Vertical line test for functions

Definition

Mathematical restatement: For all $a \in \text{domain}$, there exists at most one $b$ such that $(a,b) \in R$.

Why This Works: Derivation from the Function Definition

Start with the formal definition of a function: $f: A \to B \text{ is a function } \iff \forall x \in A, \exists! \, y \in B \text{ such that } y=f(x)$

The symbol $\exists!$ means "there exists exactly one."

Step 1: What does the graph of a relation look like?

• The graph is the set of all points $(x,y)$ where the relation holds.
• For a function $f$, the graph is $\{(x, f(x)) : x \in \text{domain}\}$.

Step 2: What does a vertical line represent?

• A vertical line at $x=a$ is the set $\{(a, y) : y \in \mathbb{R}\}$.
• It fixes the $x$-coordinate and varies $y$.

Step 3: What does intersection mean?

• The vertical line $x=a$ intersects the graph at all points $(a, y)$ that lie on the graph.
• If the graph represents a function, there can be at most one $y$ for each $x=a$ (by the definition of function).
• One intersection → exactly one $y$ for that $x$ → function behavior✓
• Zero intersections → $a$ not in domain → still okay (function just undefined there)
• Two or more intersections → multiple $y$ values for $x=a$ → violates function definition ✗

Conclusion: The vertical line test is a visual encoding of the uniqueness condition in the function definition.

Worked Examples

Graph: $x^2 + y^2 = 25$ (circle centered at origin, radius 5)

Test: Draw vertical line at $x=3$. $x=3 \implies 9 + y^2 = 25 \implies y^2 = 16 \implies y = \pm 4$

Result: The line crosses at $(3, 4)$ and $(3, -4)$ → two points.

Why this step? We're checking if one $x$ gives multiple $y$ values. Herex=3$gives both$y=4$and$y=-4$, so it's not a function.

Conclusion: ✗ Not a function. The relation is multivalued at many $x$ coordinates.

Graph: $y = x^2$ (standard parabola opening upward)

Test: Draw vertical line at any $x=a$. $x=a \implies y = a^2$

Result: For each $x=a$, there is exactly one $y = a^2$. The vertical line crosses at only one point $(a, a^2)$.

Why this step? We're verifying that every input $x$ produces exactly one output $y$. No matter which $a$ we pick, $y=a^2$ is unique.

Conclusion: ✓ Passes the test → $y=x^2$ is a function.

Graph: $y^2 = x$ (sideways parabola)

Test: Draw vertical line at $x=4$. $x=4 \implies y^2 = 4 \implies y = \pm 2$

Result: Two intersections at $(4, 2)$ and $(4, -2)$.

Why this step? We're looking for any $x$ that produces multiple $y$ values. For $x=4$, we get $y=2$ and $y=-2$, breaking the function rule.

Conclusion: ✗ Not a function. (Though $y = \sqrt{x}$ and $y = -\sqrt{x}$ individually are functions.)

Graph: $f(x) = \begin{cases} x+1 & x < 0 \\ x^2 & x \geq 0 \end{cases}$

Test:

• For $x=-2$: $y = -2+1 = -1$ (one point)
• For $x=0$: $y = 0^2 = 0$ (one point)
• For $x=2$: $y = 2^2 = 4$ (one point)

Result: Every vertical line intersects at most once.

Why this step? Even though the rule changes at $x=0$, eachx$still maps to exactly one$y$. The "jump" doesn't create multiple outputs for a single input.

Conclusion: ✓ Passes → It's a function.

Common Mistakes

Mnemonic

Connections 2.1.01-Definition-of-a-function — Formal definition that the vertical line test visualizes

• 2.2.01-Graph-of-a-function — How functions appear graphically
• 2.2.05-Horizontal-line-test-for-injective-functions — Complementary test for one-to-one functions
• 2.3.02-Domain-and-rangefrom-graphs — Using graphs to determine where functions are defined
• 3.1.01-Inverse-functions — Requires passing both vertical and horizontal line tests

#flashcards/maths

What is the vertical line test? :: A relation is a function if and only if every vertical line intersects its graph at at most one point.

Concept Map

Hinglish (regional understanding)