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Additional course content; not assessed on the AP Precalculus Exam.

AP Precalculus · Lesson 4.2

Implicitly Defined Curves

How it works

An implicit equation relates coordinates without necessarily giving one output for each input. Test whether a vertical line can meet the curve more than once. Choosing a branch can create a function but also restricts its range.

F(x,y)=0,x2+y2=9 ⇒ y=±9−x2F(x,y)=0,\quad x^2+y^2=9\ \Rightarrow\ y=\pm\sqrt{9-x^2}

Worked examples

Example 1

Does x2+y2=9\displaystyle x^2+y^2=9 define y as a single function of x?

No. At x=0\displaystyle x=0, both y=3\displaystyle y=3 and y=−3\displaystyle y=-3 satisfy the equation.

Example 2

Write the upper semicircle as a function.

y=9−x2\displaystyle y=\sqrt{9-x^2} with domain [−3,3]\displaystyle [-3,3] and range [0,3]\displaystyle [0,3].

Problems

Problem 4.2.1

For x2+y2=9\displaystyle x^2+y^2=9, find all y-values when x=0\displaystyle x=0 and when x=3\displaystyle x=3.

Circle centered at the origin with radius 3.
The full circle fails the vertical-line test for a function of x.
Show worked solution

At x=0\displaystyle x=0, y=±3\displaystyle y=\pm3. At x=3\displaystyle x=3, y=0\displaystyle y=0. The first input fails the uniqueness requirement for a function.

Circle centered at the origin with radius 3. The solution adds labeled relationships.
At x = 0 the two outputs are 3 and -3.

Problem 4.2.2

Solve xy=6\displaystyle xy=6 for y and state the domain.

Show worked solution

y=6/x\displaystyle y=6/x, with x≠0\displaystyle x\ne0. This relation does define a function of x on that domain.

Problem 4.2.3

Does x=y2\displaystyle x=y^2 define y as a function of x over all its real points?

Show worked solution

No. For x>0\displaystyle x>0 there are two outputs ±x\displaystyle \pm\sqrt{x}. Restricting to y≥0\displaystyle y\ge0 gives one branch.

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