Vertical line test for functions: Question bank

Quick vocabulary refresher so no word here is unearned:

input = the $x$ -coordinate = which vertical line we are standing on.
output = the $y$ -coordinate = where the graph sits on that line.
passes the test = every vertical line hits the graph at most once (0 or 1 times).

True or false — justify

A graph that oscillates up and down forever cannot be a function.

False. Shape is irrelevant; only the count of

y

's per

x

matters.

y = sin x

waves endlessly yet every vertical line hits it exactly once, so it passes.

If a vertical line misses the graph entirely at some

x = a

, the graph fails the test.

False. Zero intersections just means

a

is not in the domain — the function is undefined there, which is perfectly legal. "At most one" includes "none".

A relation that passes the vertical line test must also pass the horizontal line test.

False. Those tests check different directions.

y = x^{2}

passes vertical (a function) but fails horizontal at

y = 4

, since both

x = 2

and

x = - 2

land there — see 2.2.05-Horizontal-line-test-for-injective-functions.

If a graph is one single connected curve, it is automatically a function.

False. A full circle is one connected curve but a vertical line through its middle stabs it twice. Connectedness says nothing about outputs-per-input.

A function's graph must be defined for every real

x

False.

f (x) = 1/ x

is undefined at

x = 0

yet is a genuine function everywhere else. The test only judges the

x

-values where the graph actually exists.

Two different inputs sending to the same output breaks the function rule.

False. That is allowed.

x = 2

and

x = - 2

both giving

y = 4

is fine; the rule only forbids one input giving two outputs.

A vertical segment (a small vertical stroke) drawn in the plane can be part of a function's graph.

False. A vertical piece means a single

x

-value carries a whole range of

y

's — infinitely many outputs for one input, the ultimate failure.

If every horizontal line hits a graph at most once, the graph is definitely a function.

False. The horizontal test measures injectivity, not function-hood. A sideways parabola can be arranged so horizontals behave, yet it still fails the vertical test.

Spot the error

y^{2} = x

passes because

y = x

is a function." — What is wrong?

The claim swaps the relation for one of its halves. The full relation

y^{2} = x

gives

y = \pm x

, so

x = 4

yields both

2

and

- 2

— two outputs, so it fails. Only each branch separately is a function.

"The circle

x^{2} + y^{2} = 25

fails only at

x = 3

." — Fix it.

It fails at every

x

strictly between

- 5

and

5

, since each gives

y = \pm 25 - x^{2}

. Picking

x = 3

was just one convenient witness, not the sole point of failure.

"A graph with a hole (open point) at

x = 2

fails the vertical line test there." — Correct this.

A hole means the graph has zero points at

x = 2

, i.e. zero intersections. Zero is ≤ 1, so it passes there; the function is simply undefined at that

x

f (x) = ⌊ x ⌋

(step function) fails because it jumps." — Where is the mistake?

Jumps are not multiple outputs. At each

x

the step function still returns exactly one value; the vertical line meets it once. Discontinuity ≠ test failure.

"Since

y = x^{2}

maps

2

and

- 2

both to

4

, it fails the vertical line test." — Diagnose.

That reasoning describes two inputs sharing an output, which is a horizontal-test concern. The vertical test asks the opposite question, and

y = x^{2}

passes it cleanly.

"An arrow diagram where input A points to both 5 and 7 is still a function if A is the only such input." — Refute.

One violation is enough. The definition demands every input have at most one output, so a single input with two arrows already disqualifies it — this ties straight to 2.1.01-Definition-of-a-function.

Why questions

Why do we sweep a vertical line and not a horizontal one to test for a function?

A vertical line fixes the input

x

and lets

y

vary, so it literally reads off all outputs assigned to that one input — exactly the quantity the function rule limits.

Why is "at most one" used instead of "exactly one" in the test's statement?

Because an

x

outside the domain contributes zero intersections and must not be counted as a failure. "At most one" covers both the defined case (one hit) and the undefined case (no hit).

Why can a broken, gappy graph still be a function while a smooth circle cannot?

Function-hood depends only on outputs-per-input, not on continuity or smoothness. The gappy graph never doubles up an

x

; the circle doubles up almost every

x

Why does passing the vertical line test connect to having an inverse?

An inverse needs the original to pass both line tests. Vertical guarantees it is a function; horizontal guarantees each output came from one input, so the mapping can be reversed — see 3.1.01-Inverse-functions.

Why does the test tell us about the domain "for free"?

The

x

-values a vertical line can hit at least once are exactly the domain; where it hits nothing, that

x

is excluded — the graphical reading in 2.3.02-Domain-and-rangefrom-graphs.

Edge cases

A single isolated point

(3, 5)

alone in the plane — function?

Yes. Only the vertical line

x = 3

meets it, once, and all other vertical lines miss — that's ≤ 1 everywhere, so it passes (a tiny function with a one-element domain).

The empty graph (no points at all) — function?

Yes, vacuously. Every vertical line intersects zero times, satisfying "at most one" trivially. It is a function whose domain is empty.

A constant graph

y = 4

(a full horizontal line) — function?

Yes. Each vertical line meets it exactly once, at height

4

. Many inputs share the output

4

, which is allowed; it only fails the horizontal test.

The vertical line

x = 4

itself, viewed as a graph — function?

No. It contains points

(4, y)

for all

y

, so the vertical line

x = 4

coincides with it and meets infinitely many points — one input, unlimited outputs.

A graph touching a vertical line at a single tangent point (like a parabola's vertex) — pass or fail there?

Passes there. A tangent touch is still exactly one intersection point. One is allowed; only two or more break the rule.

A graph that meets some vertical line at exactly two points but every other line once — overall verdict?

Fails overall. A single offending vertical line is enough to disqualify the whole relation from being a function; passing "almost everywhere" is not passing.

Recall One-line summary to carry away

The test fails only when some vertical line scores 2 or more hits. Zero hits, one hit, gaps, jumps, waves, repeated outputs — all fine.

Connections

Parent: Vertical line test — the core rule these traps stress-test.
2.1.01-Definition-of-a-function — the "at most one output" law behind every answer.
2.2.05-Horizontal-line-test-for-injective-functions — the complementary test people confuse this with.
2.3.02-Domain-and-rangefrom-graphs — why zero intersections just marks an out-of-domain $x$ .
3.1.01-Inverse-functions — needing to pass both tests.