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Sketching Slope Fields

Compute and draw local slope-field directions from both coordinates, leaving undefined locations unfilled.

This is shared AB/BC instruction. Open the AB edition of Topic 7.3. Your BC practice and progress stay in this book.

College Board Topic 7.3 · CED p. 134 (PDF p. 139) · CED effective Fall 2020; project course outline 2026–27.

For y′=F(x,y), a slope field places a short line segment at selected points (x,y), with slope F(x,y). Calculate the slope from both coordinates, then draw the segment through the point. A segment shows local direction, not a displacement arrow or an entire solution. Horizontal segments indicate zero slope.

Slope field for y prime equals yFive columns repeat the same slopes: at y equals minus two, minus one, zero, one and two the slopes are respectively minus two, minus one, zero, one and two. Positive solutions rise, negative solutions fall, and y equals zero is an equilibrium.xy0
For y′ = y, the slope depends on height, so every horizontal row repeats.

If F depends only on y, each horizontal row repeats; if it depends only on x, each vertical column repeats. If F=x−y, equal slopes lie on diagonal lines x−y=constant. Mark undefined points rather than inventing a slope. A curve following the field must be tangent to those local directions.

Let's work through it

  1. For y′=x−y at (1,2), compute 1−2=−1. Reason: Substitute both point coordinates.
  2. Draw a short segment falling one unit for each unit to the right. Reason: A slope of −1 describes the segment’s direction.
  3. At (2,2), draw a horizontal segment. Reason: The derivative is 2−2=0 there.

Follow directions to estimate a function

A slope field gives local tangent directions at a finite set of points. For y′ = 2 − y, every plotted point at the same height has the same slope: at y = 0, 1, 2, 3 and 4 the slopes are 2, 1, 0, −1 and −2. A small segment is neither a whole solution nor a step that must be joined to the next segment. To estimate a solution, mark its initial point and draw a smooth curve tangent to nearby directions, continuously checking how the slope changes as the curve changes height.

From (0, 0.5), the selected curve rises, but its positive slopes shrink as it approaches y = 2. From (0, 3.5), another selected curve falls with negative slopes shrinking in magnitude. The constant function y = 2 is an equilibrium because its derivative and the equation’s right side are zero everywhere. The completed graphical key shows three functions from the same family. Its approximate heights near x = 1 are about 1.45 and 2.55 for the lower and upper curves. These are readings of a smooth sketch, not exact values established by a finite field. No fixed-step numerical algorithm is needed for this qualitative estimate.

Slope field for y′ = 2 − y on 0 to 3 in x and 0 to 4 in y. Each horizontal row repeats slopes 2 minus y; slopes are positive below y equals 2, zero at 2 and negative above 2. Initial points are (0,0.5) and (0,3.5). Completed curves rise concave down from 0.5 and fall concave up from 3.5, approaching equilibrium y equals 2. At x equals 1 their heights are about 1.45 and 2.55.
y′ = 2 − y completed graphical key
Inspect diagram labels

The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.

Slope field for y′ = 2 − y on 0 to 3 in x and 0 to 4 in y. Each horizontal row repeats slopes 2 minus y; slopes are positive below y equals 2, zero at 2 and negative above 2. Initial points are (0,0.5) and (0,3.5). Completed curves rise concave down from 0.5 and fall concave up from 3.5, approaching equilibrium y equals 2. At x equals 1 their heights are about 1.45 and 2.55.

View diagram in a separate tab ↗

FUN-7.CFUN-7.C.1FUN-7.C.2FUN-7.C.3

Check direction and concavity before estimating height

Use the supplied y′ = 2 − y field to estimate the curve through (0, 0.5) for 0 ≤ x ≤ 3 and compare it with the curve through (0, 3.5).

Slope field for y′ = 2 − y on 0 to 3 in x and 0 to 4 in y. Each horizontal row repeats slopes 2 minus y; slopes are positive below y equals 2, zero at 2 and negative above 2. Initial points are (0,0.5) and (0,3.5). Completed curves rise concave down from 0.5 and fall concave up from 3.5, approaching equilibrium y equals 2. At x equals 1 their heights are about 1.45 and 2.55.
y′ = 2 − y completed graphical key
Inspect diagram labels

The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.

Slope field for y′ = 2 − y on 0 to 3 in x and 0 to 4 in y. Each horizontal row repeats slopes 2 minus y; slopes are positive below y equals 2, zero at 2 and negative above 2. Initial points are (0,0.5) and (0,3.5). Completed curves rise concave down from 0.5 and fall concave up from 3.5, approaching equilibrium y equals 2. At x equals 1 their heights are about 1.45 and 2.55.

View diagram in a separate tab ↗

  1. At (0,0.5), the initial slope is 1.5; nearby positive segments make the lower curve rise.
  2. As the lower curve rises, 2 − y gets smaller, so its slope decreases and it bends downward. The equation gives y″ = −y′ < 0 below 2.
  3. The upper curve starts with slope −1.5, falls and bends upward because y″ = −y′ > 0 above 2.
  4. Sketch both approaching the horizontal equilibrium y = 2 without splicing field segments into a polygon. Read approximate y(1) values about 1.45 and 2.55; a coarse field does not certify exact coordinates.

Answer and check

The completed key shows a rising concave-down curve from 0.5 and a falling concave-up curve from 3.5. Both approach 2 as x increases. For checking the drawing, y = 2 ± 1.5e^(−x) satisfies the equation and gives about 1.448 and 2.552 at x = 1; finding these formulas is not required to construct the qualitative sketch.

FUN-7.CFUN-7.C.2FUN-7.C.3

Try it from memory

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Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.

Question 1 of 20 self-assessed

Question 1

For y′=y, do horizontal or vertical rows repeat slopes?

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Retrieval 1

For y′=y, do horizontal or vertical rows repeat slopes?

Worked answer and reason

Horizontal rows.

Every point with the same y has the same slope.

Retrieval 2

What does a horizontal segment mean?

Worked answer and reason

The solution’s instantaneous derivative is zero there.

Slope is derivative, not necessarily the function’s value.

Catch the tempting mistake

A slope field consists of a collection of solution points.
It supplies directions at selected points. A solution curve must follow those directions and pass through its initial point.
Calculator workspace and limits

For calculus, set the angle selector to Radians. This scientific calculator evaluates arithmetic and functions. It does not graph arbitrary functions, find numerical roots, evaluate numerical derivatives, or compute definite integrals. Those tasks need an approved graphing tool; the May 2027 exam permits approved handheld graphing calculators and Bluebook’s built-in Desmos on calculator-required parts. A graph or decimal estimate helps you check a conclusion; your setup, hypotheses, units and justification still do the explaining.

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Estimate two members of a solution family

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The supplied field represents y′ = 2 − y on 0 ≤ x ≤ 3 and 0 ≤ y ≤ 4. At every plotted x, slopes at y = 0, 1, 2, 3 and 4 are respectively 2, 1, 0, −1 and −2; intermediate half-unit rows follow the same rule. Two marked initial points are (0,0.5) and (0,3.5). Use qualitative graphical estimation, not a fixed-step numerical method.

Slope field for y′ = 2 − y on 0 to 3 in x and 0 to 4 in y. Each horizontal row repeats slopes 2 minus y; slopes are positive below y equals 2, zero at 2 and negative above 2. Initial points are (0,0.5) and (0,3.5).
y′ = 2 − y supplied field
Inspect diagram labels

The diagram keeps its full size here. Focus the view and use the arrow keys, or scroll across it, to inspect every label.

Slope field for y′ = 2 − y on 0 to 3 in x and 0 to 4 in y. Each horizontal row repeats slopes 2 minus y; slopes are positive below y equals 2, zero at 2 and negative above 2. Initial points are (0,0.5) and (0,3.5).

View diagram in a separate tab ↗

(a) Calculate slopes at (0,0.5), (1,2) and (2,3.5), and draw a short tangent-direction segment at each. Explain what the finite set of segments represents and why a segment is not itself a complete solution function.

FUN-7.C.1

(b) Draw a smooth estimated solution curve through (0,0.5) for 0 ≤ x ≤ 3 on the supplied field. Label its initial point, the equilibrium line and its forward direction. Estimate y(1) from your sketch and justify the curve’s monotonicity and concavity. Explain why your graphical reading is approximate.

FUN-7.CFUN-7.C.2

(c) On the same field, sketch the solution through (0,3.5) and compare its direction and concavity with the lower solution. Explain why one differential equation can describe a family of functions, why the initial conditions choose different members, and why y = 2 is a genuine equilibrium rather than merely one horizontal tangent.

FUN-7.CFUN-7.C.2FUN-7.C.3

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For the teacher
  • Provide a textual slope table alongside the SVG. Let students compute several segments by hand before using a visual field. Check undefined locations explicitly.

Unit planning and formative guidance →

Glossary and deeper reading · 5 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Function

    A mathematical function assigns exactly one output to each input in its domain. Inputs can share an output, but one input cannot have two different outputs. For example, f(x)=x² assigns the output 9 to both 3 and −3. Function notation names the rule and its input; f(3) means the output when the input is 3.

  • Derivative

    A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.

    Read more: 2.2 · Derivative →

  • Concavity

    Concavity describes how a graph’s slope changes as the input increases. A graph is concave up where its derivative increases and concave down where its derivative decreases. For example, x² is concave up because its derivative 2x increases. A positive second derivative establishes concave-up behavior on an interval, while a negative one establishes concave-down behavior.

    Read more: 5.6 · Concavity →

  • Differential equation

    A differential equation relates an unknown function to one or more of its derivatives. A solution is a function that makes the equation true throughout an appropriate interval. For example, y′=2x has solutions y=x²+C. An initial condition can select a particular member of a family of solutions.

    Read more: 7.1 · Differential equation →

  • Initial condition

    An initial condition specifies a solution’s value at a particular input. It can determine the constant in a family of solutions to a differential equation. For example, y′=2x with y(0)=3 selects y=x²+3. The solution must satisfy both the equation and the stated condition on a suitable interval.

    Read more: 7.7 · Initial condition →

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