📝 Vertical line test for functions (12 MCQs)

📖 From Calculus • 1. Basics before calculus • 12 questions available

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What is Vertical line test for functions?

Definition:
The vertical line test is a graphical method to determine if a curve represents a function, stating that if any vertical line intersects the graph at more than one point, then the relation is not a function.

Example:
For the circle $x^2 + y^2 = 1$ , a vertical line at $x = 0$ intersects at $(0,1)$ and $(0,-1)$ , so it fails the test and is not a function.

Reason:
This test quickly verifies the uniqueness condition of functions, saving time in identifying non-functions, especially useful for graphs obtained from experiments or complex equations.

Easy

Medium

Hard

📝 All Vertical line test for functions MCQs

Q1. What does the vertical line test determine about a curve in the xy‑plane?

A.Whether the curve is a function of x ✅

B.Whether the curve is continuous

C.Whether the curve has a derivative at every point

D.Whether the curve encloses an area

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The vertical line test checks if any vertical line intersects the curve more than once. If it does, the curve fails the test and is not a function of x. This directly answers what the test is used to determine.

Q2. A vertical line drawn at x = 2 meets a curve at three distinct points. What conclusion follows?

A.The curve represents a function on that interval

B.The curve fails the vertical line test and is not a function ✅

C.The curve is continuous but not differentiable at x = 2

D.The curve is symmetric about the y‑axis

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Since a single vertical line intersects the curve three times, the relation violates the definition of a function (one output for each input). Hence the curve fails the vertical line test and cannot be a function.

Q3. Curve A is a parabola opening upward; Curve B is a circle centered at the origin. Which curve definitely passes the vertical line test?

A.Both Curve A and Curve B

B.Only Curve A ✅

C.Only Curve B

D.Neither Curve A nor Curve B

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A parabola opening upward is the graph of $y = x^{2}$ (or a similar expression) and each vertical line meets it at most once, so it passes the test. A full circle fails because vertical lines near the center intersect it twice.

Q4. Consider the piecewise relation $f(x)=\begin{cases}x^{2}&x\le 0\\ \sqrt{x}&x>0\end{cases}$ . Which statement best describes its compliance with the vertical line test?

A.It fails for

x<0

because

x^{2}

is not one‑to‑one

B.It passes everywhere because each piece is a function on its domain ✅

C.It fails at

x=0

where the two pieces meet

D.It passes for

x\neq0

but fails at

x=0

due to a jump

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Both pieces are individually functions, and the definition at $x=0$ is consistent ( $0^{2}=0$ and $\sqrt{0}=0$ ). No vertical line meets more than one point, so the whole piecewise relation passes the vertical line test.

Q5. A graph shows a curve that is smooth everywhere except at $x=1$ , where a vertical tangent appears. What does the vertical line test say about this curve?

A.It fails at

x=1

because the tangent is vertical

B.It passes because a vertical tangent does not create multiple intersections ✅

C.It fails everywhere because vertical tangents indicate non‑function behavior

D.It passes only for

x<1

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A vertical tangent means the slope is infinite, but a vertical line still intersects the curve at a single point. The vertical line test concerns the number of intersection points, not the slope, so the curve still passes the test.

Q6. Which of the following statements is true regarding the relationship between the vertical line test and the concept of a function’s domain?

A.If a curve passes the test, its domain must be all real numbers

B.Passing the test guarantees a unique output for each input in the domain ✅

C.A curve can fail the test yet still have a well‑defined domain

D.The test determines the range, not the domain

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The vertical line test ensures that for every x‑value in the domain, there is at most one corresponding y‑value. This is the definition of a function’s domain‑to‑range mapping, making the statement about uniqueness correct.

Q7. Given the implicit relation $x^{2}+y^{2}=9$ , which modification would make the resulting graph pass the vertical line test?

A.Restrict the relation to

y\ge 0

only ✅

B.Replace the constant 9 with 4

C.Swap x and y variables

D.Add a term

-2x

to the equation

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Limiting the circle to its upper semicircle ( $y\ge0$ ) ensures that any vertical line meets the graph at most once, thereby satisfying the vertical line test.

Q8. If a function fails the vertical line test only for x‑values greater than 5, what can be deduced about restricting its domain to $(-\infty,5]$ ?

A.The restriction makes the relation a function on the new domain ✅

B.The restriction does not affect the failure because the test is global

C.The function becomes invertible after restriction

D.The function loses continuity after restriction

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: By removing the portion where the test fails (x>5), the remaining graph no longer has any vertical line intersecting more than once. Hence the restricted relation satisfies the vertical line test on $(-\infty,5]$ .

Q9. Compare the vertical line test with the horizontal line test for a relation that passes the former but fails the latter. Which scenario exemplifies this?

A.A strictly increasing function defined on all real numbers

B.A function that is not one‑to‑one, such as

f(x)=x^{2}

✅

C.A relation that is a circle

D.A piecewise function that swaps x and y

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: $f(x)=x^{2}$ passes the vertical line test because each x yields a single y, but horizontal lines intersect the graph at two points for positive y-values, so it fails the horizontal line test. This contrast highlights the different purposes of the two tests.

Q10. Design a curve that passes the vertical line test but is discontinuous at $x=0$ . Which description fits?

A.The graph of

f(x)=\begin{cases}1&x<0\\2&x\ge0\end{cases}

✅

B.A parabola with a hole at the origin

C.A sine wave with a vertical asymptote at

x=0

D.A circle missing a point at

x=0

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The piecewise constant function jumps from 1 to 2 at $x=0$ . Each vertical line meets the graph at exactly one point, satisfying the vertical line test, while the jump creates a discontinuity at $x=0$ .

Q11. For the implicitly defined relation $xy=4$ , which statement about the vertical line test is correct?

A.It fails because the line

x=0

meets the curve at infinitely many points

B.It passes because each vertical line intersects the hyperbola at most once ✅

C.It fails for

x>0

but passes for

x<0

D.It passes only for integer values of x

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The hyperbola $xy=4$ can be written as $y=4/x$ . For any non‑zero x, there is exactly one y, so any vertical line (except $x=0$ , which is not part of the curve) meets the graph once, satisfying the vertical line test.

Q12. A relation is defined by $f(x)=\frac{1}{x-2}$ for all real x except $x=2$ . Which observation about the vertical line test is accurate?

A.It fails at

x=2

because the function is undefined there

B.It passes because every vertical line that intersects the graph does so exactly once ✅

C.It fails for all

x<2

due to the asymptote

D.It passes only for integer values of x

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The function has a vertical asymptote at $x=2$ , but the asymptote itself is not part of the graph. Any vertical line that actually meets the curve (i.e., any line with $x\neq2$ ) intersects it at a single point, so the relation passes the vertical line test.

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