AP Calculus BC Quiz: Sketching Slope Fields

20 questions · exam conditions

0:00

Sketching Slope FieldsQuestion 1 of 20

Which slope field matches $\dfrac{dy}{dx}=y(1-y)$ for a population fraction $y$ over time $x$ ?

Slopes are zero on

y=0

and

y=1

; slopes are positive for

0<y<1

and negative for

y>1

y<0

Slopes are zero on

x=0

and

x=1

; slopes are positive for

0<x<1

and negative for

x>1

x<0

Slopes depend on both variables with zero slopes on

y=x

; slopes change sign across

y=x

Slopes are constant on diagonals

y-x=c

; slopes increase as

c

increases.

Slopes are zero only at the origin; slopes are positive in Quadrants I and III and negative in Quadrants II and IV.

← Back to quizzes

AP Calculus BC Quiz

AP Calculus BC Quiz: Sketching Slope Fields

Practice Sketching Slope Fields in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

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

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.

All questions

Question 1

Which slope field matches $\dfrac{dy}{dx}=y(1-y)$ for a population fraction $y$ over time $x$ ?

Slopes are zero on $y=0$ and $y=1$ ; slopes are positive for $0<y<1$ and negative for $y>1$ or $y<0$ . (correct answer)
Slopes are zero on $x=0$ and $x=1$ ; slopes are positive for $0<x<1$ and negative for $x>1$ or $x<0$ .
Slopes depend on both variables with zero slopes on $y=x$ ; slopes change sign across $y=x$ .
Slopes are constant on diagonals $y-x=c$ ; slopes increase as $c$ increases.
Slopes are zero only at the origin; slopes are positive in Quadrants I and III and negative in Quadrants II and IV.

Explanation: Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = y(1 - y). To verify, at y = 0, such as (any x, 0), dy/dx = 0(1-0) = 0, and at y = 1, dy/dx = 1(1-1) = 0, confirming zero slopes. Between 0 < y < 1, at (x, 0.5), dy/dx = 0.5(0.5) = 0.25 > 0, positive, while for y > 1 at (x, 2), dy/dx = 2(-1) = -2 < 0, and for y < 0 at (x, -1), dy/dx = -1(2) = -2 < 0, matching choice A. These points show slopes are independent of x, constant along horizontal lines. A tempting distractor like choice B fails because slopes depend on y, not x; at x=0.5, dy/dx varies with y, not constant along vertical lines. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.

Question 2

Which slope field matches $\dfrac{dy}{dx}=\dfrac{1}{1+y^2}$ for all real $x$ and $y$ ?

Slopes depend only on $y$ , always positive, and decrease toward $0$ as $|y|$ increases. (correct answer)
Slopes depend only on $x$ , always positive, and decrease toward $0$ as $|x|$ increases.
Slopes are negative for $y>0$ and positive for $y<0$ .
Slopes are zero along $y=0$ and increase with $|y|$ .
Slopes are undefined on $y=0$ and very steep near that line.

Explanation: This question tests understanding slope fields for rational functions with restricted ranges. For dy/dx = 1/(1+y²), the denominator 1+y² is always positive and at least 1, so slopes are always positive but never exceed 1. The slope equals 1 when y = 0 (since 1/(1+0²) = 1) and decreases toward 0 as |y| increases (since 1+y² grows without bound). At y = 2, slope = 1/(1+4) = 0.2; at y = -3, slope = 1/(1+9) = 0.1. Choice C incorrectly claims slopes are negative for y > 0, but 1/(1+y²) is always positive. For bounded slope fields, identify maximum and minimum values and where they occur.

Question 3

Which slope field matches the differential equation $\dfrac{dy}{dx}=xy$ ?

Slopes are zero along both axes; positive in Quadrants I and III and negative in Quadrants II and IV. (correct answer)
Slopes are zero along both axes; positive in Quadrants II and IV and negative in Quadrants I and III.
Slopes depend only on $x$ and are identical across each vertical column.
Slopes depend only on $y$ and are identical across each horizontal row.
Slopes are zero along $y=x$ only and constant along lines parallel to $y=x$ .

Explanation: This question requires sketching the slope field for dy/dx = xy, which depends on both x and y. At (0,0), the slope is 0·0=0; at (1,1), the slope is 1·1=1; at (-1,1), the slope is (-1)·1=-1; at (1,-1), the slope is 1·(-1)=-1. The slopes are zero along both coordinate axes (where x=0 or y=0), positive in Quadrants I and III (where x and y have the same sign), and negative in Quadrants II and IV (where x and y have opposite signs). Choice E incorrectly claims slopes are zero only along y=x, but xy=0 when either x=0 or y=0. For products in differential equations, analyze the sign of each factor to determine the overall sign pattern.

Question 4

Which slope field matches the differential equation $\dfrac{dy}{dx}=xy$ ?

Slopes are zero along both axes; slopes are positive in Quadrants I and III and negative in Quadrants II and IV. (correct answer)
Slopes are zero along $y=x$ ; slopes are positive when $y<x$ and negative when $y>x$ .
Slopes depend only on $x$ ; each vertical line has constant slope with zeros on $x=0$ .
Slopes depend only on $y$ ; each horizontal line has constant slope with zeros on $y=1$ .
Slopes are always negative and become steeper as $x$ increases.

Explanation: This question involves sketching slope fields for separable equations with xy dependence. For dy/dx = xy, slopes equal zero when either x = 0 or y = 0, giving horizontal tangents along both coordinate axes. In Quadrant I (x > 0, y > 0) and Quadrant III (x < 0, y < 0), the product xy is positive; in Quadrants II and IV, one factor is negative, making xy negative. Choice B incorrectly places zero slopes along y = x instead of the coordinate axes. To analyze dy/dx = xy, check the sign of the product in each quadrant by considering the signs of x and y separately.

Question 5

Which slope field matches $\dfrac{dy}{dx}=\cos y$ for a curve $y(x)$ where slopes repeat as $y$ changes?

Along any horizontal line $y=c$ , all segments share the same slope; slopes vary periodically with $y$ . (correct answer)
Along any vertical line $x=c$ , all segments share the same slope; slopes vary periodically with $x$ .
Slopes are zero along $y=x$ ; slopes are negative above it and positive below it.
Slopes are zero along $y=0$ and undefined along $x=0$ .
Slopes are zero along $y=1$ and increase steadily as $y$ increases.

Explanation: Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = cos y. To verify, along horizontal y=c, dy/dx = cos c, constant regardless of x. For y=0, cos0=1; y=π/2, cos(π/2)=0; y=π, cosπ=-1; y=3π/2,0; repeating every 2π in y, periodic, matching choice A. These show independence from x, varying with y. A tempting distractor like choice B fails because slopes depend on y, not x; along vertical x=c, dy/dx=cos y varies with y, not constant. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.

Question 6

Which slope field matches the differential equation $\dfrac{dy}{dx}=\dfrac{y}{x}$ for $x\ne 0$ ?

Slopes are constant on circles centered at the origin; slopes are undefined on $x^2+y^2=0$ .
Slopes are constant on rays $y=mx$ ; slopes are undefined along $x=0$ and equal $m$ on $y=mx$ . (correct answer)
Slopes are constant on horizontal lines; slopes are undefined along $y=0$ .
Slopes are constant on vertical lines; slopes are undefined along $x=0$ and equal $x$ elsewhere.
Slopes are zero along $y=x$ ; slopes are positive above $y=x$ and negative below.

Explanation: Sketching slope fields is a key skill in understanding differential equations by visualizing the slopes they dictate at various points. To verify the slope field for dy/dx = y/x (x≠0), note along a ray y=mx, slope=m everywhere on that ray, constant. For example, along y=2x, at (1,2)=2/1=2, at (2,4)=4/2=2; along y=-x, (-1,1)=1/(-1)=-1, (2,-2)=-2/2=-1. Undefined along x=0, and each ray has its own constant slope m. A tempting distractor like choice E fails because along y=x, y/x=1, not zero; zero would require y=0, the x-axis. Always pick test points along proposed lines of zero slope and constant y or x lines to confirm patterns in slope fields.

Question 7

Which slope field matches the differential equation $\dfrac{dy}{dx}=x-y$ on the $xy$ -plane?

Slopes are zero along $y=x$ ; along any horizontal line, slopes decrease as $x$ increases.
Slopes are constant on vertical lines; slopes are zero along $x=0$ and increase with $y$ .
Slopes are zero along $y=x$ ; along any horizontal line, slopes increase as $x$ increases. (correct answer)
Slopes are zero along $y=-x$ ; along any horizontal line, slopes increase as $x$ increases.
Slopes depend only on $y$ ; along any horizontal line, all segments have the same slope.

Explanation: Sketching slope fields is a key skill in understanding differential equations by visualizing the slopes they dictate at various points. To verify the slope field for dy/dx = x - y, check points along y = x, such as (1,1) where the slope is 1-1=0, and (2,2) where it's 2-2=0, confirming zero slopes there. Along a horizontal line like y=1, at x=0 the slope is 0-1=-1, at x=1 it's 1-1=0, and at x=2 it's 2-1=1, showing slopes increase with x. Similarly, along y=0, slopes equal x, increasing positively as x grows. A tempting distractor like choice A fails because it claims slopes decrease as x increases along horizontal lines, but actually they increase since the x term adds positively. Always pick test points along proposed lines of zero slope and constant y or x lines to confirm patterns in slope fields.

Question 8

Which slope field matches $\dfrac{dy}{dx}=e^{-x}y$ emphasizing how slope changes as $x$ increases?

Slopes depend only on $x$ and approach $0$ as $x$ increases, regardless of $y$ .
Slopes depend on both $x$ and $y$ ; for fixed $y$ , slopes decrease in magnitude as $x$ increases; slopes are zero on $y=0$ . (correct answer)
Slopes depend only on $y$ ; for fixed $x$ , slopes decrease as $x$ increases.
Slopes are zero along $x=0$ ; slopes increase with $|x|$ .
Slopes are constant along lines $y=-x$ ; horizontal on $y=x$ .

Explanation: This question examines how slopes change with position in exponential decay slope fields. For dy/dx = e^(-x)y, the slope depends on both x and y as a product. When y = 0, the slope is always zero regardless of x, giving a horizontal nullcline. For fixed y ≠ 0, as x increases, e^(-x) decreases toward 0, so the magnitude of the slope decreases. At (0,1), slope = e^0 × 1 = 1, while at (2,1), slope = e^(-2) × 1 ≈ 0.135, confirming the decrease. Choice A incorrectly claims slopes depend only on x, ignoring the factor of y. When analyzing slope fields with products, examine how each factor affects the slope independently.

Question 9

Which slope field matches $\dfrac{dy}{dx}=x-y$ for a solution passing through $(0,0)$ ?

Slopes are zero along $y=x$ ; above that line slopes are negative, below it slopes are positive. (correct answer)
Slopes are zero along $x=0$ ; slopes are positive for $x>0$ and negative for $x<0$ .
Slopes depend only on $y$ , with zero slopes along $y=0$ .
Slopes are zero along $y=-x$ ; above that line slopes are positive, below it slopes are negative.
All slope segments are horizontal everywhere.

Explanation: This question requires sketching a slope field for a linear differential equation. For dy/dx = x - y, slopes are zero when x = y, which is the line y = x. At point (2,1), the slope is 2-1 = 1 (positive), while at (1,2), the slope is 1-2 = -1 (negative). This confirms that above the line y = x (where y > x), we have x - y < 0, giving negative slopes, and below the line (where y < x), we have x - y > 0, giving positive slopes. Choice D incorrectly identifies y = -x as the zero-slope line, but substituting shows that dy/dx = x - (-x) = 2x, which is only zero when x = 0. When sketching slope fields, always verify your nullclines by substituting back into the differential equation.

Question 10

Which slope field matches the differential equation $\dfrac{dy}{dx}=y(1-y)$ near the equilibrium solutions?

Slopes depend only on $x$ , with zero slopes along $x=0$ and $x=1$ .
Slopes depend only on $y$ , with zero slopes along $y=0$ and $y=1$ ; positive for $0<y<1$ , negative otherwise. (correct answer)
Slopes are constant everywhere, equal to $1$ .
Slopes depend on $x+y$ , with zero slopes along the line $y=-x$ .
Slopes are zero along $y=x$ and increase with distance from that line.

Explanation: This question tests your ability to sketch slope fields for autonomous differential equations. For dy/dx = y(1-y), the slope depends only on y, not on x, so slopes are constant along horizontal lines. Setting dy/dx = 0 gives y = 0 and y = 1 as equilibrium solutions where slopes are zero. For 0 < y < 1, we have y > 0 and (1-y) > 0, so dy/dx > 0 (positive slopes). For y < 0 or y > 1, the product y(1-y) is negative, giving negative slopes. Choice A incorrectly suggests slopes depend on x, when they actually depend only on y. To verify slope fields, always check: where are slopes zero, what determines the slope value, and what are the signs in different regions.

Question 11

Which slope field matches the differential equation $\dfrac{dy}{dx}=\sin x$ on $-\pi\le x\le \pi$ ?

Slopes depend only on $y$ , repeating periodically as $y$ changes.
Slopes depend only on $x$ , with zero slopes at $x=-\pi,0,\pi$ and positive on $(0,\pi)$ , negative on $(-\pi,0)$ . (correct answer)
Slopes are zero along $y=x$ and $y=-x$ .
Slopes are constant and equal to $\sin(1)$ everywhere.
Slopes are zero along $x=0$ only, and increase with $|y|$ .

Explanation: This question tests recognizing slope fields for trigonometric differential equations. For dy/dx = sin x, the slope depends only on x, not on y, so slopes are constant along vertical lines. The sine function equals zero at x = -π, 0, and π, giving horizontal slope segments at these x-values. Between x = 0 and x = π, sin x > 0, so slopes are positive; between x = -π and x = 0, sin x < 0, so slopes are negative. Choice A incorrectly suggests slopes depend on y, but sin x is independent of y. To sketch trigonometric slope fields, recall the key features of the trig function: zeros, sign changes, and periodicity.

Question 12

Which slope field matches $\dfrac{dy}{dx}=y^2-x$ based on where the slope segments are horizontal?

Horizontal segments occur along $y=\pm\sqrt{x}$ for $x\ge 0$ ; slopes increase as $y^2-x$ increases. (correct answer)
Horizontal segments occur along $x=\pm\sqrt{y}$ for $y\ge 0$ ; slopes depend only on $x$ .
Horizontal segments occur along $y=x$ only; slopes depend only on $x-y$ .
Horizontal segments occur along $y=0$ and $y=1$ only; slopes depend only on $y$ .
No horizontal segments occur anywhere; slopes are never zero.

Explanation: This question focuses on finding nullclines (where slopes are zero) in slope fields. For dy/dx = y² - x, setting the slope to zero gives y² = x, which means x = y² or equivalently y = ±√x for x ≥ 0. These are two parabolic curves where slope segments are horizontal. Above these curves (where y² > x), slopes are positive; below them (where y² < x), slopes are negative. Choice C incorrectly identifies y = x as the only nullcline, but substituting gives dy/dx = x² - x = x(x-1), which is zero at x = 0 and x = 1, not along the entire line. When finding nullclines, solve the equation dy/dx = 0 algebraically and verify by checking specific points.

Question 13

Which slope field matches $\dfrac{dy}{dx}=\sin(x)$ for $-\pi\le x\le \pi$ ?

Slopes depend only on $y$ ; along each horizontal line, all segments match.
Slopes depend only on $x$ ; slopes are zero at $x=-\pi,0,\pi$ and positive on $(0,\pi)$ . (correct answer)
Slopes depend only on $x$ ; slopes are zero at $x=-\pi,0,\pi$ and negative on $(0,\pi)$ .
Slopes are zero along $y=x$ ; slopes are positive above that line and negative below it.
Slopes are constant on diagonal lines $y-x=c$ ; slopes are zero along $y=x$ .

Explanation: Sketching slope fields is a key skill in understanding differential equations by visualizing the slopes they dictate at various points. To verify the slope field for dy/dx = sin(x), note it depends only on x, so vertical lines have constant slopes, like at x=0 where sin(0)=0 everywhere. At x=π/2, sin(π/2)=1>0, constant for all y; at x=3π/2, sin(3π/2)=-1<0, also constant. Zeros occur at x=-π,0,π as sin vanishes there, and positive on (0,π) where sin(x)>0. A tempting distractor like choice C fails because it claims negative slopes on (0,π), but sin(x) is actually positive in that interval. Always pick test points along proposed lines of zero slope and constant y or x lines to confirm patterns in slope fields.

Question 14

Which slope field matches the differential equation $\dfrac{dy}{dx}=xy$ ?

Slopes are zero along both axes; slopes are positive in Quadrants I and III and negative in Quadrants II and IV. (correct answer)
Slopes are zero along both axes; slopes are positive in Quadrants I and II and negative in Quadrants III and IV.
Slopes are zero along $y=x$ ; slopes are positive above $y=x$ and negative below.
Slopes depend only on $x$ ; along any vertical line, all segments match and change sign at $x=0$ .
Slopes depend only on $y$ ; along any horizontal line, all segments match and change sign at $y=0$ .

Explanation: Sketching slope fields is a key skill in understanding differential equations by visualizing the slopes they dictate at various points. To verify the slope field for dy/dx = xy, check along x=0 where slopes=0y=0, and along y=0 where x0=0, confirming zeros on axes. In Quadrant I (x>0,y>0), xy>0; in III (x<0,y<0), xy>0; in II (x<0,y>0), xy<0; in IV (x>0,y<0), xy<0, matching signs. For example, at (1,1)=1>0, (-1,-1)=1>0, (-1,1)=-1<0. A tempting distractor like choice B fails because it claims positive in I and II, but in II xy is negative due to opposite signs. Always pick test points along proposed lines of zero slope and constant y or x lines to confirm patterns in slope fields.

Question 15

Which slope field matches the differential equation $\dfrac{dy}{dx}=e^{-x}$ ?

Slopes depend only on $y$ ; slopes decrease as $y$ increases and are zero at $y=0$ .
Slopes depend only on $x$ ; all slopes are positive and decrease as $x$ increases. (correct answer)
Slopes depend only on $x$ ; all slopes are negative and increase as $x$ increases.
Slopes are zero along $x=0$ ; slopes are positive for $x>0$ and negative for $x<0$ .
Slopes are constant on circles centered at the origin; slopes are steepest near the origin.

Explanation: Sketching slope fields is a key skill in understanding differential equations by visualizing the slopes they dictate at various points. To verify the slope field for dy/dx = e^{-x}, note it depends only on x, so vertical lines have constant positive slopes, like at x=0 where e^0=1. At x=1, e^{-1}≈0.37<1, decreasing; at x=-1, e^{1}≈2.7>1, larger. As x increases, slopes decrease toward 0 but remain positive. A tempting distractor like choice C fails because it claims negative slopes, but e^{-x} is always positive for real x. Always pick test points along proposed lines of zero slope and constant y or x lines to confirm patterns in slope fields.

Question 16

Which slope field matches $\dfrac{dy}{dx}=\dfrac{x}{1+y^2}$ for a system where slope weakens as $|y|$ grows?

Slopes are constant along vertical lines; sign depends on $x$ , and magnitudes decrease as $|y|$ increases. (correct answer)
Slopes are constant along horizontal lines; sign depends on $y$ , and magnitudes decrease as $|x|$ increases.
Slopes are zero along $y=x$ ; slopes are positive below $y=x$ and negative above it.
Slopes are zero along $y=0$ and $y=1$ ; slopes are positive between them and negative outside.
Slopes are undefined on $y=0$ and become steeper as $|y|$ increases.

Explanation: Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = x / (1 + y²). To verify, along vertical x=c, slope is c / (1 + y²), constant sign as c, but magnitude decreases as |y| increases, e.g., at x=1, y=0: 1/1=1; y=1:1/2=0.5; y=2:1/5=0.2. For x=-1, negative slopes similarly decreasing in magnitude with |y|. These confirm constant along verticals, sign by x, weakening with |y|, matching choice A. A tempting distractor like choice B fails because slopes depend on both, not constant along horizontals; at fixed y=1, dy/dx = x / 2 varies with x. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.

Question 17

Which slope field matches $\dfrac{dy}{dx}=y-x^2$ for a temperature deviation $y$ versus time $x$ ?

Slopes are zero along the curve $y=x^2$ ; slopes are positive above it and negative below it. (correct answer)
Slopes are zero along the curve $x=y^2$ ; slopes are positive to the right and negative to the left.
Slopes depend only on $x$ and are always nonnegative, increasing as $|x|$ increases.
Slopes depend only on $y$ and are zero on $y=0$ and $y=1$ .
Slopes are zero along the line $y=-x$ ; slopes change sign across $y=-x$ .

Explanation: Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = y - x². To verify, along y = x², at (1,1), dy/dx = 1 - 1 = 0, zero slope. Above it, at (1,2), dy/dx = 2 - 1 = 1 > 0, positive; below at (1,0), dy/dx = 0 - 1 = -1 < 0, negative, matching choice A. These points hold for other x, like (2,4) zero, (2,5) positive, (2,3) negative. A tempting distractor like choice C fails because slopes depend on both x and y, not just x; at fixed x=1, dy/dx = y -1 varies with y, not constant. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.

Question 18

Which slope field corresponds to $\dfrac{dy}{dx}=\sin x$ for a curve $y(x)$ modeling vertical displacement?

Along any vertical line $x=c$ , all segments have the same slope; slopes vary periodically with $x$ . (correct answer)
Along any horizontal line $y=c$ , all segments have the same slope; slopes vary periodically with $y$ .
Slopes are zero on $y=x$ ; slopes are positive below $y=x$ and negative above $y=x$ .
Slopes are zero on $x=0$ and increase with distance from the $y$ -axis in both directions.
Slopes are zero on both axes; slopes are positive in Quadrants I and III and negative in Quadrants II and IV.

Explanation: Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = sin x. To verify, along a vertical line like x = 0, dy/dx = sin 0 = 0, constant zero slope. At x = π/2, dy/dx = sin(π/2) = 1, constant positive, and at x = π, dy/dx = 0 again, showing periodic variation with x, matching choice A. These points confirm slopes are the same for all y at fixed x, independent of y. A tempting distractor like choice B fails because slopes depend on x, not y; along y = c, dy/dx = sin x varies with x, not constant. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.

Question 19

Which slope field matches $\dfrac{dy}{dx}=\dfrac{1}{y}$ for a process where slope depends on current level $y$ ?

Slopes are undefined on $y=0$ ; slopes are positive for $y>0$ , negative for $y<0$ , and flatten as $|y|$ increases. (correct answer)
Slopes are undefined on $x=0$ ; slopes are positive for $x>0$ , negative for $x<0$ , and flatten as $|x|$ increases.
Slopes are zero on $y=0$ ; slopes become steeper as $|y|$ increases.
Slopes are constant along diagonals $y-x=c$ and increase as $c$ increases.
Slopes are zero on $y=1$ ; slopes are positive for $y>1$ and negative for $y<1$ .

Explanation: Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = 1/y. To verify, on y = 0, dy/dx is undefined due to division by zero, as stated. For y > 0, at (x,1), dy/dx = 1/1 = 1 > 0, and at (x,2), dy/dx = 1/2 = 0.5 > 0 but flatter; for y < 0, at (x,-1), dy/dx = -1 < 0, and at (x,-2), dy/dx = -0.5 < 0 but less steep, matching choice A. These points show slopes flatten as |y| increases and are independent of x. A tempting distractor like choice C fails because on y=0 it's undefined, not zero, and slopes flatten, not steepen, with |y|. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.

Question 20

Which slope field matches the differential equation $\dfrac{dy}{dx}=y-x^2$ ?

Slopes are zero along the parabola $y=x^2$ ; positive above it and negative below it. (correct answer)
Slopes are zero along the parabola $x=y^2$ ; positive to the right and negative to the left.
Slopes are zero along the line $y=x$ ; positive below it and negative above it.
Slopes depend only on $x$ and are symmetric about the $y$ -axis with both positive and negative values.
Slopes are constant everywhere and equal to $0$ .

Explanation: This question tests sketching slope fields for dy/dx = y - x². To find zero slopes, solve y - x² = 0, giving y = x², a parabola. At (1,2), the slope is 2-1²=1 (positive, above parabola); at (2,3), the slope is 3-4=-1 (negative, below parabola); at (0,0), the slope is 0-0=0 (on parabola). The slopes are positive above the parabola y=x² and negative below it. Choice C incorrectly identifies the zero-slope curve as the line y=x, but setting y-x²=0 clearly gives the parabola y=x². When the derivative equals zero along a curve, that curve divides the plane into regions with different slope signs.