Identifying Graphs of Inverse Cosine

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| Questions: 20 | Updated: Oct 14, 2025

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1) Which pair gives the domain and range of y = arccos(x)?

Domain (−∞, ∞), range (−∞, ∞)

Domain [−1, 1], range [0, π]

Domain [0, π], range [−1, 1]

Domain [−π, π], range [−1, 1]

The domain of y = arccos(x) is all x-values where cosine is defined, which is from −1 to 1. The range is the possible angle outputs, which are between 0 and π radians.

Domain is [−1, 1], Range is [0, π].

Explanation

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About This Quiz

Identifying Graphs Of Inverse Cosine - Quiz

Ready to explore the world of inverse trig graphs? In this quiz, you’ll dive into the shape, domain, and range of y = arccos(x) and learn how it connects to the cosine function. You’ll identify key points, analyze symmetry, and understand how reflections and transformations affect the graph. Step by... see morestep, you’ll gain confidence recognizing and interpreting inverse cosine graphs with clarity! see less

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2) Which point lies on the graph of y = arccos(x)?

(-1, 0)

(1, π)

(0, π/2)

(0, 0)

The point (0, π/2) lies on the graph because arccos(0) equals π/2.

The cosine of π/2 is 0.

Explanation

The point (0, π/2) lies on the graph because arccos(0) equals π/2.

The cosine of π/2 is 0.

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3) Which description matches the overall shape of y = arccos(x)?

Decreasing from (−1, π) to (1, 0)

Increasing from (−1, 0) to (1, π)

Increasing from (−1, π) to (1, 0)

Decreasing from (−1, 0) to (1, π)

The graph of y = arccos(x) starts at (−1, π) and decreases smoothly to (1, 0).

As x increases from −1 to 1, the angle whose cosine equals x decreases from π to 0.

Explanation

The graph of y = arccos(x) starts at (−1, π) and decreases smoothly to (1, 0).

As x increases from −1 to 1, the angle whose cosine equals x decreases from π to 0.

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4) Which statement about symmetry is true for y = arccos(x)?

It is even

It is odd

It is symmetric about y = π/2

It is neither even nor odd and satisfies arccos(−x) = π − arccos(x)

The function is neither even nor odd but follows the rule arccos(−x) = π − arccos(x).

The graph is not symmetric about the y-axis or origin.

Explanation

The function is neither even nor odd but follows the rule arccos(−x) = π − arccos(x).

The graph is not symmetric about the y-axis or origin.

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5) Which x-intercept information correctly identifies the graph of y = arccos(x)?

No x-intercepts

X-intercepts at x = −1 and x = 1

One x-intercept at (1, 0)

One x-intercept at (−1, 0)

The graph crosses the x-axis at x = 1 because arccos(1) equals 0.

There is one x-intercept at (1, 0).

Explanation

The graph crosses the x-axis at x = 1 because arccos(1) equals 0.

There is one x-intercept at (1, 0).

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6) Which transformation description is correct?

Y = arccos(x) + 2 shifts the graph up 2 units

Y = arccos(x + 2) shifts the graph up 2 units

Y = arccos(x) + 2 shifts the graph right 2 units

Y = arccos(x − 2) shifts the graph up 2 units

Adding a constant outside the function moves the graph vertically.

y = arccos(x) + 2 shifts the graph up by 2 units.

Explanation

Adding a constant outside the function moves the graph vertically.

y = arccos(x) + 2 shifts the graph up by 2 units.

Submit

7) Which statement about end behavior at the domain endpoints is correct for y = arccos(x)?

The graph has horizontal tangents at x = ±1

The graph has oblique asymptotes

The graph is undefined at x = ±1

The graph has vertical tangents at x = −1 and x = 1

At x = −1 and x = 1, the graph becomes vertical.

The slopes approach infinity, giving vertical tangents at both endpoints.

Explanation

At x = −1 and x = 1, the graph becomes vertical.

The slopes approach infinity, giving vertical tangents at both endpoints.

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8) Which inverse relationship correctly identifies the graph of y = arccos(x)?

It is the reflection of y = cos(x) across y = x for x ∈ [−π, π]

It is the reflection of y = cos(x) across y = x restricted to x ∈ [0, π]

It is the reflection of y = cos(x) across the x-axis

It is the reflection of y = cos(x) across the y-axis

The inverse cosine is the mirror image of the cosine function across the line y = x, but cosine must be restricted to x-values between 0 and π.

The graph is the reflection of y = cos(x) for x in [0, π].

Explanation

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9) Which statement about concavity is correct for y = arccos(x)?

Concave up on (0, 1), concave down on (−1, 0)

Concave down on (−1, 1)

Concave up on (−1, 0), concave down on (0, 1)

Concave up on (−1, 1)

For y = arccos(x), the second derivative changes sign at x = 0.

It is positive (concave up) on the left half (-1, 0) and negative (concave down) on the right half (0,

Explanation

For y = arccos(x), the second derivative changes sign at x = 0.

It is positive (concave up) on the left half (-1, 0) and negative (concave down) on the right half (0,

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10) Which interval shows where the graph of y = arccos(x) exists horizontally?

(−∞, ∞)

[0, π]

[−π, π]

[−1, 1]

The graph exists horizontally only between x = −1 and x = 1.

The inverse cosine is only defined for cosine outputs in this range.

Explanation

The graph exists horizontally only between x = −1 and x = 1.

The inverse cosine is only defined for cosine outputs in this range.

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11) What is the y-intercept of y = arccos(x)?

(0, π/2)

(0, 0)

(0, π)

It has no y-intercept]

When x = 0, arccos(0) equals π/2.

The y-intercept is at (0, π/2).

Explanation

When x = 0, arccos(0) equals π/2.

The y-intercept is at (0, π/2).

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12) Which formula describes reflecting the base graph y = arccos(x) across the horizontal line y = π/2?

Y = arccos(x) − π/2

Y = π − arccos(x)

Y = arccos(−x)

Y = arccos(x) + π

Reflecting across the line y = π/2 changes every y value to π minus the original y value.

The new equation is y = π − arccos(x).

Explanation

Reflecting across the line y = π/2 changes every y value to π minus the original y value.

The new equation is y = π − arccos(x).

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13) Which set of key points identifies y = arccos(x)?

(−1, 0), (0, 0), (1, π/2)

(−1, 0), (0, π/2), (1, π)

(−1, π), (0, π/2), (1, 0)

(−1, π/2), (0, π), (1, 0)

The base graph passes through (−1, π), (0, π/2), and (1, 0).

These are the standard key points for the inverse cosine graph.

Explanation

The base graph passes through (−1, π), (0, π/2), and (1, 0).

These are the standard key points for the inverse cosine graph.

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14) Which statement correctly identifies y = arccos(2x)?

Domain [−1, 1], Range [0, π], increasing

Domain [−2, 2], Range [0, π], decreasing

Domain [−1, 1], Range [0, 2π], decreasing

Domain [−1/2, 1/2], Range [0, π], decreasing

In y = arccos(2x), the value inside must still be between −1 and 1, so x must be between −1/2 and 1/2.

Domain is [−1/2, 1/2], Range is [0, π], and the function is decreasing.

Explanation

In y = arccos(2x), the value inside must still be between −1 and 1, so x must be between −1/2 and 1/2.

Domain is [−1/2, 1/2], Range is [0, π], and the function is decreasing.

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15) Which description identifies the base graph y = arccos(x) using intercepts and monotonicity?

Passes through (0, π/2) and (1, 0), strictly decreasing on [−1, 1]

Passes through (0, 0) and (1, π), strictly increasing on [−1, 1]

Passes through (0, π) and (−1, 0), strictly decreasing on [−1, 1]

Passes through (0, π/2) and (−1, 0), strictly increasing on [−1, 1]

The graph passes through (0, π/2) and (1, 0) and decreases across the interval from −1 to 1.

It is strictly decreasing over its domain.

Explanation

The graph passes through (0, π/2) and (1, 0) and decreases across the interval from −1 to 1.

It is strictly decreasing over its domain.

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16) Which relation correctly identifies the inverse pairing of x and y on the graph of y = arccos(x)?

Y = cos(x) with y ∈ [−1, 1]

X = cos(y) with y ∈ [0, π]

X = cos(y) with y ∈ [−π, π]

Y = cos(x) with x ∈ [0, π]

For y = arccos(x), x and y are related by x = cos(y), where y lies between 0 and π.

That’s the inverse relationship of the cosine function.

Explanation

For y = arccos(x), x and y are related by x = cos(y), where y lies between 0 and π.

That’s the inverse relationship of the cosine function.

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17) Which limit statement identifies the right endpoint behavior of y = arccos(x)?

Lim{x→1−} arccos(x) = π/2

Lim{x→1−} arccos(x) = π

Lim{x→1+} arccos(x) = 0

Lim{x→1−} arccos(x) = 0

As x approaches 1 from the left, arccos(x) approaches 0.

The graph approaches its minimum y-value at the right endpoint.

Explanation

As x approaches 1 from the left, arccos(x) approaches 0.

The graph approaches its minimum y-value at the right endpoint.

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18) Which transformation identifies y = arccos(−x) relative to y = arccos(x)?

Reflection across the line y = π/2

Reflection across the x-axis

Shift right by 1 unit

Shift up by π units

Replacing x with −x reflects the graph across the vertical line y = π/2.

arccos(−x) = π − arccos(x).

Explanation

Replacing x with −x reflects the graph across the vertical line y = π/2.

arccos(−x) = π − arccos(x).

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19) Which equation corresponds to shifting the base graph y = arccos(x) right by 1/2 unit?

Y = arccos(x) + 1/2

Y = arccos(x − 1/2)

Y = arccos(x + 1/2)

Y = arccos(2x)

Replacing x with (x − 1/2) shifts the graph to the right by 1/2 unit.

Subtracting inside the function moves the graph horizontally to the right.

Explanation

Replacing x with (x − 1/2) shifts the graph to the right by 1/2 unit.

Subtracting inside the function moves the graph horizontally to the right.

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20) For x in [0, 1], which describes the image under y = arccos(x)?

π/2 ≤ y ≤ π

−π/2 ≤ y ≤ π/2

0 ≤ y ≤ π/2

−π ≤ y ≤ 0

When x is between 0 and 1, arccos(x) gives angles from 0 to π/2.

The cosine of small angles near 0 is close to 1.

Explanation

When x is between 0 and 1, arccos(x) gives angles from 0 to π/2.

The cosine of small angles near 0 is close to 1.

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All (20)
Unanswered ()
Answered ()

Which pair gives the domain and range of y = arccos(x)?

Which point lies on the graph of y = arccos(x)?

Which description matches the overall shape of y = arccos(x)?

Which statement about symmetry is true for y = arccos(x)?

Which x-intercept information correctly identifies the graph of y =...

Which transformation description is correct?

Which statement about end behavior at the domain endpoints is correct...

Which inverse relationship correctly identifies the graph of y =...

Which statement about concavity is correct for y = arccos(x)?

Which interval shows where the graph of y = arccos(x) exists...

What is the y-intercept of y = arccos(x)?

Which formula describes reflecting the base graph y = arccos(x) across...

Which set of key points identifies y = arccos(x)?

Which statement correctly identifies y = arccos(2x)?

Which description identifies the base graph y = arccos(x) using...

Which relation correctly identifies the inverse pairing of x and y on...

Which limit statement identifies the right endpoint behavior of y =...

Which transformation identifies y = arccos(−x) relative to y =...

Which equation corresponds to shifting the base graph y = arccos(x)...

For x in [0, 1], which describes the image under y = arccos(x)?

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