Graph of y = arcsin(x): Domain, Range, Shape, Symmetry, Transformations

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

Domain (−∞, ∞), Range (−∞, ∞)

Domain [−1, 1], Range [−π/2, π/2]

Domain [0, π], Range [−1, 1]

Domain [−π, π], Range [−1, 1]

arcsin is only defined for x in [−1,1] and outputs y in [−π/2, π/2].

Explanation

arcsin is only defined for x in [−1,1] and outputs y in [−π/2, π/2].

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

(−1, 0)

(1, π/2)

(0, π/2)

(0, π/4)

arcsin(1) = π/2.

Explanation

arcsin(1) = π/2.

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

Increasing from (−1, −π/2) to (1, π/2)

Decreasing from (−1, −π/2) to (1, π/2)

Increasing from (−1, π/2) to (1, −π/2)

Decreasing from (−1, π/2) to (1, −π/2)

arcsin is strictly increasing on its domain.

Explanation

arcsin is strictly increasing on its domain.

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

It is even

It is symmetric about y = π/2

It is odd and satisfies arcsin(−x) = −arcsin(x)

It is neither even nor odd

arcsin is an odd function: arcsin(−x) = −arcsin(x).

Explanation

arcsin is an odd function: arcsin(−x) = −arcsin(x).

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

No x-intercepts

X-intercepts at x = −1 and x = 1

One x-intercept at (1, 0)

One x-intercept at (0, 0)

arcsin(0) = 0, so the graph passes through (0,0).

Explanation

arcsin(0) = 0, so the graph passes through (0,0).

6) Which transformation description is correct?

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

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

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

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

Adding outside ( +2 ) translates the graph up.

Explanation

Adding outside ( +2 ) translates the graph up.

7) Which statement about the domain endpoints is correct for y = arcsin(x)?

The graph has horizontal tangents at x = ±1

The graph has oblique asymptotes

The graph is defined at x = ±1 with values y = ±π/2

The graph is undefined at x = ±1

arcsin(±1) = ±π/2; the derivative blows up (vertical tangents), but values exist.

Explanation

arcsin(±1) = ±π/2; the derivative blows up (vertical tangents), but values exist.

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

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

It is the reflection of y = sin(x) across y = x restricted to x ∈ [−π/2, π/2]

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

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

To invert sine, restrict to x ∈ [−π/2, π/2]; then reflect across y = x.

Explanation

To invert sine, restrict to x ∈ [−π/2, π/2]; then reflect across y = x.

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

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

Concave down on (−1, 1)

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

Concave up on (−1, 1)

Second derivative x/(1−x²)^{3/2} is negative on (−1,0) and positive on (0,1).

Explanation

Second derivative x/(1−x²)^{3/2} is negative on (−1,0) and positive on (0,1).

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

(−∞, ∞)

[0, π]

[−π, π]

[−1, 1]

Domain in x is [−1,1].

Explanation

Domain in x is [−1,1].

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

(0, π/2)

(0, 0)

(0, π)

It has no y-intercept

arcsin(0)=0.

Explanation

arcsin(0)=0.

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

Y = arcsin(x) − π/2

Y = arcsin(−x)

Y = −arcsin(x)

Y = π − arcsin(x)

Reflection across y=k maps y → 2k − y; here 2(π/2) − arcsin(x) = π − arcsin(x).

Explanation

Reflection across y=k maps y → 2k − y; here 2(π/2) − arcsin(x) = π − arcsin(x).

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

(−1, −π/2), (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)

Standard key points: (−1,−π/2), (0,0), (1,π/2).

Explanation

Standard key points: (−1,−π/2), (0,0), (1,π/2).

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

Domain [−1, 1], Range [−π/2, π/2], increasing

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

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

Domain [−1/2, 1/2], Range [−π/2, π/2], increasing

Require 2x ∈ [−1,1] → x ∈ [−1/2,1/2]; arcsin is increasing with same range.

Explanation

Require 2x ∈ [−1,1] → x ∈ [−1/2,1/2]; arcsin is increasing with same range.

15) Which description identifies the base graph y = arcsin(x)?

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

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

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

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

Key point (1,π/2); the function increases across its domain.

Explanation

Key point (1,π/2); the function increases across its domain.

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

Y = sin(x) with y ∈ [−1, 1]

X = sin(y) with y ∈ [−π/2, π/2]

X = sin(y) with y ∈ [−π, π]

Y = sin(x) with x ∈ [−π/2, π/2]

For y = arcsin(x), equivalently x = sin(y) with principal range y ∈ [−π/2, π/2].

Explanation

For y = arcsin(x), equivalently x = sin(y) with principal range y ∈ [−π/2, π/2].

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

Lim{x→1−} arcsin(x) = π

Lim{x→1+} arcsin(x) = 0

Lim{x→1−} arcsin(x) = 0

Lim{x→1−} arcsin(x) = π/2

As x approaches 1 from the left, arcsin(x) approaches π/2.

Explanation

As x approaches 1 from the left, arcsin(x) approaches π/2.

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

π/2 ≤ y ≤ π

−π/2 ≤ y ≤ π/2

0 ≤ y ≤ π/2

−π ≤ y ≤ 0

arcsin maps [0,1] to [0, π/2].

Explanation

arcsin maps [0,1] to [0, π/2].

19) Which transformation identifies y = arcsin(−x) relative to y = arcsin(x)?

Reflection across the origin

Reflection across the x-axis only

Shift right by 1 unit

Shift up by π units

arcsin(−x) = −arcsin(x): a reflection through the origin (odd symmetry).

Explanation

arcsin(−x) = −arcsin(x): a reflection through the origin (odd symmetry).

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

Y = arcsin(x) + 1/2

Y = arcsin(x − 1/2)

Y = arcsin(x + 1/2)

Y = arcsin(2x)

Replace x with (x − 1/2) for a right shift by 1/2.

Explanation

Replace x with (x − 1/2) for a right shift by 1/2.

Identifying Graphs of Inverse Sine

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

2)

What first name or nickname would you like us to use?

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

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

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

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

6) Which transformation description is correct?

7) Which statement about the domain endpoints is correct for y = arcsin(x)?

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

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

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

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

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

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

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

15) Which description identifies the base graph y = arcsin(x)?

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

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

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

19) Which transformation identifies y = arcsin(−x) relative to y = arcsin(x)?

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