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Reasoning Using Slope Fields

Reason about a solution family, equilibrium, direction, and concavity from a slope field and equation.

College Board Topic 7.4 · CED p. 135 (PDF p. 140) · CED effective Fall 2020; project course outline 2026–27.

A solution curve follows the field; an initial condition picks its starting point. Sketch a smooth curve tangent to nearby segments, and infer where it increases, decreases, or levels off. For y′=2−y, the horizontal line y=2 is an equilibrium solution. Below 2, slopes are positive; above 2, slopes are negative. Solutions move toward 2 as x increases.

A slope field represents a family of possible functions, not one answer. To infer concavity, differentiate the differential equation using the ordinary product or chain rule as needed. For y′=2−y, differentiating both sides gives y″=−y′, so a rising solution below 2 is concave down. In another example, y′=xy gives y″=y+xy′ by the product rule because y depends on x. A sketch is an estimate; do not report an exact coordinate from a coarse field unless the model gives it.

Let's work through it

  1. For y′=2−y with y(0)=1, the initial slope is 1. Reason: Use the initial point in the field rule.
  2. The curve increases while below 2 and bends downward because y″=−y′<0. Reason: The equation gives both direction and concavity.
  3. It approaches the equilibrium y=2 in the forward direction. Reason: The smooth model has positive slopes below 2 that shrink toward zero.

A zero-slope location is not necessarily an equilibrium

In y′ = x − y, zero-slope segments lie on the diagonal y = x. That diagonal is a collection of locations where a solution would have a horizontal tangent; it is not itself a solution. The function y = x has derivative 1, while x − y would be 0. No constant y = c is an equilibrium on an interval, because its derivative 0 cannot equal x − c for every x.

A selected solution through (0,1) initially falls. The supplied field and completed curve show it flattening near (0.7,0.7), then rising. Its approximate height at x = 3 is 2.1. At a horizontal tangent y′ = 0, differentiating the model gives y″ = 1 − y′ = 1, so the curve has a local minimum there. The initial condition selects one function from a family; changing it changes the curve. Keep qualitative estimates distinct from exact coordinates or a formula verified separately.

Slope field for y′ = x − y on 0 to 3 in x and 0 to 3 in y. The slope is x minus y. Initial point is (0,1); zero-slope locations lie on y equals x. The completed selected curve falls concave up to a minimum near (0.69,0.69), then rises concave up to about (3,2.10). The dashed zero-slope line is distinct from the solution.
y′ = x − y completed graphical key
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Slope field for y′ = x − y on 0 to 3 in x and 0 to 3 in y. The slope is x minus y. Initial point is (0,1); zero-slope locations lie on y equals x. The completed selected curve falls concave up to a minimum near (0.69,0.69), then rises concave up to about (3,2.10). The dashed zero-slope line is distinct from the solution.

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

Verify a family without confusing it with a field

A candidate family for y′ = x − y is y = x − 1 + Ce^(−x). Verify the family and select its members through (0,1) and (0,2).

  1. Differentiate a candidate member: y′ = 1 − Ce^(−x).
  2. Substitute it into the differential equation: x − y = 1 − Ce^(−x), agreeing with y′ for every x.
  3. At x = 0 the value is −1 + C. The two initial conditions therefore give C = 2 and C = 3.
  4. These formulas verify two solution functions. A field shows their tangent directions, while an initial condition selects one member; a verified exact formula is different from a numerical reading of a sketch.

Answer and check

Both members satisfy the differential equation: y = x − 1 + 2e^(−x) passes through (0,1), and y = x − 1 + 3e^(−x) passes through (0,2). Their initial slopes are −1 and −2. The completed graphical key illustrates the first member; its coordinates read from the drawing remain approximate.

FUN-7.C.3

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Question 1 of 20 self-assessed

Question 1

For y′=2−y starting above 2, does y increase or decrease?

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These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.

Retrieval 1

For y′=2−y starting above 2, does y increase or decrease?

Worked answer and reason

Decrease.

The slope is negative when y>2.

Retrieval 2

Is y=2 a solution of y′=2−y?

Worked answer and reason

Yes.

A constant has derivative zero, and the right side is zero at y=2.

Catch the tempting mistake

A zero slope at one point means the whole solution is constant.
An equilibrium requires the constant function to satisfy the equation for all relevant x, not just one point.
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Follow a turning curve through a two-coordinate field

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The supplied slope field represents y′ = x − y on 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3. For example, slopes at (0,1), (1,1) and (2,1) are −1, 0 and +1. The selected solution passes through (0,1). Use qualitative estimation; an exact formula or numerical stepping algorithm is not required.

Slope field for y′ = x − y on 0 to 3 in x and 0 to 3 in y. The slope is x minus y. Initial point is (0,1); zero-slope locations lie on y equals x.
y′ = x − 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′ = x − y on 0 to 3 in x and 0 to 3 in y. The slope is x minus y. Initial point is (0,1); zero-slope locations lie on y equals x.

View diagram in a separate tab ↗

(a) Locate the line of horizontal field segments. Decide whether that line is itself a solution by checking its derivative against the differential equation. Decide whether any constant function y = c is an equilibrium on an interval, and explain your conclusion.

FUN-7.C.1FUN-7.C.3

(b) Sketch a smooth estimated solution through (0,1) over the supplied domain. Mark its approximate minimum and estimate y(3). Explain from the field why it first decreases and then increases, and why the plotted coordinate estimates cannot be stated as exact.

FUN-7.CFUN-7.C.2

(c) Differentiate the differential equation to find y″ in terms of y′. Calculate y′ and y″ at the initial point and y″ at a horizontal tangent of the selected solution. Explain how these values support the shape of your sketch and how an initial condition selects one function from the solution family.

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

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For the teacher
  • Ask students to explain their sketch with derivative signs and concavity. Use the same equation at two different initial heights to show a family of solutions.

Unit planning and formative guidance →

Glossary and deeper reading · 8 refreshers

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

  • 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 →

  • Slope field

    A slope field displays short line segments with slopes specified by a first-order differential equation. For y′=F(x,y), the segment at (x,y) has slope F(x,y). For example, y′=y gives horizontal segments along y=0 and positive slopes above that axis. Solution curves follow the local slopes, and an initial point helps choose the relevant curve.

    Read more: 7.3 · Slope field →

  • 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 →

  • 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 →

  • Local extremum

    A local extremum is a value at least as large as, or at least as small as, all values at nearby allowed inputs. A local maximum compares nearby heights above, while a local minimum compares them below. For example, x² has a local minimum at 0. A local extremum need not be the greatest or least value over the entire domain.

    Read more: 5.2 · Local extremum →

  • 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 →

  • Interval

    An interval is a set of real numbers containing every number between any two of its members. Parentheses exclude an endpoint, while square brackets include it. For example, (1,4] includes 4 and every number between 1 and 4, but excludes 1. An infinite endpoint always uses a parenthesis because infinity is not a real endpoint.

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