Identifying Graphs of Inverse Tangent

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

(−∞, ∞), Range (−π/2, π/2)

[−1, 1], Range [0, π]

[0, π], Range [−1, 1]

[−π, π], Range [−1, 1]

The arctangent function has domain (−∞, ∞) and range (−π/2, π/2).

Explanation

The arctangent function has domain (−∞, ∞) and range (−π/2, π/2).

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

(−1, 0)

(1, π/2)

(1, π/4)

(0, π/2)

(1, π/4) lies on the arctangent graph since tan(π/4) = 1.

Explanation

(1, π/4) lies on the arctangent graph since tan(π/4) = 1.

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

Decreasing S-curve with vertical asymptotes

Increasing S-curve, crosses (0, 0), horizontal asymptotes y = ±π/2

Increasing line with slope 1 and no asymptotes

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

The arctangent graph is an increasing S-curve crossing (0,0) with horizontal asymptotes y = ±π/2.

Explanation

The arctangent graph is an increasing S-curve crossing (0,0) with horizontal asymptotes y = ±π/2.

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

It is even

It is symmetric about y = π/2

It is neither even nor odd

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

The arctangent function is odd since arctan(−x) = −arctan(x).

Explanation

The arctangent function is odd since arctan(−x) = −arctan(x).

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

One x-intercept at (0, 0)

No x-intercepts

X-intercepts at x = −1 and x = 1

One x-intercept at (1, 0)

The function arctan(x) = 0 when x = 0, giving one x-intercept at (0, 0).

Explanation

The function arctan(x) = 0 when x = 0, giving one x-intercept at (0, 0).

6) Which transformation description is correct?

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

Y = arctan(x) − 2 shifts the graph right 2 units

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

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

Adding +2 outside the function moves the graph up 2 units.

Explanation

Adding +2 outside the function moves the graph up 2 units.

7) Which statement about asymptotes is correct for y = arctan(x)?

Vertical asymptotes at x = ±1

Horizontal asymptotes y = ±π/2

No asymptotes of any kind

Vertical asymptotes at x = 0 and x = π

The graph approaches y = ±π/2 as x → ±∞.

Explanation

The graph approaches y = ±π/2 as x → ±∞.

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

Reflection of y = cos(x) across y = x with x ∈ [0, π]

Reflection of y = tan(x) across y = x with x ∈ (−π/2, π/2)

Reflection of y = sin(x) across the x-axis

Reflection of y = tan(x) across the y-axis

y = arctan(x) is the reflection of y = tan(x) over y = x restricted to (−π/2, π/2).

Explanation

y = arctan(x) is the reflection of y = tan(x) over y = x restricted to (−π/2, π/2).

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

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

Concave down on (−1, 1)

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

Concave up on (−1, 1)

The arctangent graph is concave down on (−1, 0) and concave up on (0, 1).

Explanation

The arctangent graph is concave down on (−1, 0) and concave up on (0, 1).

10) On which interval of x does y = arctan(x) exist?

[0, π]

[−π, π]

(−∞, ∞)

[−1, 1]

The arctangent function is defined for all real x values.

Explanation

The arctangent function is defined for all real x values.

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

(0, π/2)

(0, 0)

(0, π)

It has no y-intercept

The graph crosses the origin at (0, 0).

Explanation

The graph crosses the origin at (0, 0).

12) Which formula shifts the base graph y = arctan(x) up by 1 unit?

Y = arctan(x − 1)

Y = arctan(x + 1)

Y = 1 − arctan(x)

Y = arctan(x) + 1

Adding +1 outside the function shifts the graph up 1 unit.

Explanation

Adding +1 outside the function shifts the graph up 1 unit.

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

(−1, −π/4), (0, 0), (1, π/4)

(−1, −π/2), (0, 0), (1, π/2)

(−1, 0), (0, 0), (1, π/2)

(−1, −π/4), (0, π/4), (1, 0)

y = arctan(x) passes through (−1, −π/4), (0, 0), and (1, π/4).

Explanation

y = arctan(x) passes through (−1, −π/4), (0, 0), and (1, π/4).

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

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

Domain (−∞, ∞), Range [−π/2, π/2], increasing

Domain (−∞, ∞), Range (−π/2, π/2), increasing

Domain (−∞, ∞), Range (−π, π), decreasing

The graph is increasing for all x with range (−π/2, π/2).

Explanation

The graph is increasing for all x with range (−π/2, π/2).

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

Passes through (0, 0), strictly decreasing, horizontal asymptotes y = ±π/2

Passes through (0, 0), strictly increasing, horizontal asymptotes y = ±π/2

Passes through (0, π/2), strictly increasing, no asymptotes

Passes through (0, 0), constant on (−1, 1)

The arctangent function is strictly increasing and approaches ±π/2 horizontally.

Explanation

The arctangent function is strictly increasing and approaches ±π/2 horizontally.

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

Y = tan(x) with y ∈ (−π/2, π/2)

Y = tan(x) with x ∈ (−π/2, π/2)

X = tan(y) with y ∈ (−π, π)

X = tan(y) with y ∈ (−π/2, π/2)

y = arctan(x) means x = tan(y) for y ∈ (−π/2, π/2).

Explanation

y = arctan(x) means x = tan(y) for y ∈ (−π/2, π/2).

17) Which limit statement identifies the right-end behavior of y = arctan(x)?

Lim{x→∞} arctan(x) = π/2

Lim{x→∞} arctan(x) = 0

Lim{x→∞} arctan(x) = −π/2

Lim{x→∞} arctan(x) does not exist

As x → ∞, arctan(x) → π/2.

Explanation

As x → ∞, arctan(x) → π/2.

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

−π/2 ≤ y ≤ 0

0 ≤ y ≤ π/4

π/4 ≤ y ≤ π/2

−π ≤ y ≤ 0

For 0 ≤ x ≤ 1, y ranges from 0 to π/4.

Explanation

For 0 ≤ x ≤ 1, y ranges from 0 to π/4.

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

Reflection across the y-axis only

Reflection across the origin

Shift right by 1 unit

Shift up by π units

Since the function is odd, y = arctan(−x) reflects across the origin.

Explanation

Since the function is odd, y = arctan(−x) reflects across the origin.

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

Y = arctan(x) + 1/2

Y = arctan(2x)

Y = arctan(x + 1/2)

Y = arctan(x − 1/2)

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

Explanation

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