AP Calculus AB Quiz: Reasoning Using Slope Fields

What this quiz covers

This quiz focuses on Reasoning Using Slope Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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. $y=-2$ only (correct answer)
2. $y=2$ only
3. Both $y=-2$ and $y=2$
4. Neither $y=-2$ nor $y=2$
5. All constant functions are stable

Explanation: This question uses slope field reasoning to identify stable equilibrium solutions. For $dy/dx = y^2 - 4 = (y-2)(y+2)$, equilibrium occurs when $dy/dx = 0$, giving $y = 2$ and $y = -2$. For stability, we check the sign of $dy/dx$ near each equilibrium. Near $y = -2$: slightly above gives $(-1.9)^2 - 4 \approx -0.39 < 0$ (decreasing toward $-2$), slightly below gives $(-2.1)^2 - 4 \approx 0.41 > 0$ (increasing toward $-2$). Near $y = 2$: slightly below gives $(1.9)^2 - 4 \approx -0.39 < 0$ (moving away), slightly above gives $(2.1)^2 - 4 \approx 0.41 > 0$ (moving away). Only $y = -2$ attracts nearby solutions, making it stable. Choice B fails because $y = 2$ is unstable as solutions move away from it. To identify stable equilibria in slope fields, find where $dy/dx = 0$ and check if nearby slopes point toward that equilibrium.

Question 2

1. They are decreasing for all $x$.
2. They have slope between $0$ and $1$ everywhere. (correct answer)
3. They have vertical tangents when $y=0$.
4. They are constant when $y\neq 0$.
5. They oscillate up and down periodically.

Explanation: This question involves slope field reasoning for equations with bounded derivatives. For $\frac{dy}{dx} = \frac{1}{1+y^2}$, the denominator $1 + y^2$ is always $\geq 1$ for any real $y$, so the slope is always between 0 and 1. Since $1 + y^2 \geq 1$, we have $\frac{1}{1+y^2} \leq 1$, and since $1 + y^2 > 0$, we have $\frac{dy}{dx} > 0$. Therefore, all solution curves have slopes strictly between 0 and 1 everywhere. Choice A fails because solutions are always increasing (positive slope), never decreasing. When analyzing slope fields, examine the range of possible values for $\frac{dy}{dx}$ to understand the constraints on solution behavior and ensure slopes stay within predictable bounds.

Question 3

1. vertical tangents for all $y$.
2. horizontal tangents for all $y$. (correct answer)
3. slope $1$ for all $y$.
4. negative slope for all $y$.
5. undefined slope for all $y$.

Explanation: This question requires slope field reasoning about slopes along vertical lines. For dy/dx = x/(1+y²) at x = 0, we have dy/dx = 0/(1+y²) = 0 for all y values. Since the numerator is zero when x = 0, the slope equals zero regardless of the y-coordinate, making all slope segments horizontal along the vertical line x = 0. The denominator 1+y² is always positive, so the fraction is well-defined and equals zero. Choice A fails because slopes are zero (horizontal), not infinite (vertical), along x = 0. When analyzing slope fields along vertical lines where the differential equation has x in the numerator, substitute the x-value to see how slopes vary with y.

Question 4

1. increasing because slope is positive.
2. decreasing because slope is negative. (correct answer)
3. constant because slope is zero.
4. undefined because slope is infinite.
5. increasing because slope is zero.

Explanation: This question requires slope field reasoning about local solution behavior at specific points. For $\frac{dy}{dx} = \frac{2}{1+x^2} - y^2$ at point (0,2), we evaluate: $\frac{dy}{dx} = \frac{2}{1+0^2} - 2^2 = 2 - 4 = -2 < 0$. A negative slope indicates the solution is locally decreasing at (0,2). The slope magnitude of 2 shows the solution decreases at a moderate rate near this point. Choice A fails because the slope is negative, not positive, indicating decreasing rather than increasing behavior. When analyzing local solution behavior in slope fields, substitute the point's coordinates into the differential equation to determine the slope sign and magnitude at that location.

Question 5

1. All solutions decrease and are concave down for all $x$.
2. All solutions increase with slopes that decrease as $x$ increases. (correct answer)
3. All solutions are constant functions.
4. All solutions have slopes that increase as $x$ increases.
5. Solutions have negative slope for $x>0$.

Explanation: This question involves slope field reasoning for solutions of equations depending only on x. For dy/dx = e^(-x), the slope depends only on x, not on y, so all solutions are vertical translations of each other. Since e^(-x) > 0 for all x, all solutions are increasing. As x increases, e^(-x) decreases toward 0, so slopes decrease but remain positive, meaning solutions increase at a decreasing rate (concave down). Choice A fails because solutions increase, not decrease, and choice D fails because slopes decrease, not increase. When analyzing slope fields for dy/dx = f(x), remember all solutions have identical slope behavior at each x-value, differing only by vertical shifts.

Question 6

1. It is decreasing because slopes are negative at $x=0$
2. It has a horizontal tangent because slope is $0$ at $x=0$ (correct answer)
3. It has a vertical tangent because denominator is $0$
4. It is increasing because slopes are positive at $x=0$
5. It is constant because slope depends only on $y$

Explanation: This problem tests slope field evaluation at specific points. For dy/dx = x/(1+y²), we evaluate the slope at the point (0,2). Substituting x = 0 into the differential equation gives dy/dx = 0/(1+2²) = 0/5 = 0. A slope of 0 means the solution has a horizontal tangent at this point. The answer "vertical tangent" might seem plausible if one confuses the roles of numerator and denominator, but the denominator 1+y² is never zero. When evaluating slopes from differential equations, substitute the exact coordinates and simplify carefully to determine the tangent line's behavior.

Question 7

1. They move away from $y=2$ for all $x>0$.
2. They remain below $y=2$ and increase toward $y=2$ for $x>0$. (correct answer)
3. They cross $y=2$ repeatedly for $x>0$.
4. They decrease without bound for $x>0$.
5. They are constant for all $x$.

Explanation: This question involves slope field reasoning to predict long-term solution behavior. For dy/dx = x(2-y), when y < 2 and x > 0, we have (positive)(positive) = positive slope, so solutions increase. As these solutions approach y = 2, the factor (2-y) approaches 0, making the slope approach 0, so the solution levels off approaching y = 2. Since y < 2 throughout this process, solutions never actually reach or cross y = 2. Choice A incorrectly suggests solutions move away from y = 2, but positive slopes when y < 2 drive solutions toward this horizontal asymptote. When reading slope fields, identify equilibrium lines where dy/dx = 0 and determine whether solutions approach or diverge from these lines based on nearby slope signs.

Question 8

1. Slopes are always negative.
2. Slopes are always between $-\tfrac12$ and $\tfrac12$. (correct answer)
3. Slopes are always greater than $1$.
4. Slopes are undefined when $y=0$.
5. Slopes depend on $x$ only.

Explanation: This question involves slope field reasoning for bounded slope behavior. For dy/dx = y/(y²+1), we can analyze the range of possible slopes. Taking the derivative of f(y) = y/(y²+1), we get f'(y) = (1-y²)/(y²+1)², which equals 0 when y = ±1. Evaluating f(±1) = ±1/2, and noting f(0) = 0, f(y) → 0 as y → ±∞, the function achieves maximum 1/2 at y = 1 and minimum -1/2 at y = -1. Therefore, slopes are always between -1/2 and 1/2. Choice A fails because slopes can be positive when y > 0. To analyze slope bounds in slope fields, find the range of the differential equation by examining its critical points and limiting behavior.

Question 9

1. It initially decreases and later stays constant.
2. It initially increases and is concave up near $x=0$. (correct answer)
3. It is constant for all $x$.
4. It initially decreases and is concave down near $x=0$.
5. It initially increases and is concave down near $x=0$.

Explanation: This question requires using slope field reasoning to analyze solution behavior. For dy/dx = y - x at point (0,1), the slope is 1 - 0 = 1, so the curve is initially increasing. Moving slightly right from (0,1) to small positive x values, the slopes remain positive and even increase since y > x in that region, indicating the curve is concave up. Nearby points show slopes that increase as we move right and up, confirming concave up behavior. Choice A fails because the solution increases rather than decreases initially. To read slope fields effectively, first evaluate the slope at the given point, then examine how slopes change in the immediate neighborhood to determine concavity.

Question 10

1. All solutions have negative slope.
2. Solutions with $y>0$ are increasing. (correct answer)
3. Solutions with $-1<y<0$ are increasing.
4. All solutions are constant.
5. Slopes depend only on $x$.

Explanation: This question requires slope field reasoning for logarithmic differential equations. For dy/dx = ln(1+y) with y > -1, we need to analyze when ln(1+y) is positive, negative, or zero. Since ln(1+y) > 0 when 1+y > 1, i.e., when y > 0, solutions with y > 0 have positive slope and are increasing. When -1 < y < 0, we have 0 < 1+y < 1, so ln(1+y) < 0, making these solutions decreasing. When y = 0, ln(1) = 0, giving a horizontal tangent. Choice C fails because solutions with -1 < y < 0 are decreasing, not increasing. To analyze slope fields involving logarithmic functions, determine where the logarithmic expression is positive, negative, or zero based on its argument.

Question 11

1. It is increasing at $x=0$.
2. It is decreasing at $x=0$. (correct answer)
3. It has a horizontal tangent at $x=0$.
4. It is constant for all $x$.
5. It has a vertical tangent at $x=0$.

Explanation: This question uses slope field reasoning about exponential-type behavior. For dy/dx = y - 1 at point (0,0), we have dy/dx = 0 - 1 = -1 < 0. A negative slope indicates the solution is decreasing at x = 0. Moving slightly right from (0,0), y becomes more negative while x remains small, so dy/dx = y - 1 becomes more negative, maintaining the decreasing trend. This creates exponential decay behavior moving away from the equilibrium y = 1. Choice A fails because the slope is negative, indicating decreasing rather than increasing behavior. When analyzing solution behavior at specific points in slope fields, evaluate the differential equation there and consider how the slope changes as the solution evolves.

Question 12

1. All solutions are decreasing.
2. All solutions are increasing or flat, with horizontal tangents at $x=3$. (correct answer)
3. All solutions have slope $0$ only when $y=0$.
4. Slopes depend only on $y$.
5. Solutions must cross at $(3,0)$.

Explanation: This question requires slope field reasoning for equations depending only on x. For dy/dx = (x-3)², the slope depends only on x, not on y, so all solutions have the same slope at each x-value. Since (x-3)² ≥ 0 for all real x, all slopes are non-negative, making solutions either increasing or flat. The slope equals 0 only when x = 3, giving horizontal tangents there. For x ≠ 3, slopes are positive, so solutions increase. All solution curves are vertical translations of each other. Choice A fails because solutions increase (or are flat), never decrease, since slopes are non-negative. When analyzing slope fields for dy/dx = f(x) with f(x) ≥ 0, all solutions are non-decreasing and have horizontal tangents where f(x) = 0.

Question 13

1. It decreases toward $0$ as $x$ increases.
2. It increases toward $1$ as $x$ increases. (correct answer)
3. It stays constant at its initial value.
4. It increases without bound as $x$ increases.
5. It crosses $y=0$ for some $x>0$.

Explanation: This question uses slope field reasoning to analyze logistic growth behavior. For dy/dx = y(1-y) with initial value 0 < y(0) < 1, we have (positive)(positive) = positive slope, so the solution increases. As y approaches 1, the factor (1-y) approaches 0, making dy/dx approach 0, so the solution approaches y = 1 asymptotically. The solution remains between 0 and 1 throughout since it starts there and the slope becomes zero as it nears y = 1. Choice A fails because the solution increases toward 1, not decreases toward 0. To interpret slope fields for autonomous equations, identify where dy/dx = 0 to find equilibrium values, then analyze the sign of dy/dx between equilibria to determine solution behavior.

Question 14

1. The solution remains $y=0$ for all $x$. (correct answer)
2. The solution becomes positive for $x>0$.
3. The solution becomes negative for $x>0$.
4. The solution is undefined at $x=\pi/2$.
5. The solution oscillates between $-1$ and $1$.

Explanation: This question uses slope field reasoning about solutions through the origin. For dy/dx = y cos x, if a solution passes through (0,0), then y(0) = 0. Since dy/dx = y cos x and y = 0 along this solution, we have dy/dx = 0 × cos x = 0 for all x. A differential equation with dy/dx = 0 everywhere along a solution means that solution is constant. Therefore, y remains 0 for all x. Choice B fails because if y = 0 initially, it cannot become positive since dy/dx = 0 prevents any change. When analyzing solutions through equilibrium points in slope fields, check if the equilibrium persists throughout the domain by examining whether dy/dx remains zero.

Question 15

1. increasing because slopes are positive.
2. decreasing because slopes are negative. (correct answer)
3. constant because slopes are zero.
4. undefined because slopes are infinite everywhere.
5. alternating between increasing and decreasing with $x$.

Explanation: This question uses slope field reasoning to analyze solutions in specific quadrants. For dy/dx = x/y in Quadrant IV, we have x > 0 and y < 0, so dy/dx = (positive)/(negative) < 0. Negative slopes indicate that solution curves are locally decreasing in Quadrant IV. The magnitude depends on the ratio |x/y|, but the sign is consistently negative throughout this quadrant. Choice A fails because slopes are negative, not positive, making solutions decrease rather than increase. To determine solution behavior in different quadrants using slope fields, analyze the signs of variables in the differential equation for each quadrant separately.

Question 16

1. All solutions move away from $y=2$.
2. All solutions approach $y=2$. (correct answer)
3. All solutions become negative.
4. All solutions are periodic.
5. All solutions have slope $0$.

Explanation: This question requires slope field reasoning about long-term solution behavior. For $\frac{dy}{dx} = 2 - y$, we can rewrite this as $\frac{dy}{dx} = -(y - 2)$. When $y > 2$, we have $\frac{dy}{dx} < 0$ (decreasing toward $y = 2$); when $y < 2$, we have $\frac{dy}{dx} > 0$ (increasing toward $y = 2$). When $y = 2$, $\frac{dy}{dx} = 0$ (equilibrium). Therefore, all solutions approach the horizontal line $y = 2$ as $x$ increases, regardless of their initial values. Choice A fails because solutions move toward $y = 2$, not away from it. To analyze long-term behavior in slope fields, identify horizontal equilibria by setting $\frac{dy}{dx} = 0$ and determine whether nearby solutions converge to or diverge from these equilibria.

Question 17

1. decrease toward $y=-1$.
2. increase toward $y=1$. (correct answer)
3. remain constant.
4. increase without bound.
5. oscillate between $-1$ and $1$.

Explanation: This question involves slope field reasoning for solutions between equilibria. For $dy/dx = x(1-y^2)$ with $x > 0$ and $|y| < 1$, we have $1-y^2 > 0$ (since $y^2 < 1$), so $dy/dx = (\text{positive})(\text{positive}) > 0$. This means solutions increase. As y approaches $\pm 1$, the factor $(1-y^2)$ approaches 0, so $dy/dx$ approaches 0, meaning solutions approach these boundary values asymptotically. For initial values with $|y| < 1$, solutions increase toward $y = 1$ if they start positive, or increase toward $y = 1$ from negative values. Choice A fails because solutions increase toward equilibria, not decrease. To analyze solutions between equilibria in slope fields, determine the sign of the differential equation and identify which equilibrium the solution approaches based on the initial condition location.

Question 18

1. $x=-1$
2. $y=2$ (correct answer)
3. $y=x+1$
4. $y=-2$
5. $x=2$

Explanation: This question requires slope field reasoning to identify equilibrium solutions. For $\frac{dy}{dx} = (x+1)^2(y-2)$, equilibrium solutions occur when $\frac{dy}{dx} = 0$ for all x. Since $(x+1)^2 \geq 0$ for all real x and equals 0 only when $x = -1$, the only way to have $\frac{dy}{dx} = 0$ everywhere along a horizontal line is if $y-2 = 0$, giving $y = 2$. Along the line $y = 2$, we have $\frac{dy}{dx} = (x+1)^2 \times 0 = 0$ for all x, making it an equilibrium solution. Choice A fails because $x = -1$ is a vertical line where slopes are zero only at the single point where it intersects $y = 2$. To find equilibrium solutions in slope fields, identify horizontal lines where the differential equation equals zero for all x-values.

Question 19

1. All solutions are horizontal lines.
2. All solutions have the same slope at a given $x$-value. (correct answer)
3. All solutions intersect each other infinitely often.
4. Slopes depend only on $y$, not on $x$.
5. Solutions are undefined when $x=0$.

Explanation: This question requires slope field reasoning about equations depending only on x. For $\frac{dy}{dx} = \sin x$, the slope depends only on the x-coordinate, not on y. This means at any fixed x-value, all points (x,y) have the same slope $\sin x$ regardless of their y-coordinate. Therefore, all solution curves have identical slopes at each x-value, differing only by vertical shifts (constants of integration). Choice D incorrectly states slopes depend only on y, when they actually depend only on x. When the differential equation has the form $\frac{dy}{dx} = f(x)$, all solutions are vertical translations of each other, making slope fields show horizontal rows of identical slope segments.

Question 20

1. horizontal for all $y$. (correct answer)
2. vertical for all $y$.
3. positive for all $y$.
4. negative for all $y$.
5. zero only when $y=0$.

Explanation: This question requires slope field reasoning about slopes along vertical lines. For $\frac{dy}{dx} = y \sin x$ at $x = \pi$, we have $\frac{dy}{dx} = y \sin(\pi) = y \times 0 = 0$ for all y values. Since $\sin(\pi) = 0$, the slope equals zero regardless of the y-coordinate, making all slope segments horizontal along the vertical line $x = \pi$. This creates a vertical line of horizontal slope segments. Choice B fails because slopes are finite (zero), not infinite, so segments are horizontal, not vertical. When analyzing slope fields along specific vertical lines, substitute the x-value into the differential equation and observe how slopes vary with y along that line.