Implicitly Defined Functions — AP® Precalculus Notes

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

So far most functions have come as formulas: $y = f(x)$ , or $x(t)$ and $y(t)$ . Many curves are more natural as a single equation tying $x$ and $y$ together. A circle is the clearest case: $x^2 + y^2 = 25$ says "the distance from the origin is 5" without ever saying $y = \ldots$ .

Explicit and implicit

Form	Example	How you get points
Explicit	$y = x^2 - 3$	put in $x$ , read off $y$
Parametric	$(t^2, 2t)$	put in $t$ , read off $x$ and $y$
Implicit	$x^2 + xy + y^2 = 7$	put in $x$ , solve for every $y$

The graph of an implicitly defined function is the set of all points $(x, y)$ that make its equation true. To check whether a point is on the curve, substitute both coordinates: $(1, 2)$ is on $x^2 + xy + y^2 = 7$ because $1 + 2 + 4 = 7$ , and $(2, 2)$ is not because $4 + 4 + 4 = 12$ .

Finding points

Worked example 1. Find every point of $x^2 + xy + y^2 = 7$ with $x = 1$ .

Substitute: $1 + y + y^2 = 7$ , so $y^2 + y - 6 = 0$ .

Factor: $(y - 2)(y + 3) = 0$ , so $y = 2$ or $y = -3$ .

The points are $(1, 2)$ and $(1, -3)$ . One $x$ , two $y$ -values: this curve is not the graph of a function of $x$ .

Splitting a curve into functions

A curve that fails the vertical line test can often be cut into pieces that pass it. For the circle $x^2 + y^2 = 25$ , solving for $y$ gives $y = \pm\sqrt{25 - x^2}$ : the upper half $y = \sqrt{25 - x^2}$ and the lower half $y = -\sqrt{25 - x^2}$ are each functions of $x$ .

The circle x squared plus y squared = 25. Its upper half, y = the square root of 25 minus x squared, is drawn in green and its lower half, y = minus the square root of 25 minus x squared, in blue. A dashed vertical line at x = 3 meets the circle twice, at (3, 4) and (3, -4). — **Figure 1.** The circle $x^2 + y^2 = 25$ as two functions of $x$ . LevlPrep original diagram.

Worked example 2. Solve $x^2 + xy + y^2 = 7$ for $y$ .

Treat it as a quadratic in $y$ : $y^2 + (x)y + (x^2 - 7) = 0$ , with $a = 1$ , $b = x$ , $c = x^2 - 7$ .

The quadratic formula: $y$ $= \frac{-x \pm \sqrt{x^2 - 4(x^2 - 7)}}{2}$ $= \frac{-x \pm \sqrt{28 - 3x^2}}{2}$ .

The "+" sign gives the upper piece and the "−" sign the lower piece. Check with $x = 1$ : $\frac{-1 \pm \sqrt{25}}{2}$ gives $2$ and $-3$ , as before.

The square root also tells you how wide the curve is. Real values of $y$ exist only when $28 - 3x^2 \ge 0$ , that is $-\sqrt{\frac{28}{3}} \le x \le \sqrt{\frac{28}{3}}$ . At the two ends the square root is $0$ and the two pieces meet.

Symmetry from the equation

You can read symmetry without graphing. Swapping $x$ and $y$ in $x^2 + xy + y^2 = 7$ gives the same equation, so if $(a, b)$ is on the curve, so is $(b, a)$ : the curve is symmetric about the line $y = x$ . Replacing $x$ with $-x$ and $y$ with $-y$ also gives the same equation, so $(-a, -b)$ is on it too.

Implicit equations that are functions

Some implicit equations do give one $y$ for each $x$ . In $y^3 + y = x$ , the left side increases whenever $y$ increases, so two different $y$ -values can never give the same $x$ . This defines $y$ as a function of $x$ , even though solving for $y$ by hand is hard. A table is easy, though: choose $y$ , compute $x$ . From the points $(2, 1)$ and $(10, 2)$ , the average rate of change of $y$ with respect to $x$ is $\frac{2 - 1}{10 - 2} = \frac{1}{8}$ , the slope of the segment joining them.