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What is a parametric function, and how do x and y depend on a third variable?

Topic 4.1 Parametric Functions: define a parametric function giving x and y as functions of a parameter t, and graph and interpret the curve it traces.

A focused answer to AP Precalculus Topic 4.1, covering how a parametric function defines x and y each as a function of a parameter t, how to build a table and graph the curve, the direction of motion, and eliminating the parameter.

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

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A parametric function defines a curve by giving each coordinate as a separate function of a parameter $t$ : $x = x(t)$ and $y = y(t)$ , often written $f(t) = (x(t), y(t))$ . As $t$ increases, the point $(x(t), y(t))$ moves through the plane, tracing a curve with a definite direction of travel (its orientation), usually shown with arrows. To graph one, build a table of $t$ values with the matching $x$ and $y$ , plot the points, and join them in order of increasing $t$ . To eliminate the parameter and recover a relationship between $x$ and $y$ directly, solve one equation for $t$ and substitute into the other. Parametric form carries extra information that a plain $y = f(x)$ graph cannot: the direction and timing of the motion, and curves that fail the vertical line test.

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What this topic is asking
The parametric definition
Building a table and graphing
Eliminating the parameter
Try this

What this topic is asking

The College Board (Topic 4.1) wants you to understand a parametric function, in which both coordinates $x$ and $y$ are given as functions of a third variable, the parameter $t$ . Written $f(t) = (x(t), y(t))$ , it traces a curve in the plane as $t$ varies. You should build a table, plot the curve, note its direction of travel, and eliminate the parameter when possible.

The parametric definition

The parameter $t$ often represents time, so the curve is the path of a moving point and $t$ records when the point is where. This makes parametric form the natural language for motion.

Building a table and graphing

The order of the points matters: two parametric functions can trace the same shape but in opposite directions, and they are different functions.

Graphing and orienting a parametric curve

Graph $x(t) = t^2$ , $y(t) = t$ for $t$ in $[-2, 2]$ , and describe the orientation.

step 1 Build a table

At $t = -2$ : $(4, -2)$ . At $t = -1$ : $(1, -1)$ . At $t = 0$ : $(0, 0)$ . At $t = 1$ : $(1, 1)$ . At $t = 2$ : $(4, 2)$ .

step 2 Plot the points

The points lie on a sideways parabola opening to the right, symmetric about the $x$ -axis.

step 3 Mark the orientation

As $t$ increases from $-2$ to $2$ , the curve is traced from the bottom point $(4, -2)$ up through the vertex $(0, 0)$ to the top point $(4, 2)$ : the direction of travel is upward.

step 4 Note the failed vertical line test

The curve is not the graph of a single function $y = f(x)$ , since $x = 4$ corresponds to both $y = -2$ and $y = 2$ . Parametric form handles this naturally, because each $t$ gives exactly one point.

Eliminating the parameter

To convert to a direct relationship between $x$ and $y$ , solve one component for $t$ and substitute into the other. In the worked example, $y = t$ gives $t = y$ , and $x = t^2$ becomes $x = y^2$ : the sideways parabola, now as a single equation. Eliminating the parameter recovers the shape but discards the direction and the timing, so it is a one-way simplification.

A point worth stating once is that parametric form carries strictly more information than a plain Cartesian equation. The same set of points can be parametrised in infinitely many ways, traced at different speeds or in different directions, and each parametrisation is a distinct function even though they share a graph. This is why orientation and the parameter domain must always be stated; dropping them loses exactly the motion information that makes parametric form useful for the planar-motion models of Topic 4.2.

Try this

Q1. For $x(t) = t - 3$ , $y(t) = 2t$ , find the point at $t = 1$ . [1 point]

Cue. $x(1) = -2$ , $y(1) = 2$ , so the point is $(-2, 2)$ .

Q2. Eliminate the parameter for $x = t$ , $y = t + 4$ . [1 point]

Cue. Since $t = x$ , substitute to get $y = x + 4$ , a line.

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). A parametric function is given by

x(t) = t + 1

and

y(t) = t^2

. What point corresponds to

t = 2

? (A)

(2, 4)

(B)

(3, 4)

(C)

(3, 2)

(D)

(2, 2)

Show worked answer →

The correct answer is (B), $(3, 4)$ .

Substitute $t = 2$ into each component: $x(2) = 2 + 1 = 3$ and $y(2) = 2^2 = 4$ . The point is $(x, y) = (3, 4)$ . Choice (A) forgets to add $1$ to $x$ ; the parameter feeds both components.

AP 2025 (style)3 marksSection II (free response, no calculator). A curve is defined by

x(t) = 2t

and

y(t) = t^2 - 1

for

t

[-2, 2]

. (a) Make a table of

(x, y)

t = -2, 0, 2

. (b) Eliminate the parameter to write

y

as a function of

x

Show worked answer →

A 3-point question on parametric tables and parameter elimination.

(a) Table (1 point): at $t = -2$ , $(x, y) = (-4, 3)$ ; at $t = 0$ , $(0, -1)$ ; at $t = 2$ , $(4, 3)$ .
(b) Eliminate (2 points): from $x = 2t$ , solve $t = \frac{x}{2}$ . Substitute into $y = t^2 - 1$ : $y = \left(\frac{x}{2}\right)^2 - 1 = \frac{x^2}{4} - 1$ . The curve is a parabola opening upward.

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Sources & how we know this

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