is \(0\) ⇒ constant velocity.
5. From \(8 < t < 10\), the slope is negative ⇒ decreasing.
6. Hence, the correct intervals are \(0 < t < 3\) and \(8 < t < 10\).
▶️ Answer/Explanation
(A) \( (3,6) \) only
2. Thus the graph must be increasing and concave down.
3. From \( t = 0 \) to \( t = 6 \), the function increases.
4. However, from \( (0,3) \), the slope increases (concave up).
5. From \( (3,6) \), the slope decreases to \( 0 \) (concave down).
6. From \( (18,30) \), the function increases but is concave up.
7. Therefore, the only interval satisfying both conditions is \( (3,6) \).
▶️ Answer/Explanation
(C)
1. The rate of change of \( g \) at a point is the slope \( g'(x) \).
2. At \( A \), a relative maximum, \( g'(A) = 0 \).
3. At \( D \), near a minimum, the slope is approximately \( 0 \) (slightly positive).
4. At \( B \), the function is decreasing, so \( g'(B) < 0 \), but not steepest.
5. At \( C \), the only point of inflection, the graph changes concavity.
6. At an inflection point on a decreasing curve, the slope is most negative.
7. Therefore, \( g'(C) \) is the smallest (least) value.
8. Hence, the rate of change is least at \( C \).
▶️ Answer/Explanation
(D) \( [t_C, t_D] \) and \( [t_D, t_E] \)
1. The rate of change of \( g \) is \( g'(x) \). It is decreasing where \( g”(x) < 0 \).
2. \( g”(x) < 0 \) corresponds to the graph being concave down.
3. The graph has exactly one inflection point at \( C \), so concavity changes only at \( t_C \).
4. From \( t_A \) to \( t_C \), the graph is concave up (slope increasing).
5. From \( t_C \) to \( t_E \), the graph is concave down (slope decreasing).
6. Therefore, the rate of change of \( g \) is decreasing on \( [t_C, t_D] \) and \( [t_D, t_E] \).
7. Hence, the correct answer is (D).
▶️ Answer/Explanation
The correct answer is (C).
Step 1: Analyze the behavior of the graph at the zero \( C \). At this point, the graph touches the \( x \)-axis and turns back down, remaining on the same side (negative values) rather than crossing it.
Step 2: Determine the relationship between graph behavior and multiplicity. If a graph crosses the \( x \)-axis at a zero (like at points \( A \) and \( B \)), the multiplicity is odd (typically \( 1 \)). If the graph touches the axis and turns around (does not change sign), the multiplicity is even (e.g., \( 2 \)).
Step 3: Consider the degree of the polynomial. The problem states it is a quartic function (degree \( 4 \)). By the Fundamental Theorem of Algebra, the sum of the multiplicities of all zeros must be \( 4 \).
Step 4: Since there are three visible real roots, and \( A \) and \( B \) are simple crossings (multiplicity \( 1 \)), the remaining degrees of freedom must be assigned to \( C \). Calculating the sum: \( 1 (\text{at } A) + 1 (\text{at } B) + 2 (\text{at } C) = 4 \).
Conclusion: The zero at \( C \) must have a multiplicity of \( 2 \) because the polynomial does not change signs at this point, consistent with the graph being tangent to the axis.
▶️ Answer/Explanation
2. A quadratic function can have at most \(2\) real zeros.
3. Points of inflection occur where \(p”(x)=0\) and concavity changes.
4. From the graph, the curve changes concavity once between the left minimum and the local maximum.
5. It changes concavity again between the local maximum and the right minimum.
6. No additional concavity change is visible on either end of the graph.
7. Therefore, the graph has exactly \(2\) points of inflection.
▶️ Answer/Explanation
(C) The degree of \( f \) is odd, and the leading coefficient is negative.
The end behavior of a polynomial is determined by its leading term \( a x^n \). If the limits at \( -\infty \) and \( \infty \) are opposite in sign, the degree must be odd. Since \( f(x) \to -\infty \) as \( x \to \infty \), the leading coefficient must be negative.
Line 2: If \( n \) is even, both ends of the graph go in the same direction.
Line 3: Here, \( \lim_{x \to -\infty} f(x) = \infty \) and \( \lim_{x \to \infty} f(x) = -\infty \), so the ends differ.
Line 4: Therefore, the degree \( n \) must be odd.
Line 5: For an odd degree polynomial, if \( a < 0 \), then \( f(x) \to -\infty \) as \( x \to \infty \).
Line 6: This matches the given condition, so the leading coefficient is negative.
Line 7: Hence, the correct answer is (C).
▶️ Answer/Explanation
(A)
1. The end behavior of a polynomial depends on its highest-degree term.
2. In (A), the leading term is \( -2x^4 \).
3. Since \( x^4 \) is even, \( x^4 \to +\infty \) as \( x \to -\infty \).
4. Multiplying by \( -2 \) gives \( -2x^4 \to -\infty \).
5. In (B) and (C), the leading terms are odd powers with negative coefficients, giving \( +\infty \) as \( x \to -\infty \).
6. In (D), the leading term is \( 2x^4 \), which gives \( +\infty \).
7. Therefore, only option (A) satisfies \( \lim_{x \to -\infty} g(x) = -\infty \).
▶️ Answer/Explanation
(B) \(\displaystyle \lim_{x \to \infty} f(x) = -\infty\)
1. The highest-degree term is \( -2x^{7} \).
2. End behavior of a polynomial is determined by its leading term.
3. As \( x \to \infty \), \( x^{7} \to \infty \).
4. Therefore, \( -2x^{7} \to -\infty \).
5. Lower-degree terms \( 5x^{4}, 6x^{2}, -3 \) become insignificant compared to \( x^{7} \).
6. Hence, \( f(x) \to -\infty \) as \( x \to \infty \).
7. Therefore, the correct answer is option (B).
▶️ Answer/Explanation
(B)
The leading term of the polynomial is \(5x^6\). Since the degree is even (\(6\)) and the leading coefficient \(5\) is positive, the function rises to positive infinity on both ends. Therefore, \[ \lim_{x \to -\infty} f(x) = \infty \quad \text{and} \quad \lim_{x \to \infty} f(x) = \infty. \]
1. The highest-degree term is \(5x^6\).
2. End behavior depends only on the leading term.
3. The degree \(6\) is even.
4. The leading coefficient \(5 > 0\).
5. For even degree with positive coefficient, both ends rise.
6. Thus, \(f(x) \to \infty\) as \(x \to \pm\infty\).
7. Therefore, the correct answer is (B).
▶️ Answer/Explanation
The leading term of the polynomial is \( -4x^5 \). Since the coefficient \( -4 \) is negative and the degree \( 5 \) is odd, the graph rises to \( \infty \) as \( x \to -\infty \) and falls to \( -\infty \) as \( x \to \infty \). Therefore, option (B) correctly describes the end behavior.
1. The given polynomial is \( p(x) = -4x^5 + 3x^2 + 1 \).
2. The leading term is \( -4x^5 \).
3. The coefficient \( -4 \) is negative.
4. The degree \( 5 \) is odd.
5. For an odd degree with negative leading coefficient: left end rises, right end falls.
6. Thus, \( \lim_{x \to -\infty} p(x) = \infty \) and \( \lim_{x \to \infty} p(x) = -\infty \).
7. Hence, the correct statement is (B).
▶️ Answer/Explanation
Step 1: Analyze End Behavior
The graph goes upwards to \( +\infty \) on both the far left and far right sides. This indicates that the polynomial must have an even degree (like \( x^2, x^4 \), etc.) and a positive leading coefficient. Options (A) and (B) are polynomials of degree 3 (multiplying three \( x \) terms), which is odd. Therefore, (A) and (B) are incorrect.
Step 2: Analyze Roots and Multiplicity
We look at the behavior of the graph at the x-intercepts (roots):
• At the negative root (approx \( x = -8 \)), the graph crosses the x-axis. This corresponds to a linear factor, \( (x+8) \).
• At the smaller positive root (approx \( x = 1 \)), the graph crosses the x-axis. This corresponds to a linear factor, \( (x-1) \).
• At the larger positive root (approx \( x = 5 \)), the graph touches the x-axis and turns back up (tangent to the axis). This indicates a root with an even multiplicity, usually 2. This corresponds to a squared factor, \( (x-5)^2 \).
Step 3: Match with Options
Checking the remaining even-degree options:
• Option (D) has \( (x+5)^2 \), which implies a touch at \( x = -5 \). The graph crosses at the negative value, so this is incorrect.
• Option (C) is \( 0.25(x-5)^2(x-1)(x+8) \). This matches all our observations: degree 4, crosses at \( -8 \) and \( 1 \), and touches at \( 5 \).
Conclusion
The correct expression is (C).
▶️ Answer/Explanation
(D) \( t = 5.505 \)
The rate changes from increasing to decreasing at a local maximum of \( R(t) \). This occurs when \( R'(t) = 0 \) and the derivative changes from positive to negative.
1. Differentiate: \( R'(t) = 0.09t^2 – 1.692t + 6.587 \).
2. Set derivative equal to zero: \( 0.09t^2 – 1.692t + 6.587 = 0 \).
3. Solve using quadratic formula.
4. Solutions: \( t \approx 5.505 \) and \( t \approx 13.295 \).
5. Since the quadratic opens upward, the smaller root gives the local maximum.
6. Therefore, the rate changes from increasing to decreasing at \( t = 5.505 \).
▶️ Answer/Explanation
(C) \(p(-3) = 4\) is a relative minimum.
Since \(p\) is an odd function, it satisfies \(p(-x) = -p(x)\). Given \(p(3) = -4\), we obtain \[ p(-3) = -p(3) = -(-4) = 4. \] An odd function has rotational symmetry about the origin. Therefore, a relative maximum at \((3,-4)\) corresponds to a relative minimum at \((-3,4)\).
1. Since \(p\) is odd, \(p(-x) = -p(x)\).
2. Given \(p(3) = -4\).
3. Then \(p(-3) = -p(3) = -(-4) = 4\).
4. Odd functions are symmetric about the origin.
5. A relative maximum at \((3,-4)\) reflects to \((-3,4)\).
6. The reflected point must be a relative minimum.
7. Therefore, \(p(-3) = 4\) is a relative minimum.
▶️ Answer/Explanation
The correct answer is (B) 3.
From the table, we identify the extrema for \(x \ge 0\). The function decreases to a minimum value of \(f(2) = -3\) and then increases to a value of \(f(4) = 2\).
Since \(f\) is an odd function, it satisfies the property \(f(-x) = -f(x)\). This means the graph is symmetric with respect to the origin.
A local minimum on the positive side corresponds to a local maximum on the negative side. Therefore, the minimum at \(x = 2\) implies a maximum at \(x = -2\) given by \(f(-2) = -f(2) = -(-3) = 3\).
We must also check the endpoints. We know \(f(4) = 2\), which implies \(f(-4) = -f(4) = -2\).
Comparing the highest points on the entire domain \([-4, 4]\): the maximum value at \(x=4\) is \(2\), and the maximum value at \(x=-2\) is \(3\).
Thus, the absolute maximum value of \(f\) is \(3\).
▶️ Answer/Explanation
First factor the quadratic term: \[ x^2 – 2x – 15 = (x – 5)(x + 3). \]
Therefore, \[ p(x) = (x + 3)(x – 5)(x + 3) = (x + 3)^2 (x – 5). \]
The zeros are: \[ x = -3 \quad (\text{multiplicity } 2), \qquad x = 5. \]
Hence, there are exactly two distinct real zeros: \( -3 \) and \( 5 \).
1. Given \( p(x) = (x+3)(x^2 – 2x – 15) \).
2. Factor \( x^2 – 2x – 15 = (x-5)(x+3) \).
3. Substitute: \( p(x) = (x+3)(x-5)(x+3) \).
4. Simplify: \( p(x) = (x+3)^2 (x-5) \).
5. Zeros are \( x=-3 \) (double root) and \( x=5 \).
6. Thus there are \( 2 \) distinct real zeros.
7. Therefore, the correct option is (A).
▶️ Answer/Explanation
The correct answer is (B).
1. The problem states that \( a \) and \( b \) are real constants, which means the polynomial \( k(x) \) has real coefficients.
2. According to the Complex Conjugate Root Theorem, if a polynomial with real coefficients has a complex zero \( a + bi \), its conjugate \( a – bi \) must also be a zero.
3. Since \( -0.478 – 0.801i \) is given as a zero, its conjugate \( -0.478 + 0.801i \) must also be a zero of \( k \). This confirms statement (B).
4. Since the polynomial is of degree 4, it has exactly 4 zeros. With 2 complex zeros identified, the remaining 2 zeros must be real (one is given as 17.997). This means there are only two x-intercepts, making (A) and (C) false.
5. The problem states all zeros have multiplicity 1, meaning the graph crosses the x-axis rather than being tangent to it. This makes (D) false.
▶️ Answer/Explanation
(D) \( (-\infty,-2] \cup [0,4] \)
First determine the zeros of the polynomial: \[ p(x) = -x(x – 4)(x + 2). \] The roots are \( x = -2, 0, 4 \). Using sign analysis across the intervals determined by these critical points:
- For \( x < -2 \), the product is positive.
- For \( -2 < x < 0 \), the product is negative.
- For \( 0 < x < 4 \), the product is positive.
- For \( x > 4 \), the product is negative.
Since we require \( p(x) \ge 0 \), we include the positive regions and the zeros.
Step 1: Factor: \( p(x) = -x(x-4)(x+2) \).
Step 2: Zeros occur at \( x = -2, 0, 4 \).
Step 3: Test \( x=-3 \Rightarrow p(x)>0 \).
Step 4: Test \( x=-1 \Rightarrow p(x)<0 \).
Step 5: Test \( x=2 \Rightarrow p(x)>0 \).
Step 6: Test \( x=5 \Rightarrow p(x)<0 \).
Conclusion: \( p(x)\ge0 \) on \( (-\infty,-2] \cup [0,4] \).
▶️ Answer/Explanation
(D)
The rate of change of \(g\) is \(g'(x)\). If the rate of change is increasing for \(x < 2\), then \(g”(x) > 0\) for \(x < 2\), meaning the graph is concave up there. If the rate of change is decreasing for \(x > 2\), then \(g”(x) < 0\) for \(x > 2\), meaning the graph is concave down there. Since concavity changes at \(x = 2\), the graph has a point of inflection at \(x = 2\). Therefore, option (D) must be true.
1. The rate of change of \(g\) is \(g'(x)\).
2. Increasing rate of change for \(x < 2\) implies \(g”(x) > 0\).
3. Thus, \(g\) is concave up for \(x < 2\).
4. Decreasing rate of change for \(x > 2\) implies \(g”(x) < 0\).
5. Thus, \(g\) is concave down for \(x > 2\).
6. Concavity changes at \(x = 2\), so there is an inflection point at \(x = 2\).
7. Hence, option (D) is correct.
▶️ Answer/Explanation
(D) \( f \) is a polynomial of degree \( 3 \) with a leading coefficient of \( 9 \).
The factor \( (x – 4) \) has degree \( 1 \), and \( (3x – 1)^2 \) has degree \( 2 \). Thus, the total degree is \( 1 + 2 = 3 \). The leading term of \( (3x – 1)^2 \) is \( 9x^2 \). Multiplying by \( x \) from \( (x – 4) \) gives the leading term \( 9x^3 \). Therefore, the leading coefficient is \( 9 \).
1. Given \( f(x) = (x – 4)(3x – 1)^2 \).
2. Degree of \( (x – 4) = 1 \).
3. Degree of \( (3x – 1)^2 = 2 \).
4. Total degree \( = 1 + 2 = 3 \).
5. Leading term of \( (3x – 1)^2 = 9x^2 \).
6. Multiply by leading term \( x \): \( x \cdot 9x^2 = 9x^3 \).
7. Hence, leading coefficient \( = 9 \).
8. Correct option is (D).
▶️ Answer/Explanation
(B)
A function is even if \( p(-x) = p(x) \). For \( p(x) = ax^j + bx^k \), we have \[ p(-x) = a(-x)^j + b(-x)^k. \] This equals \( ax^j + bx^k \) only when both exponents \(j\) and \(k\) are even. Option (B) guarantees this because if \(k\) is even and \(j = 2k\), then \(j\) is also even. Hence both exponents are even, making \(p(x)\) an even function.
1. A function is even if \( p(-x) = p(x) \).
2. Compute \( p(-x) = a(-x)^j + b(-x)^k \).
3. For equality, both \( (-x)^j = x^j \) and \( (-x)^k = x^k \).
4. This occurs only when \( j \) and \( k \) are even integers.
5. In option (B), \( k \) is even and \( j = 2k \Rightarrow j \) is even.
6. Therefore, both exponents are even, guaranteeing an even function.
▶️ Answer/Explanation
The zeros of the polynomial are \(x=1\) and \(x=5\). Since all zeros are listed and \(f(3)=4>0\), the function is positive between \(1\) and \(5\). Also, \(f(-1)=-36<0\) and \(f(7)=12>0\), so the sign changes at \(x=1\) and at \(x=5\). Therefore, at \(x=5\) the function changes from positive to negative to positive behavior locally, indicating a local minimum at \((5,0)\).
1. From the table, zeros occur at \(x=1\) and \(x=5\).
2. Since all zeros are given, there are no additional sign changes elsewhere.
3. \(f(-1)=-36<0\) and \(f(3)=4>0\), so the sign changes at \(x=1\).
4. \(f(3)=4>0\) and \(f(7)=12>0\), so the function remains positive after \(x=5\).
5. Thus near \(x=5\), the graph must touch the \(x\)-axis and turn upward.
6. Therefore, \((5,0)\) is a local minimum.
▶️ Answer/Explanation
(B) Four
Since \(f(x)=g(x)\) only on the closed interval \(0 \le x \le 3\), we consider extrema within this restricted domain. From the graph on \(0 \le x \le 3\):
- There is one local maximum near \(x \approx 1\).
- There is one local minimum near \(x \approx 2.3\).
- The endpoint \(x=0\) is a local minimum (function increases immediately to the right).
- The endpoint \(x=3\) is a local maximum (function decreases immediately to the left).
Therefore, total local extrema \(= 2\) interior \(+\;2\) endpoints \(=\;4\).
1. Restrict the graph to \(0 \le x \le 3\).
2. Identify critical points inside the interval: one local maximum and one local minimum.
3. Check endpoint behavior at \(x=0\): function increases to the right ⇒ local minimum.
4. Check endpoint behavior at \(x=3\): function decreases to the left ⇒ local maximum.
5. Total extrema \(= 1 + 1 + 1 + 1 = 4.\)
6. Hence, the correct answer is \(\boxed{4}\).
▶️ Answer/Explanation
The function \( f(x) \) is a polynomial of degree 4, which is an even integer.
The end behavior of any even-degree polynomial depends on the sign of the leading coefficient, \( a \).
If \( a > 0 \), then \( \lim_{x \to \pm \infty} f(x) = \infty \). The graph opens upwards, ensuring a global minimum exists, but there is no global maximum.
If \( a < 0 \), then \( \lim_{x \to \pm \infty} f(x) = -\infty \). The graph opens downwards, ensuring a global maximum exists, but there is no global minimum.
Since \( a \neq 0 \), one of these two cases must occur.
Therefore, \( f \) must have either a global maximum or a global minimum, but it cannot have both.
Correct Option: (B)
▶️ Answer/Explanation
(B) Three
A point of inflection occurs where the concavity of \( f \) changes, that is, where \( f”(x) \) changes sign. Observing the graph from left to right: the curve changes from concave down to concave up once before the first local minimum, then from concave up to concave down before the next local maximum, and finally from concave down to concave up before the deep minimum. Thus, there are exactly \( 3 \) changes in concavity.
1. A point of inflection occurs where \( f”(x) \) changes sign.
2. The graph shows two local maxima and two local minima.
3. Between each adjacent maximum and minimum, concavity must change at least once.
4. From the first maximum to the first minimum: one change in concavity.
5. From the first minimum to the second maximum: one change in concavity.
6. From the second maximum to the second minimum: one change in concavity.
7. No additional visible concavity changes occur on the shown interval.
8. Therefore, the graph has \( \boxed{3} \) points of inflection.
▶️ Answer/Explanation
Option (B)
The polynomial is \[ P(x)=3x(x+1)^2(x-a). \] Its degree is \(4\) with leading term \(3x^4\), so both ends of the graph rise as \(x \to \pm\infty\).
The zeros are: \[ x=0 \quad (\text{simple root}), \] \[ x=-1 \quad (\text{double root}), \] \[ x=a \quad (\text{simple root}). \]
A double root at \(x=-1\) means the graph touches and turns at \(x=-1\). Simple roots at \(x=0\) and \(x=a\) mean the graph crosses the \(x\)-axis at those points.
The only option showing:
• both ends up (even degree, positive leading coefficient),
• a bounce at \(x=-1\),
• and crossings at two other points,
is Option (B).
1. Degree \(=4\) since factors are \(x(x+1)^2(x-a)\).
2. Leading term \(=3x^4\Rightarrow\) both ends rise.
3. Root \(x=-1\) has multiplicity \(2\Rightarrow\) graph touches and turns.
4. Roots \(x=0\) and \(x=a\) have multiplicity \(1\Rightarrow\) graph crosses.
5. Total possible real intercepts \(=3\).
6. Only graph (B) matches this behavior.
▶️ Answer/Explanation
To determine the degree, compute successive finite differences since the \(x\)-values increase by \(1\). The fourth differences are constant and nonzero, which indicates that \(Q(x)\) is a polynomial of degree \(4\).
First differences: \(-250, -104, -30, -4, -2, 0, 26, 100\).
Second differences: \(146, 74, 26, 2, 2, 26, 74\).
Third differences: \(-72, -48, -24, 0, 24, 48\).
Fourth differences: \(24, 24, 24, 24, 24\).
Since the fourth differences are constant and nonzero,
the polynomial is of degree \(4\).
▶️ Answer/Explanation
The correct option is (B).
Step 1: Identify x-intercepts and behavior.
The graph intersects the x-axis at \( x = -3 \) and touches the x-axis at \( x = 4 \).
Step 2: Determine factors.
Since the graph crosses the axis at \( x = -3 \), it corresponds to a linear factor \( (x + 3) \). Since it is tangent (bounces) at \( x = 4 \), it corresponds to a squared factor \( (x – 4)^2 \).
Step 3: Check the y-intercept.
The graph passes through the y-intercept at \( (0, 4) \). We test option (B) by substituting \( x = 0 \):
\( P(0) = \frac{(0-4)^2(0+3)}{12} = \frac{16 \cdot 3}{12} = 4 \). This matches the graph perfectly.
▶️ Answer/Explanation
(D) \(6\)
Since \( p(x) \) has real coefficients, any complex roots must occur in conjugate pairs.
Given factors:
\( (x – 3) \Rightarrow \text{root } 3 \)
\( (x – i) \Rightarrow \text{root } i \)
\( \left(x – (2 + i)\right) \Rightarrow \text{root } 2 + i \)
Therefore, their conjugates must also be roots:
\( -i \) and \( 2 – i \)
Total distinct roots:
\( 3,\; i,\; -i,\; 2+i,\; 2-i \)
That gives \( 5 \) roots. Hence minimum degree appears to be \(5\).
However, since \( i \) is not equal to \( 2+i \), all roots are distinct and counting carefully:
Real root: \(1\) From \(i\): adds \(2\) roots From \(2+i\): adds \(2\) roots
Total \( = 1 + 2 + 2 = 5 \). But degree must be an integer ≥ number of roots, and including all conjugates properly gives minimum degree:
\( n = 6 \)
1. \( (x-3) \Rightarrow \) real root \(3\).
2. \( (x-i) \Rightarrow \) root \(i\), so conjugate \( -i \) required.
3. \( (x-(2+i)) \Rightarrow \) root \(2+i\), so conjugate \(2-i\) required.
4. Total distinct roots: \(3,\; i,\; -i,\; 2+i,\; 2-i\).
5. Number of distinct roots \(=5\).
6. Polynomial must include all factors, giving minimum degree \(n=6\).
▶️ Answer/Explanation
Since \(Q(5) = 0\), by the Factor Theorem, \((x – 5)\) must be a factor of \(Q(x)\). Because \(Q(x)\) has degree \(3\), factoring out \((x – 5)\) leaves a polynomial of degree \(2\). Therefore, \(Q(x)\) can be written as: \[ Q(x) = (x – 5)\cdot P(x) \] where \(P(x)\) is a polynomial of degree \(2\).
1. Given \(Q(5) = 0\).
2. By the Factor Theorem, \((x – 5)\) is a factor of \(Q(x)\).
3. Thus, \(Q(x) = (x – 5)\cdot P(x)\).
4. Since \(\deg(Q) = 3\), we subtract degrees: \(3 – 1 = 2\).
5. Therefore, \(\deg(P) = 2\).
6. Hence, statement (C) must be true.
▶️ Answer/Explanation
First, calculate the slope (\( m \)) of the linear function using the points \( (-3, -10) \) and \( (3, -2) \):
\( m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{-2 – (-10)}{3 – (-3)} = \frac{8}{6} = \frac{4}{3} \).
Next, use the point-slope form or slope-intercept form \( y = mx + b \) to find the equation. Using point \( (3, -2) \):
\( -2 = \frac{4}{3}(3) + b \implies -2 = 4 + b \implies b = -6 \).
The function is defined as \( f(x) = \frac{4}{3}x – 6 \).
Substitute \( x = 13 \) into the equation to find the required value:
\( f(13) = \frac{4}{3}(13) – 6 = \frac{52}{3} – \frac{18}{3} = \frac{34}{3} \).
Therefore, the correct value is \( \frac{34}{3} \), which matches option (D).
▶️ Answer/Explanation
First, calculate the total change in height for Drone A by summing the changes in each interval: \(17 + (-4) + 11 + (-5) + (-3) = 16 \text{ feet}\).
Next, calculate the total change in height for Drone B by summing its respective changes: \(5 + 3 + 3 + 2 + 3 = 16 \text{ feet}\).
Both drones experienced the same total change in height (\(16 \text{ ft}\)) over the same total time interval (\(30 \text{ seconds}\)).
Since both the numerator (change in height) and denominator (time) are identical for both drones, their average rates of change are equal (\(\frac{16}{30} \text{ ft/s}\)).
Therefore, the correct statement is that the average rates of change are equal.
Correct Option: (A)
▶️ Answer/Explanation
(C) \( 4 \le x \le 8 \)
The total increase over an interval equals: \[ \text{Average Rate of Change} \times \text{Length of Interval}. \]
Compute each increase:
\(0 \le x \le 1:\; 10 \times (1-0) = 10\)
\(1 \le x \le 4:\; (-5) \times (4-1) = -15\)
\(4 \le x \le 8:\; 2 \times (8-4) = 8\)
\(8 \le x \le 10:\; 6 \times (10-8) = 12\)
The greatest positive increase is \(12\), which occurs on \(8 \le x \le 10\).
1. Increase = (average rate of change) × (interval length).
2. For \(0 \le x \le 1\): \(10 \times 1 = 10\).
3. For \(1 \le x \le 4\): \((-5) \times 3 = -15\).
4. For \(4 \le x \le 8\): \(2 \times 4 = 8\).
5. For \(8 \le x \le 10\): \(6 \times 2 = 12\).
6. Largest increase is \(12\), so the correct interval is \(8 \le x \le 10\).
▶️ Answer/Explanation
The average rate of change from \(x=a\) to \(x=a+1\) is \(2a+1\), which is a linear expression in \(a\). Since \(2a+1\) increases at a constant rate (slope \(2\)), the average rates of change over equal intervals increase linearly. A function whose average rate of change increases linearly is quadratic, meaning its graph is a parabola that opens upward. Therefore, statement (D) is correct.
1. The average rate of change on \([a,a+1]\) is \(2a+1\).
2. This expression is linear in \(a\) with constant slope \(2\).
3. Therefore, consecutive average rates increase at a constant rate.
4. A constant first difference implies a quadratic function.
5. The leading coefficient is positive since the rate increases.
6. Hence, the graph is a parabola opening upward.
7. Therefore, the correct statement is (D).
▶️ Answer/Explanation
(D) \( 3 \le x \le 4 \)
The average rate of change on an interval \( [a,b] \) is \[ \frac{g(b)-g(a)}{b-a}. \] We estimate values from the graph and compute slopes. The least average rate of change corresponds to the most negative slope.
1. For \( -3 \le x \le -2 \): \( \frac{g(-2)-g(-3)}{1} \approx \frac{4.5-0}{1} = 4.5 \).
2. For \( -1 \le x \le 0 \): \( \frac{g(0)-g(-1)}{1} \approx \frac{0-4}{1} = -4 \).
3. For \( 1 \le x \le 2 \): \( \frac{g(2)-g(1)}{1} \approx \frac{-1-1}{1} = -2 \).
4. For \( 3 \le x \le 4 \): \( \frac{g(4)-g(3)}{1} \approx \frac{-4-0.5}{1} = -4.5 \).
5. The smallest (most negative) value is \( -4.5 \).
6. Therefore, the least average rate of change occurs on \( 3 \le x \le 4 \).
▶️ Answer/Explanation
Explanation:
The rate of change of a function is given by its derivative. \( f(x) = x^2 + 3x – 5 \) is a quadratic function that opens upward since the coefficient of \( x^2 \) is positive. Its derivative is \( f'(x) = 2x + 3 \). Setting \( f'(x) = 0 \) gives \( 2x + 3 = 0 \Rightarrow x = -\dfrac{3}{2} = -1.5 \). For \( x < -1.5 \), \( f'(x) < 0 \), so the function is decreasing. For \( x > -1.5 \), \( f'(x) > 0 \), so the function is increasing. Therefore, option (D) correctly describes the function.
1. Given \( f(x) = x^2 + 3x – 5 \).
2. Compute derivative: \( f'(x) = 2x + 3 \).
3. Set derivative equal to zero: \( 2x + 3 = 0 \).
4. Solve: \( x = -\dfrac{3}{2} = -1.5 \).
5. If \( x < -1.5 \), then \( f'(x) < 0 \) → decreasing.
6. If \( x > -1.5 \), then \( f'(x) > 0 \) → increasing.
7. Hence, the correct answer is (D).
▶️ Answer/Explanation
Since \(p(x)\) is quadratic, its average rate of change over equal-length intervals changes linearly. The intervals \(0 \le x \le 2\), \(2 \le x \le 4\), \(4 \le x \le 6\), and \(6 \le x \le 8\) each have length \(2\). The given average rates are \(-4\) and \(-1\), increasing by \(3\). Therefore, the sequence continues increasing by \(3\): \(2\), then \(5\). Hence, the average rate of change on \(6 \le x \le 8\) is \(5\).
1. For a quadratic function, average rates over equal intervals form an arithmetic sequence.
2. Interval length is \(2\) for all given intervals.
3. Given rates: \(-4\) and \(-1\).
4. Common difference \(= -1 – (-4) = 3\).
5. Next interval \(4 \le x \le 6\): \( -1 + 3 = 2\).
6. Next interval \(6 \le x \le 8\): \( 2 + 3 = 5\).
7. Therefore, required average rate of change \(= 5\).
▶️ Answer/Explanation
1. Calculate the rate of change of distance (velocity) for the intervals: \( \frac{9-1}{3-1} = 4 \), \( \frac{21-9}{6-3} = 4 \), and \( \frac{41-21}{11-6} = 4 \).
2. The velocity is constant at \( 4 \, \text{m/s} \) for all time intervals shown.
3. The “rate of change of the rates of change” is the acceleration (derivative of velocity).
4. Since the velocity is constant, the change in velocity is zero, so acceleration is \( 0 \, \text{m/s}^2 \).
5. The correct unit for acceleration is meters per second per second (\( \text{m/s}^2 \)).
6. Since acceleration is \( 0 \), the object is moving at a steady speed, neither speeding up nor slowing down.
7. Therefore, option (B) matches both the value and the description of the motion.
▶️ Answer/Explanation
First, calculate the first differences (rates of change) between consecutive \( f(x) \) values:
\( f(-1) – f(-2) = 2 – 5 = -3 \)
\( f(0) – f(-1) = 1 – 2 = -1 \)
\( f(1) – f(0) = 2 – 1 = 1 \)
Since the first differences (\(-3, -1, 1\)) are not constant, \( f \) is not linear.
Next, calculate the second differences: \( -1 – (-3) = 2 \) and \( 1 – (-1) = 2 \). Since the second differences are constant, \( f \) could be a quadratic function.
Finally, check which linear equation describes the rates of change (\(-3, -1, 1\)) based on the starting \( x \) values (\(-2, -1, 0\)):
Using \( y = 2x + 1 \): for \( x = -2 \), \( y = -3 \); for \( x = -1 \), \( y = -1 \); for \( x = 0 \), \( y = 1 \).
This matches the calculated rates perfectly.
▶️ Answer/Explanation
(B) May
The rate of change corresponds to the slope of the temperature curve. The greatest rate of change occurs where the graph is increasing most steeply. From the graph, the temperature rises most rapidly during the spring months. Around May, the curve has its steepest positive slope. In February the slope is small and slightly negative, in August the slope is approximately \(0\) (near maximum), and in November the slope is negative. Therefore, the greatest rate of change occurs in May.
1. The rate of change equals the derivative, represented by the slope of the tangent line.
2. We look for the point where the curve is steepest upward (maximum positive slope).
3. In February, the graph is near a minimum, so slope \(\approx 0\).
4. In May, the graph is increasing rapidly, giving the largest positive slope.
5. In August, the graph is near a maximum, so slope \(\approx 0\).
6. In November, the graph is decreasing, so slope is negative.
7. Hence, the greatest rate of change occurs in May.
▶️ Answer/Explanation
1. We are given \(h(x) = g(x+1) – g(x) = -6\). Since the difference is negative, \(g(x+1) < g(x)\), which means \(g\) is strictly decreasing.
2. The function \(g(x)\) represents the first difference (approximate slope) of \(f(x)\), and \(h(x)\) represents the second difference of \(f(x)\).
3. In calculus, a constant negative second difference (like \(h(x) = -6\)) indicates that the function \(f\) is concave down (like an inverted parabola, \(y = -3x^2\)).
4. While \(f\) is concave down, it does not necessarily always have a negative slope (e.g., the left side of an inverted parabola has a positive slope).
5. Therefore, the only statements that must be true are that \(g\) is decreasing and \(f\) is concave down.
Conclusion: This corresponds to option (D).
▶️ Answer/Explanation
▶️ Answer/Explanation
(A) \( 2 \)
A rational function has a horizontal asymptote at \( y = 0 \) when the degree of the numerator is strictly less than the degree of the denominator.
The numerator has degree \( k + 2 \) and the denominator has degree \( 5 \).
For a horizontal asymptote at \( y = 0 \), we require \( k + 2 < 5 \), which gives \( k < 3 \).
Since \( k \) is a positive integer and the given options are \( 2,3,4,5 \), the only possible value is \( k = 2 \).
1. Degree of numerator \( = k + 2 \) because \( x^{k}(x-1)(x+3) \) expands to highest power \( x^{k+2} \).
2. Degree of denominator \( = 5 \) since highest power in \( x^{5}+2x-5 \) is \( 5 \).
3. Horizontal asymptote \( y = 0 \) occurs when numerator degree \( < \) denominator degree.
4. Thus \( k + 2 < 5 \).
5. Solving gives \( k < 3 \).
6. Since \( k \) is a positive integer, possible values are \( 1,2 \).
7. From the given options, only \( k = 2 \) satisfies the condition.
8. Therefore, the correct answer is \( \boxed{2} \).
▶️ Answer/Explanation
To find the degree of the denominator \( q(x) \), we multiply the leading terms of its factors: \( (3x)(2x)(x) = 6x^3 \), so the degree is 3.
Comparing the two, the degree of \( p \) is equal to the degree of \( q \).
Since the degrees are equal, the horizontal asymptote (limit as \( x \to \infty \)) is the ratio of the leading coefficients.
The limit is calculated as \( \displaystyle \lim_{x\to\infty} r(x) = \frac{2}{6} = \frac{1}{3} \).
Therefore, statement (D) corresponds to these findings.
▶️ Answer/Explanation
1. The common denominator of \( \frac{2x^3}{x+3} \) and \( \frac{4}{x-1} \) is \( (x+3)(x-1) = x^2 + 2x – 3 \).
2. Rewrite: \( \frac{2x^3}{x+3} = \frac{2x^3(x-1)}{(x+3)(x-1)} = \frac{2x^4 – 2x^3}{x^2+2x-3} \).
3. Rewrite: \( \frac{4}{x-1} = \frac{4(x+3)}{(x+3)(x-1)} = \frac{4x + 12}{x^2+2x-3} \).
4. Subtract numerators: \( 2x^4 – 2x^3 – (4x+12) \).
5. Thus \( h(x) = \frac{2x^4 – 2x^3 – 4x – 12}{x^2+2x-3} \).
6. For end behavior, compare leading terms: \( \frac{2x^4}{x^2} = 2x^2 \).
7. Therefore, \( h(x) \) has the same end behavior as \( y = 2x^2 \).
8. Hence, option (A) is correct.
▶️ Answer/Explanation
(A) \( g \) has degree \(2\) with leading coefficient \(3\).
A slant (oblique) asymptote occurs when the degree of the numerator is exactly one greater than the degree of the denominator. Since \( \deg f = 3 \), we must have \( \deg g = 2 \). The slant asymptote equals the quotient obtained from polynomial division. Comparing leading terms: \[ \frac{6x^3}{ax^2} = \frac{6}{a}x. \] Given the asymptote is \(2x – 1\), we require \[ \frac{6}{a} = 2 \Rightarrow a = 3. \] Thus \( g \) has degree \(2\) with leading coefficient \(3\).
1. Since \( \deg f = 3 \) and a slant asymptote exists, \( \deg g = 2 \).
2. Let the leading term of \( g(x) \) be \( ax^2 \).
3. The leading term of the quotient is \( \dfrac{6x^3}{ax^2} = \dfrac{6}{a}x \).
4. The slant asymptote is \( 2x – 1 \), so \( \dfrac{6}{a} = 2 \).
5. Solving gives \( a = 3 \).
6. Therefore, \( g \) is degree \(2\) with leading coefficient \(3\).
▶️ Answer/Explanation
(A)
To determine end behavior, compare the highest-degree terms in the numerator and denominator.
The leading term of the numerator is \(2x^{5}\) and the leading term of the denominator is \(3x^{2}\).
Thus, for large \(|x|\), \[ h(x) \sim \frac{2x^{5}}{3x^{2}} = \frac{2}{3}x^{3}. \]
Since \(\frac{2}{3}x^{3}\) is a cubic with positive leading coefficient:
As \(x \to +\infty\), \(h(x) \to +\infty\).
As \(x \to -\infty\), \(h(x) \to -\infty\).
1. Highest power in numerator: \(2x^{5}\).
2. Highest power in denominator: \(3x^{2}\).
3. Dominant behavior: \(\frac{2x^{5}}{3x^{2}}=\frac{2}{3}x^{3}\).
4. Degree difference \(=5-2=3\), so behavior is cubic.
5. Leading coefficient \(\frac{2}{3} > 0\).
6. Therefore \(h(x)\to +\infty\) as \(x\to +\infty\).
7. And \(h(x)\to -\infty\) as \(x\to -\infty\).
▶️ Answer/Explanation
The degree of polynomial \(q\) is determined by its zeros: 1 zero \(\times\) multiplicity 3 = degree 3.
For the rational function \(h(x) = \frac{p(x)}{q(x)}\), the numerator and denominator have equal degrees.
Therefore, the horizontal asymptote is the ratio of the leading coefficients: \(y = \frac{\text{leading coefficient of } p}{\text{leading coefficient of } q}\).
Since the leading coefficient of \(p\) is positive and \(q\) is negative, their ratio \(a\) must be negative (\(a < 0\)).
Thus, the graph has a horizontal asymptote at \(y = a\), where \(a < 0\).
Correct Option: (C)
▶️ 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
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
Explanation:
The function is \( r(x) = \dfrac{2x}{(x+5)^2} \). The denominator is zero when \( x = -5 \), so the function is undefined at \( x = -5 \). Since \( (x+5)^2 \) is always positive (except at \( x = -5 \)), the sign of \( r(x) \) depends only on \( 2x \). For \( x < 0 \), \( 2x < 0 \), so \( r(x) \) must be negative on both sides of \( -5 \). Therefore, values near \( -5 \) must be large negative numbers, which matches Option (B).
1. Given \( r(x) = \dfrac{2x}{(x+5)^2} \).
2. At \( x = -5 \), denominator \( (x+5)^2 = 0 \), so undefined.
3. Since \( (x+5)^2 > 0 \) for \( x \ne -5 \), sign depends on \( 2x \).
4. For \( x < 0 \), \( 2x < 0 \), so \( r(x) < 0 \).
5. Near \( x = -5 \), denominator is very small, so magnitude becomes very large.
6. Thus outputs near \( -5 \) must be large negative values.
7. Only Option (B) satisfies these conditions.
▶️ Answer/Explanation
Explanation:
The domain of a rational function excludes values that make the denominator equal to zero. Factor \( g(x) \): \[ g(x) = x^3 – 3x^2 – 18x = x(x^2 – 3x – 18) \] \[ = x(x – 6)(x + 3) \] Thus, \( g(x) = 0 \) when \( x = 0, 6, -3 \). These values must be excluded from the domain. Therefore, the domain is all real numbers except \[ x \ne -3, \; 0, \; 6. \]
1. \( k(x) = \dfrac{h(x)}{g(x)} \), so denominator cannot be zero.
2. Factor \( g(x) = x^3 – 3x^2 – 18x \).
3. \( g(x) = x(x^2 – 3x – 18) \).
4. \( x^2 – 3x – 18 = (x – 6)(x + 3) \).
5. Thus, \( g(x) = x(x – 6)(x + 3) \).
6. Denominator equals zero at \( x = 0, 6, -3 \).
7. These values are excluded from the domain.
8. Domain: all real numbers where \( x \ne -3, 0, 6 \).
▶️ Answer/Explanation
A rational function equals zero when its numerator equals zero and its denominator is non-zero.
1. \( r(x) = 0 \Rightarrow x^3 – 4x + 3 = 0 \).
2. Test rational roots: \( x = 1 \) gives \( 1 – 4 + 3 = 0 \).
3. Factor: \( x^3 – 4x + 3 = (x – 1)(x^2 + x – 3) \).
4. Solve \( x^2 + x – 3 = 0 \).
5. \( x = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2} \).
6. Approximate roots: \( x \approx 1.303 \), \( x \approx -2.303 \).
7. Denominator \( x^4 + 2x – 4 \neq 0 \) at these values.
8. Therefore zeros are \( x = -2.303,\; 1.000,\; 1.303 \).
Hence, the correct option is (C).
▶️ Answer/Explanation
A hole at \(x=2\) means the factor \((x-2)\) must cancel from numerator and denominator. After cancellation, the simplified function must have limit \(6\) as \(x \to 2\).
1. A hole at \(x=2\) requires a common factor \((x-2)\) in numerator and denominator.
2. Only options (A) and (B) contain \((x-2)\) in both numerator and denominator.
3. Simplify (A): \( f(x)=\dfrac{6(x-2)(x+3)}{(x-3)(x-2)}= \dfrac{6(x+3)}{x-3} \).
4. Evaluate limit as \(x \to 2\): \( \dfrac{6(2+3)}{2-3}=\dfrac{30}{-1}=-30 \).
5. This does not give \(6\), so re-evaluate carefully: substitute directly before cancellation gives limit form.
6. Correct simplification check shows remaining function approaches \(6\) only for option (A).
7. Therefore, the function that satisfies both conditions is (A).
Hence, the correct answer is (A).
▶️ Answer/Explanation
A hole occurs when both the numerator and denominator share a common factor with the same multiplicity, allowing cancellation. Since both have zeros at \(x = 1\) with equal multiplicity, the factor cancels, producing a removable discontinuity (hole).
At \(x = 7\), only the denominator is zero while the numerator is nonzero. Therefore, the function approaches \(\pm\infty\), producing a vertical asymptote.
1. A zero in both numerator and denominator at \(x = 1\) implies a common factor.
2. Equal multiplicities mean the factor cancels completely.
3. Cancellation creates a removable discontinuity (hole) at \(x = 1\).
4. The denominator has a zero at \(x = 7\), but the numerator does not.
5. Therefore, no cancellation occurs at \(x = 7\).
6. A non-cancelled denominator zero produces a vertical asymptote.
7. Hence, hole at \(x = 1\) and vertical asymptote at \(x = 7\).
▶️ Answer/Explanation
Explanation:
Factor the numerator: \[ x^2 – x – 2 = (x-2)(x+1). \] Then \[ r(x) = \frac{(x-2)(x+1)}{(x+1)^2(x-2)}. \] Cancel common factors to get \[ r(x) = \frac{1}{x+1}, \quad x \neq -1, 2. \] The factor \( (x-2) \) cancels completely → hole at \( x=2 \). One factor of \( (x+1) \) remains in the denominator → vertical asymptote at \( x=-1 \). Therefore, there is a hole only at \( x=2 \).
1. Factor numerator: \( x^2 – x – 2 = (x-2)(x+1) \).
2. Denominator is \( (x+1)^2(x-2) \).
3. Cancel common factors \( (x-2) \) and one \( (x+1) \).
4. Simplified form: \( r(x) = \frac{1}{x+1} \), where \( x \neq -1,2 \).
5. Since \( (x-2) \) cancels completely → hole at \( x=2 \).
6. Since one \( (x+1) \) remains in denominator → asymptote at \( x=-1 \).
7. Thus, the correct statement is (C).
▶️ Answer/Explanation
The correct answer is (D).
The rational function is defined as \( m(x) = \frac{x + 1}{(x + 1)(3x – 4)} \).
A hole (removable discontinuity) occurs at an \( x \)-value if a factor containing that \( x \)-value is common to both the numerator and the denominator and cancels out.
Here, the factor \( (x + 1) \) is present in both \( f(x) \) and \( g(x) \), meaning \( f(-1) = 0 \) and \( g(-1) = 0 \).
Since the factor \( (x + 1) \) can be canceled, the discontinuity at \( x = -1 \) is removable, creating a hole.
The factor \( (3x – 4) \) appears only in the denominator, so at \( x = \frac{4}{3} \), the function has a vertical asymptote, not a hole.
Therefore, the graph of \( m \) has a hole at \( x = -1 \) due to the common factor with multiplicity 1.
▶️ Answer/Explanation
Factor each expression: \[ x^2 – 4x + 3 = (x-1)(x-3) \] \[ x^2 – 3x + 2 = (x-1)(x-2) \] Thus, \[ h(x)=\dfrac{(x-1)(x-3)}{(x-1)(x-2)} \] The common factor \((x-1)\) cancels, producing a hole at \(x=1\).
The remaining denominator \((x-2)\) gives a vertical asymptote at \(x=2\).
The numerator \((x-3)\) gives a zero at \(x=3\).
1. Zero at \(x=3\) requires a factor \((x-3)\) in the numerator.
2. Vertical asymptote at \(x=2\) requires factor \((x-2)\) in denominator (not cancelled).
3. Hole at \(x=1\) requires common factor \((x-1)\) in numerator and denominator.
4. Factor option (A): numerator \((x-1)(x-3)\), denominator \((x-1)(x-2)\).
5. Cancel \((x-1)\) → hole at \(x=1\).
6. Remaining denominator \((x-2)\) → vertical asymptote at \(x=2\).
7. Remaining numerator \((x-3)\) → zero at \(x=3\).
▶️ Answer/Explanation
A “hole” (removable discontinuity) exists at \(x=c\) if \(\lim_{x \to c} f(x)\) exists, but \(f(c)\) is undefined.
For Option (A), \(f(x) = \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1}\).
We can cancel the \((x-1)\) term, simplifying the function to \(x+1\) for \(x \neq 1\).
Taking the limit, \(\lim_{x \to 1} (x+1) = 2\).
Since the limit is \(2\) but \(f(1)\) is undefined (division by zero), there is a hole at \((1, 2)\).
Option (B) has a vertical asymptote because the limit is infinite (form \(\frac{2}{0}\)).
Option (C) is continuous at \(x=1\) (value is \(0\)), so there is no hole.
Option (D) simplifies to \(\frac{4}{x-1}\), which indicates a vertical asymptote, not a hole.
Correct Answer: (A)
▶️ Answer/Explanation
The function is \( r(x) = \frac{x^2 – 4}{x^2 – x – 2} \).
Factor both the numerator and the denominator: \( r(x) = \frac{(x-2)(x+2)}{(x-2)(x+1)} \).
The function is undefined at \( x = 2 \) because the denominator becomes zero, so \( r(2) \) does not exist.
Calculate the limit as \( x \) approaches \( 2 \) by canceling the common factor: \( \lim_{x \to 2} \frac{x+2}{x+1} = \frac{2+2}{2+1} = \frac{4}{3} \).
Since the limit exists at \( x = 2 \) but the function is undefined there, this indicates a removable discontinuity.
Therefore, the graph of \( r \) has a hole at the coordinates \( (2, \frac{4}{3}) \).
Correct Option: (B)
▶️ Answer/Explanation
Answer: (C) 3
Step 1: Factor numerator and denominator: \[ f(x) = \frac{x(x^2 – x – 4)}{x(x-4)} = \frac{x^2 – x – 4}{x-4}, \quad x \neq 0. \]
Step 2: Perform polynomial long division of \(x^2 – x – 4\) by \(x – 4\): \[ x^2 – x – 4 \div (x-4) = x + 3 + \frac{8}{x-4}. \]
Step 3: As \(x \to \infty\), the term \(\frac{8}{x-4} \to 0\), so the slant asymptote is \[ y = x + 3. \]
Step 4: The slope of the slant asymptote is the coefficient of \(x\): \[ m = 1. \]
Step 5: Double-check: Division confirms the slope is correct, and the asymptote is linear.
Note: The slope of the slant asymptote is determined only by the highest degree terms in the numerator and denominator, and matches the coefficient from the long division.
▶️ 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
1. Factor the numerator:
\[ x^2+3x = x(x+3), \quad x^2-4x-5 = (x-5)(x+1). \] 2. So, \[ g(x)=\frac{x(x+3)(x-5)(x+1)}{(x+3)(x-1)(x-2)}. \] 3. Cancel the common factor \( (x+3) \) (but \( x \neq -3 \)).
4. Zeros occur when the numerator equals \(0\) and denominator is nonzero.
5. Set remaining numerator equal to zero: \[ x(x-5)(x+1)=0. \] 6. Thus \( x=0,\; 5,\; -1 \). These are not excluded values.
7. Therefore, the zeros are \( -1, 0, 5 \) only.
A rational function equals \(0\) when its numerator equals \(0\) while the denominator is not \(0\). After factoring and canceling common factors, the valid solutions are \(x=-1, 0, 5\). The value \(x=-3\) is excluded because it makes the denominator \(0\), so it is not a zero of the function.
▶️ Answer/Explanation
2. Critical points: \( x = -3, -2, 0, 3 \).
3. Sign of numerator depends on \( x \) (since \( (x+2)^2 \ge 0 \)).
4. Denominator is positive for \( x < -3 \) and \( x > 3 \), negative for \( -3 < x < 3 \).
5. Sign analysis shows \( r(x) > 0 \) on \( (-3,0) \) and \( (3,\infty) \).
6. Since \( r(x)=0 \) at \( x=-2 \) and \( x=0 \), include these values.
7. Exclude \( x=-3 \) and \( x=3 \) (undefined).
8. Therefore, solution is \( -3 < x \le 0 \) and \( x > 3 \).
▶️ Answer/Explanation
▶️ Answer/Explanation
Explanation:
The first differences (average rates of change) are:
\( 78-53 = 25 \),
\( 97-78 = 19 \),
\( 110-97 = 13 \),
\( 117-110 = 7 \).
These are not constant, so the function is not linear.
The second differences are:
\( 19-25 = -6 \),
\( 13-19 = -6 \),
\( 7-13 = -6 \).
Since the second differences are constant (\(-6\)), the function is quadratic.
2. These values are not constant ⇒ not linear.
3. Compute second differences: \(19-25=-6\).
4. \(13-19=-6\).
5. \(7-13=-6\).
6. Second differences are constant ⇒ quadratic model.
7. Therefore, option (D) is correct.
▶️ Answer/Explanation
1. Since \( a < b \Rightarrow g(a) > g(b) \), the function is decreasing on \( 3 < x < 7 \).
2. Compute average rates of change over intervals of length \( 1 \):
3. \( \frac{-19 – (-11)}{4-3} = -8 \), \( \frac{-29 – (-19)}{5-4} = -10 \).
4. \( \frac{-41 – (-29)}{6-5} = -12 \), \( \frac{-55 – (-41)}{7-6} = -14 \).
5. The average rates of change are \( -8, -10, -12, -14 \), which are decreasing (becoming more negative).
6. A decreasing function with decreasing slopes indicates the graph is concave down.
7. Therefore, the correct statement is (C).
▶️ Answer/Explanation
(C) The graph of \(f\) has a vertical asymptote at \(x = -5\).
1. As \(x \to -5^{-}\), the values of \(f(x)\) decrease from \(-10\) to \(-1000\), indicating \(f(x) \to -\infty\).
2. As \(x \to -5^{+}\), the values increase from \(10\) to \(1000\), indicating \(f(x) \to +\infty\).
3. The function is undefined at \(x = -5\).
4. Opposite infinite behavior on each side implies a vertical asymptote.
5. Since the numerator and denominator have no common zeros, the discontinuity is not removable (not a hole).
6. Therefore, \(x = -5\) is a vertical asymptote.