# Implicitly defined functions

> AP Precalculus · Unit 4: Functions Involving Parameters, Vectors, and Matrices
> Source: https://www.owlsprep.com/study/ap-precalculus-u4-implicitly-defined-functions/

This module explains how an equation in two variables can implicitly describe one or more functions. You will graph such an equation by finding solution points, use the vertical line test to decide whether the whole graph is a function, solve for one variable to break a relation into function portions (such as the top and bottom halves of a circle), find output values for a given input, and describe how the two quantities vary together along the graph.

**Prerequisites:** Function definition and function notation $y=f(x)$; The vertical line test for functions; Solving equations for a chosen variable, including taking square roots

## Learning objectives

- Distinguish an explicitly defined function $y=f(x)$ from an equation in two variables that describes one or more functions implicitly
- Graph an equation in two variables by finding and plotting its solution points
- Use the vertical line test to decide whether the whole graph of an equation is a function
- Solve for one variable to break a relation into function portions (such as the top and bottom halves of a circle), find output values for a given input, and describe how the two quantities vary together

## Explicit Functions vs. Equations in Two Variables

For most of this course you have worked with explicitly defined functions, written as $y=f(x)$: the output $y$ is given directly by a rule applied to the input $x$, and each input has exactly one output. Many important curves, however, come from an equation that simply relates $x$ and $y$, such as $x^2+y^2=25$. An equation like this is a relation: its graph is the set of every ordered pair $(x,y)$ that makes the equation true. It is not written in the form $y=f(x)$, and on its own it need not be a function at all.

**Implicitly defined function** — An equation involving two variables can implicitly describe one or more functions. Even though the equation is not solved for $y$, solving it for one variable produces functions whose graphs make up part or all of the graph of the equation.

*Notation:* $x^2 + y^2 = 25$

*Example:* $x^2+y^2=25$ implicitly describes the two functions $y=\sqrt{25-x^2}$ (the top half of the circle) and $y=-\sqrt{25-x^2}$ (the bottom half).

> **info**
>
> The graph of an equation in two variables can be a complete curve — like the full circle $x^2+y^2=25$ — that is not itself a function of $x$, because some vertical lines cross it twice. Packaged inside that one relation, though, are several genuine functions.

## Graphing an Equation in Two Variables

You graph an equation in two variables the same way you graph any function: by finding solutions and plotting them. A solution is an ordered pair $(x,y)$ that satisfies the equation. Choose an input value, substitute it, solve for the matching output value or values, and mark each resulting point. The graph is the collection of all such points.

**Worked example:** Graph the circle $x^2+y^2=25$ by building a table of solutions.

1. Solve the equation for $y$ so you can compute outputs from inputs:

   $$y = \pm\sqrt{25 - x^2}$$
2. Choose inputs from $-5$ to $5$ and find every matching output. Each input with $|x|<5$ gives two outputs:

   $$x=0 \Rightarrow y=\pm 5,\quad x=\pm 3 \Rightarrow y=\pm 4,\quad x=\pm 4 \Rightarrow y=\pm 3,\quad x=\pm 5 \Rightarrow y=0$$
3. Plot every point — for example $(0,5),(0,-5),(3,4),(3,-4),(4,3),(4,-3),(5,0),(-5,0)$. Connecting them traces a circle of radius $5$ centered at the origin.

> **info**
>
> Only inputs with $-5 \le x \le 5$ give real outputs; for $|x|>5$ the value $25-x^2$ is negative and has no real square root, so the graph stops at $x=\pm 5$.

> **Exam tip:** When you solve an equation for $y$ and reach a square root, keep the $\pm$: most inputs then give two outputs, so plot both the upper and the lower point.

## The Vertical Line Test: Is the Whole Graph a Function?

A graph represents $y$ as a function of $x$ only when every vertical line meets it at most once, so that each input has a single output. The full circle $x^2+y^2=25$ fails this test: the vertical line $x=3$ meets it at both $(3,4)$ and $(3,-4)$, giving one input two outputs. So the circle, taken as a whole, is not a function of $x$ — it is a relation that bundles several functions together.

- **Not a function of $x$:** the circle $x^2+y^2=25$, the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$, the sideways parabola $x=y^2$, and the hyperbola $x^2-y^2=1$ each fail the vertical line test.
- **Already a function of $x$:** an equation such as $x^3+y=1$ can be solved to give the single rule $y=1-x^3$, so its whole graph passes the vertical line test.
- **A relation can still be a function of $y$:** the sideways parabola $x=y^2$ is not a function of $x$, but every input $y$ gives exactly one output $x$, so it is a function of $y$.

> **info**
>
> Failing the vertical line test does not make an equation useless — it just means you must describe the graph as a relation, or split it into function portions, which is the next step.

> **Exam tip:** To show a graph is not a function of $x$, you only need one vertical line that crosses it more than once — a single counterexample is enough.

## Breaking a Relation into Function Portions

To pull genuine functions out of an equation that fails the vertical line test, solve it for one variable. Each branch of the result — each sign of a square root, or each restricted piece — is a function whose graph is a portion of the full graph. Laid side by side, the portions rebuild the entire graph.

**Worked example:** Split the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ into function portions, and find the points on it where $x=2$.

1. Solve the equation for $y$:

   $$\frac{y^2}{9}=1-\frac{x^2}{16} \;\Rightarrow\; y=\pm 3\sqrt{1-\frac{x^2}{16}}$$
2. Read off the two portions: the $+$ sign gives the upper portion $y=3\sqrt{1-\frac{x^2}{16}}$ and the $-$ sign gives the lower portion $y=-3\sqrt{1-\frac{x^2}{16}}$. Each is a function with domain $-4\le x\le 4$.
3. Substitute the input $x=2$ into each portion:

   $$y=\pm 3\sqrt{1-\frac{4}{16}}=\pm 3\sqrt{\tfrac{3}{4}}=\pm\frac{3\sqrt{3}}{2}$$
4. So the input $x=2$ meets the ellipse at $\left(2,\frac{3\sqrt{3}}{2}\right)$ on the upper portion and $\left(2,-\frac{3\sqrt{3}}{2}\right)$ on the lower portion.

**Worked example:** The relation $x=y^2$ is not a function of $x$. Break it into function portions of $x$.

1. Solve for $y$:

   $$y=\pm\sqrt{x}$$
2. The upper portion $y=\sqrt{x}$ and the lower portion $y=-\sqrt{x}$ are each functions of $x$ with domain $x\ge 0$; together they form the whole sideways parabola.
3. For the input $x=9$, the two portions give $y=3$ and $y=-3$, i.e. the points $(9,3)$ and $(9,-3)$.

> **info**
>
> You do not always have to solve globally. Near any single point of the graph you can pick the one portion that passes through it; that portion behaves like an ordinary function there, which is all most problems need.

> **Exam tip:** Each choice of sign after a square root is a separate function portion. Decide which portion you want (top or bottom, right or left) before you pick the sign.

## How the Two Quantities Vary Together

Even without solving for $y$, you can describe how the two quantities on an implicitly defined graph change together by comparing two ordered pairs that lie close together. Look at the ratio of the change in the two variables — the average rate of change from one point to the next. If that ratio is positive, the two variables increase together or decrease together; if it is negative, one increases while the other decreases.

**Worked example:** On the circle $x^2+y^2=25$, describe how $x$ and $y$ vary together along the arc from $(0,5)$ to $(5,0)$.

1. List a few solution points along the arc:

   $$(0,5),\;(3,4),\;(4,3),\;(5,0)$$
2. As the input $x$ increases $0\to 3\to 4\to 5$, the output $y$ decreases $5\to 4\to 3\to 0$. One goes up while the other goes down, so on this arc the two quantities vary in opposite directions.
3. Check with the ratio of the changes between $(0,5)$ and $(3,4)$:

   $$\frac{\Delta y}{\Delta x}=\frac{4-5}{3-0}=-\frac{1}{3}<0$$
4. The negative ratio confirms that as $x$ increases, $y$ decreases along this portion.

The way the graph runs can level off. Where the output $y$ momentarily stops changing while $x$ changes — the average rate of change of $y$ with respect to $x$ is zero — the graph runs horizontally; on the circle this happens at the top $(0,5)$ and bottom $(0,-5)$. Where instead the input $x$ momentarily stops changing while $y$ changes, the graph runs vertically; on the circle this happens at the far-right $(5,0)$ and far-left $(-5,0)$ points.

**Check your understanding**

Test your understanding of implicitly defined functions with this AP-style question:

1. On the graph of $x^2+y^2=25$, which point lies on the lower-half function $y=-\sqrt{25-x^2}$?

   - $(3,4)$
   - $(4,3)$
   - $(3,-4)$
   - $(0,5)$

   *Why:* Correct: the lower-half portion uses the $-$ sign, so its outputs are negative. Substituting $x=3$ gives $y=-\sqrt{25-9}=-4$, the point $(3,-4)$.

> **Exam tip:** Read 'vary together' from two nearby points: compare the change in $y$ to the change in $x$. Same sign means they move the same way; opposite signs mean one rises as the other falls.

## Common pitfalls

- **Wrong:** Assuming every equation in $x$ and $y$ is a function and writing it as $y=f(x)$ without checking.
  - Why it fails: Students are used to functions from earlier units and expect one output per input.
  - Correct: First apply the vertical line test; if a vertical line can cross the graph twice, the whole equation is a relation, not a single function of $x$.
- **Wrong:** Dropping the $\pm$ after solving for $y$, e.g. reporting $y=\sqrt{25-x^2}$ as the whole circle.
  - Why it fails: The square-root key gives only the principal (non-negative) root, so students keep just the upper portion.
  - Correct: Keep both signs: $y=+\sqrt{25-x^2}$ is the top portion and $y=-\sqrt{25-x^2}$ is the bottom portion; together they form the full graph.
- **Wrong:** Forgetting the domain of a portion, e.g. plugging $x=6$ into $y=\pm\sqrt{25-x^2}$ for the circle of radius $5$.
  - Why it fails: Students treat the portion like a polynomial that accepts any input.
  - Correct: A portion is only defined where the expression under the root is non-negative; for $y=\pm\sqrt{25-x^2}$ that means $-5\le x\le 5$.
- **Wrong:** Concluding a relation is 'not a function' in every sense once it fails the vertical line test.
  - Why it fails: Students hear 'not a function' and stop, missing that it can still be a function of the other variable or split into portions.
  - Correct: Say precisely what fails: the whole graph is not a function of $x$, but it may be a function of $y$, and it always splits into function portions.
- **Wrong:** Mixing up which points run horizontally and which run vertically on a graph such as a circle.
  - Why it fails: Students memorize 'top and bottom' versus 'sides' without linking them to which quantity stops changing.
  - Correct: The graph runs horizontally where $y$ momentarily stops changing (top and bottom of the circle) and vertically where $x$ momentarily stops changing (left and right of the circle).

## Cheatsheet

| Concept | Idea / Formula | Notes |
| --- | --- | --- |
| Explicit function | $y=f(x)$ | One rule, one output per input; used throughout the course |
| Equation in two variables (relation) | e.g. $x^2+y^2=25$ | Graph is all $(x,y)$ that satisfy it; need not be a function |
| Graph by solving | find solution points and plot | Pick inputs, solve for outputs, mark each point |
| Vertical line test | each vertical line meets the graph $\le 1$ time | If some vertical line meets it twice, not a function of $x$ |
| Function portion | $y=+\sqrt{25-x^2}$ / $y=-\sqrt{25-x^2}$ | Each sign after a root is one function; portions rebuild the graph |
| Output for a given input | substitute $x$, solve for $y$ | Most inputs on a two-branch relation give two outputs ($\pm$) |
| Vary together (positive ratio) | $\frac{\Delta y}{\Delta x}>0$ between nearby points | $x$ and $y$ increase together or decrease together |
| Vary together (negative ratio) | $\frac{\Delta y}{\Delta x}<0$ between nearby points | One increases while the other decreases |
| Horizontal / vertical spots | $\Delta y\approx 0$ vs $\Delta x\approx 0$ | Runs horizontally where $y$ stops changing, vertically where $x$ stops changing |

## What's next

Implicitly defined functions build directly on the idea of a function and the vertical line test, and they set up the rest of Unit 4. Next you will look at conic sections — circles, ellipses, and hyperbolas — which are exactly the equations in two variables you have been splitting into portions here, and at parametric functions, which give another way to trace a whole curve (including ones that fail the vertical line test) using a single parameter. The skills from this module — graphing an equation by finding solutions, testing whether a graph is a function, solving for one variable, and describing how two quantities vary together — carry straight into those topics.

- [Parametric functions of conic sections](https://www.owlsprep.com/study/ap-precalculus-u4-parametric-functions-of-conic-sections/)
- [Parametric functions](https://www.owlsprep.com/study/ap-precalculus-u4-parametric-functions/)
- [AP Precalculus Vectors](https://www.owlsprep.com/study/ap-precalculus-u4-vectors/)

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