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What is an implicitly defined relation, and how does it differ from a function written y = f(x)?

Topic 4.5 Implicitly Defined Functions: interpret a relation given by an equation in x and y, and analyze its graph even when it is not a function of x.

A focused answer to AP Precalculus Topic 4.5, covering relations defined implicitly by an equation in x and y, why they need not pass the vertical line test, and how to analyze their graphs and extract function pieces.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

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Quick answer

An implicitly defined relation is an equation in $x$ and $y$ , such as $x^2 + y^2 = 25$ or $x^2 - y^2 = 1$ , that describes a curve without isolating $y$ . Unlike a function written $y = f(x)$ , an implicit relation can assign more than one $y$ -value to a single $x$ , so it may fail the vertical line test, as a circle does. To analyze it, you can solve for $y$ where possible, which often splits the curve into two or more function pieces (for a circle, an upper and a lower semicircle), each with its own domain. The implicit form is the natural way to write conic sections and other curves that are not functions, and it connects to parametric form, since parametrising a relation produces motion along the same curve. The key idea is that "relation" is broader than "function": every function is a relation, but not every relation is a function.

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What this topic is asking
Relations versus functions
Why implicit curves can fail the vertical line test
Reading the graph directly
Try this

What this topic is asking

The College Board (Topic 4.5) wants you to work with implicitly defined relations: a relationship between $x$ and $y$ given by an equation, such as $x^2 + y^2 = 25$ , that is not solved for $y$ . Such a relation need not be a function of $x$ ; it can fail the vertical line test. You analyze its graph, and where useful split it into function pieces by solving for $y$ .

Relations versus functions

So an implicit equation describes a curve, and whether that curve is a function depends on whether any vertical line meets it more than once.

Why implicit curves can fail the vertical line test

This is exactly why circles, ellipses and hyperbolas are written implicitly: they are not functions, so $y = f(x)$ cannot capture them in one piece.

Splitting an implicit relation into functions

Express $x^2 + y^2 = 9$ as functions of $x$ and state where each is defined.

step 1 Solve for y

$y^2 = 9 - x^2$ , so $y = \pm\sqrt{9 - x^2}$ .

step 2 Name the two pieces

$y = \sqrt{9 - x^2}$ is the upper semicircle (non-negative $y$ ); $y = -\sqrt{9 - x^2}$ is the lower semicircle (non-positive $y$ ).

step 3 Find the domain

The expression under the root must be non-negative: $9 - x^2 \ge 0$ , that is $-3 \le x \le 3$ . So each piece has domain $[-3, 3]$ .

step 4 Explain why two pieces

The full circle gives two $y$ -values for each $x$ in $(-3, 3)$ , so no single function of $x$ works. Splitting at the $x$ -axis into upper and lower halves makes each piece pass the vertical line test, recovering two genuine functions.

Reading the graph directly

You can analyze an implicit relation without solving for $y$ by testing points and using symmetry. For $x^2 + y^2 = 25$ , the intercepts are where one variable is zero ( $(\pm 5, 0)$ and $(0, \pm 5)$ ), and the symmetry in both $x$ and $y$ (since both appear squared) shows the curve is symmetric about both axes and the origin. Recognizing the standard forms, circle, ellipse, parabola, hyperbola, lets you sketch the curve from the equation, which is the work of Topic 4.6.

A point worth stating once is that switching from implicit to explicit form is sometimes impossible or messy, and that is fine. The implicit equation is itself a complete description of the curve; solving for $y$ is only a tool for analysis when it is convenient. Some relations have no clean explicit form at all, yet their graphs are perfectly well defined. Treating the implicit equation as the primary object, rather than always trying to force a $y = f(x)$ form, is the mindset that carries through the conic sections and their parametrisations in Topics 4.6 and 4.7.

Try this

Q1. Does $x = y^2$ define $y$ as a function of $x$ ? [1 point]

Cue. No: for $x = 4$ both $y = 2$ and $y = -2$ work, so it fails the vertical line test; it is a relation, not a function of $x$ .

Q2. Solve $x^2 + y^2 = 1$ for the upper half as a function of $x$ . [1 point]

Cue. $y = \sqrt{1 - x^2}$ , with domain $[-1, 1]$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2025 (style)1 marksSection I, Part A (multiple choice, no calculator). The equation

x^2 + y^2 = 25

defines which kind of relation? (A) A function of

x

(B) A circle, which is not a function of

x

Show worked answer →

The correct answer is (B), a circle, which is not a function of $x$ .

$x^2 + y^2 = 25$ is a circle of radius $5$ centered at the origin. It fails the vertical line test (for example $x = 0$ gives both $y = 5$ and $y = -5$ ), so it is not a function of $x$ ; it is an implicitly defined relation.

AP 2025 (style)3 marksSection II (free response, no calculator). Consider the implicit relation

x^2 + y^2 = 16

. (a) Solve for

y

to express the relation as two functions of

x

. (b) State the domain of these functions and explain why two are needed.

Show worked answer →

A 3-point question on splitting an implicit relation into functions.

(a) Solve (2 points): $y^2 = 16 - x^2$ , so $y = \sqrt{16 - x^2}$ (the upper semicircle) and $y = -\sqrt{16 - x^2}$ (the lower semicircle).
(b) Domain and reason (1 point): both functions have domain $[-4, 4]$ , where $16 - x^2 \ge 0$ . Two functions are needed because the full circle assigns two $y$ -values to most $x$ -values, so no single function of $x$ can describe it; splitting into upper and lower halves gives one $y$ per $x$ on each piece.

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Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)