From an Equation to a Relation
An equation involving \(x\) and \(y\) does not automatically specify \(y\) as one function of \(x\). For a given value of \(x\), there may be no corresponding \(y\), exactly one, or several. To see this clearly, begin by treating the equation as a set of points.
A function \(f:I\to\mathbb R\), where \(I\subseteq\mathbb R\), has graph \(\{(x,f(x)):x\in I\}\). Thus its graph contains exactly one point with first coordinate \(x\) for each \(x\in I\), and no points whose first coordinate is outside \(I\).
The restriction in this statement matters: it removes any points of \(R\) whose first coordinate lies outside \(I\). To say that \(R\) itself is the graph of a function with domain \(I\), we also need \(R\subseteq I\times\mathbb R\). The uniqueness condition alone does not remove extra points outside that strip.
What Is a Local Function Branch?
Suppose \((a,b)\) satisfies \(F(a,b)=0\). A local function branch through \((a,b)\) is a function \(g:I\to\mathbb R\), defined on an interval \(I\) containing \(a\), such that \(g(a)=b\) and \(F(x,g(x))=0\) for every \(x\in I\). Its graph is a selected part of the relation. The entire relation may contain other points above the same \(x\)-values.
Sometimes a branch is isolated from the rest of the relation near \((a,b)\). Precisely, this happens if there are open intervals \(I\) and \(J\), with \(a\in I\) and \(b\in J\), such that the zero set in \(I\times J\) is exactly the graph of one function defined on all of \(I\). This stronger description means that no other part of the relation enters that rectangle. At a crossing, there may be separate function branches, but the full relation may not be one function in any rectangle around the crossing.
These distinctions keep three different ideas separate: an equation, the full relation it defines, and a function branch selected from that relation. The examples below use explicit algebra to determine which situation occurs.
A Circle Has Two Branches
Worked Example: The Local Branches of a Circle
Consider the circle \(x^2+y^2=25\), and focus on the point \((3,4)\). The equation gives \(y^2=25-x^2\), so any real solution must satisfy \(|x|\leq 5\). Solving for \(y\) gives two possibilities: $$ y=\sqrt{25-x^2} \qquad\text{or}\qquad y=-\sqrt{25-x^2}. $$ For \(x=3\), these give \(y=\sqrt{25-9}=\sqrt{16}=4\) and \(y=-\sqrt{16}=-4\). Thus \((3,4)\) is on the upper branch, while \((3,-4)\) is on the lower branch.
To isolate a branch through \((3,4)\), take \(I=(2,4)\) and \(J=(0,5)\). For every \(x\in I\), \(25-x^2>25-16=9>0\), so the positive square root is defined. Also \(0<\sqrt{25-x^2}<5\). In \(I\times J\), the negative root is excluded because it is less than zero, so the zero set there is exactly $$ g(x)=\sqrt{25-x^2},\qquad x\in(2,4). $$ At \(x=3\), \(g(3)=\sqrt{16}=4\), as required. To check the branch, substitute it into the circle equation: $$ x^2+g(x)^2=x^2+\bigl(\sqrt{25-x^2}\bigr)^2=x^2+25-x^2=25. $$ The lower branch \(y=-\sqrt{25-x^2}\) also satisfies the circle equation, but it is a different function branch and is not part of the zero set restricted to this rectangle.
The example shows why the circle itself is not the graph of one function over \((-5,5)\): for instance, \(x=3\) corresponds to both \(y=4\) and \(y=-4\). The upper and lower semicircles are each graphs over that interval, but the full circle is not.
An Ellipse Also Requires a Branch Choice
Worked Example: An Upper Branch of an Ellipse
Consider the ellipse $$ \frac{x^2}{4}+\frac{y^2}{9}=1 $$ near its top point \((0,3)\). Isolate \(y^2\) by subtracting \(x^2/4\) from both sides and multiplying by \(9\): $$ \frac{y^2}{9}=1-\frac{x^2}{4}, \qquad y^2=9\left(1-\frac{x^2}{4}\right). $$ For \(|x|<2\), the expression in parentheses is positive. Therefore the two solutions are $$ y=3\sqrt{1-\frac{x^2}{4}} \qquad\text{and}\qquad y=-3\sqrt{1-\frac{x^2}{4}}. $$ At \(x=0\), the solutions are \(y=3\) and \(y=-3\); the upper one passes through \((0,3)\).
Choose \(I=(-1,1)\) and \(J=(2,4)\). When \(x\in I\), \(0\leq x^2<1\), so $$ \frac{3}{4}<1-\frac{x^2}{4}\leq 1. $$ Consequently the positive root satisfies \(3\sqrt{3/4}<y\leq 3\), which places it in \(J\); the negative root is less than zero and is excluded. The upper branch is \(g(x)=3\sqrt{1-x^2/4}\) on \(I\). Its value at \(0\) is \(g(0)=3\sqrt{1}=3\). Substitution verifies the equation: $$ \frac{x^2}{4}+\frac{g(x)^2}{9} = \frac{x^2}{4}+\frac{9(1-x^2/4)}{9} = \frac{x^2}{4}+1-\frac{x^2}{4} =1. $$ Thus this branch is a function whose graph lies on the ellipse; the lower branch is a separate part of the same relation.
Solving for \(y\) can reveal branches, but the domain and the branch choice still have to be stated. The square-root formula alone does not say whether the positive root, the negative root, or both are intended.
Intersecting Curves Can Defeat a Single Local Graph
Worked Example: Two Curves Crossing at the Origin
Consider the relation $$ (y-x)(y+x)=0. $$ A product is zero exactly when at least one factor is zero, so the relation consists of the two lines \(y=x\) and \(y=-x\). At \(x=0\), both lines give \(y=0\), and the curves intersect at \((0,0)\). At \(x=1\), however, the relation contains both \((1,1)\) and \((1,-1)\). These are distinct points with the same first coordinate, so the full relation is not the graph of a function on any interval containing \(1\).
More importantly, no rectangle around \((0,0)\) makes the entire relation a single function graph. Any open interval \(I\) around \(0\) contains some \(x\ne0\). At that \(x\), \(y=x\) and \(y=-x\) are distinct because \(x\ne -x\) when \(x\ne0\). Both corresponding points lie in every sufficiently small rectangle centered at the origin once \(x\) is chosen close enough to zero. Thus there are two relation points above such an \(x\) in the rectangle.
There are nevertheless two local function branches: \(g_1(x)=x\) and \(g_2(x)=-x\). For the first branch, substitution gives \((g_1(x)-x)(g_1(x)+x)=(0)(2x)=0\). For the second, it gives \((g_2(x)-x)(g_2(x)+x)=(-2x)(0)=0\). Each graph satisfies the equation, but together they form the crossing relation, not one function graph.
Why the Graph Criterion Works
The criterion follows directly from what it means to be the graph of a function. The existence condition ensures that every input in the proposed domain has an output. The uniqueness condition ensures that this output is not ambiguous.
For every \(x\in I\), let \(f(x)\) be the unique real number \(y\) with \((x,y)\in R\). Existence makes this assignment possible, and uniqueness makes it well-defined.
Every point of \(R\cap(I\times\mathbb R)\) has first coordinate in \(I\), and its second coordinate must equal the unique value \(f(x)\). Conversely, \((x,f(x))\) belongs to \(R\cap(I\times\mathbb R)\) for every \(x\in I\). Hence this restriction is exactly \(\{(x,f(x)):x\in I\}\).
If \(R\cap(I\times\mathbb R)\) is the graph of a function \(f:I\to\mathbb R\), then for each \(x\in I\), the point \((x,f(x))\) is in the restriction. No other point \((x,y)\) with \(y\ne f(x)\) can be in that graph. Therefore exactly one \(y\) corresponds to each \(x\in I\).
This proof also explains why a graph test must account for both existence and uniqueness. If an input has no corresponding point, it is absent from the relation's domain. If it has multiple corresponding points, the relation cannot assign it a single function value.
A Common Mistake: Calling the Whole Relation a Branch
A frequent error is to solve for \(y\), list the possible expressions, and then treat them as though the equation had defined one function. For the circle, a mistaken conclusion might read: “\(y=\pm\sqrt{25-x^2}\), so the circle is the graph of a function on \((-5,5)\).” The calculation does not establish that conclusion: at \(x=3\), the two values are \(4\) and \(-4\), so uniqueness fails.
The correction is to name the branches separately. On \((-5,5)\), the upper semicircle is the graph of \(y=\sqrt{25-x^2}\), and the lower semicircle is the graph of \(y=-\sqrt{25-x^2}\). The full circle contains both graphs and is not a single-valued function of \(x\) on that interval. A branch is a chosen function satisfying the relation, not a claim that every point of the relation belongs to that one graph.
Why Local Branches Matter
Many equations are easier to study in their original two-variable form than after solving globally for one variable. Local branches let us focus on the portion of a curve near a point and ask whether it can be represented as \(y\) depending on \(x\). The answer can differ from one point to another, and different branches of the same relation can have different behavior.
This distinction prepares the way for implicit differentiation. That technique studies how \(x\) and \(y\) vary together while the equation \(F(x,y)=0\) remains satisfied. Before differentiating such a relation, it is essential to know whether a function branch is being considered and which point on the curve is under discussion. The relation alone may not specify a unique \(y\)-value.
Check Your Understanding
Use the definitions and examples above to answer the following questions.
- For a relation \(R\) and an interval \(I\), why does the graph criterion refer to \(R\cap(I\times\mathbb R)\)?
- For the circle \(x^2+y^2=25\), what are the two \(y\)-values when \(x=4\), and why does this prevent the full circle from being a function graph over an interval containing \(4\)?
- For the ellipse \(x^2/4+y^2/9=1\), write the upper branch and check its value at \(x=0\).
- For \((y-x)(y+x)=0\), name the two function branches and explain why the full relation is not one function near the origin.
- State the difference between a relation and a local function branch.
