Implicitly Defined Functions

Unit 4 · Functions Involving Parameters, Vectors, and Matrices

What AP Precalc asks here

An implicit equation $F(x, y) = 0$ defines a curve without explicitly writing $y$ as a function of $x$ . Some implicit curves (the unit circle $x^2 + y^2 = 1$ ) can be solved for $y$ — but yield two branches. Others (like a general conic with cross terms) cannot be solved cleanly. Verifying that a point lies on the curve is straightforward: substitute and check.

Implicit definition

Form

F(x, y) = 0

On the curve

(x_0, y_0) \text{ on the curve} \iff F(x_0, y_0) = 0

AP Tip: Implicit equations often define multiple branches of

y(x)

. The unit circle has

y = +\sqrt{1 - x^2}

(upper) and

y = -\sqrt{1 - x^2}

(lower) — neither alone is the full circle.

Type 1

Verify a point on an implicit curve

Substitute and check.

Example 1

(3, 4)

on the curve

x^2 + y^2 = 25

? (A) Yes (B) No, because the LHS is 49 (C) No, because the RHS is wrong (D) Cannot determine without more information

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Type 2

Solve for y to find branches

Solve algebraically. If a square root appears, both branches are valid.

Example 2

Solve

x^2 + y^2 = 9

for

y

. (A)

y = \sqrt{9 - x^2}

(B)

y = 9 - x^2

(C)

y = \pm \sqrt{9 - x^2}

(D)

y = \sqrt{9 + x^2}

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Type 3

Recognize the shape

Match the form of the equation to standard conic shapes.

Example 3

Identify the curve

\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1

. (A) Circle of radius 3 (B) Ellipse with semi-axes 3 and 2 (C) Hyperbola (D) Parabola

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