Implicitly defined functions

An implicitly defined curve is the set of all points (x, y) that make an equation true. Membership is decided by substitution: plug the candidate point into the equation and see whether the two sides agree. For $x^2 + y^2 = 25$, the point (3, −4) qualifies because $9 + 16 = 25$; the point (2, 4) does not, because $4 + 16 = 20$.

Nothing in this definition mentions functions. The equation is a filter on points, and the curve is whatever survives the filter. Circles, ellipses, sideways parabolas, and tilted curves of every kind are all legitimate graphs under this definition, whether or not y can be written as a single function of x.

The vertical line test answers exactly one question: is y a function of x on this curve? If some vertical line crosses twice, the answer is no, because one input x would need two outputs y. That is the entire verdict.

The classic misreading promotes that verdict into "the curve is invalid" or "the curve has no graph." A circle fails the vertical line test and remains a perfectly good curve; the failure only means its points cannot all be produced by one formula of the form y = f(x). Some implicit curves even pass the test: $xy = 12$ rearranges to $y = 12/x$, a genuine function that happened to be written implicitly. Failing the test is a statement about function-ness, never about existence.

Solving an implicit equation for y can produce more than one expression. From $x^2 + y^2 = 25$: $y^2 = 25 - x^2$, so $y = \sqrt{25 - x^2}$ or $y = -\sqrt{25 - x^2}$. Each choice is a branch, and each branch on its own is an honest function (the upper and lower semicircles). The full curve is both branches together.

The discipline: when a square root appears in the solve, the ± is not decoration. Keeping only the positive root silently deletes half the curve. Write both branches, then use whichever one the problem's region calls for.

Intercepts come from the same substitution idea: set y = 0 and solve for x to find where the curve meets the x-axis, and set x = 0 for the y-axis. For $x^2 + 4y^2 = 16$: y = 0 gives $x = \pm 4$; x = 0 gives $4y^2 = 16$, so $y = \pm 2$. Note that the constant 16 itself is not an intercept of anything; it must be processed through the equation.

Locally, an implicit curve usually behaves like a function even when it globally does not: near (0, 5) the circle $x^2 + y^2 = 25$ looks exactly like the smooth cap $y = \sqrt{25 - x^2}$. The exceptions are points where the curve turns vertical, like (5, 0) on that circle: no matter how much you zoom in there, x values just inside 5 carry two nearby y values, so no single function describes the curve near that point.