Unit 3 Absolute Values Test

Give the vertex, domain, and range of the absolute value equation.

f(x) = 3│x + 3│ - 5

• Vertex: ( -3, -5)

  Domain: IR

  Range: y < -5

• Vertex: ( -3, 5)

  Domain: IR

  Range: y > 5

• Vertex: ( -3, -5)

  Domain: IR

  Range: y > -5

• Vertex: ( -3, -5)

  Domain: x > -3

  Range:IR

Tags

CCSS.HSF-IF.C.7D

Give the vertex, domain, and range of the absolute value equation.

f(x) = -2│x - 3│ + 2

• Vertex: ( 3, 2)

  Domain: IR

  Range: y < 2

• Vertex: ( 3, -2)

  Domain: IR

  Range: y > -2

• Vertex: ( -3, 2)

  Domain: IR

  Range: y > 2

• Vertex: ( -3, -2)

  Domain: IR

  Range: y > -2

Tags

CCSS.HSF-IF.C.7D

Solve the equation algebraically.

|3x + 5| = 1

• x = 2 or x = −4/3

• x = 2 or x = -4

• x = 2 or x = -2

• x = -4/3 or x = -2

Solve the equation algebraically.

|3 − 2𝑥| = 5

• x = -1 or x = 4

• x = 1 or x = -4

• x = -1

• x = -4

Solve the equation algebraically.

2|2𝑥 − 1| + 4 = 16

• x = − 3/2 or x = 5/2

• x = − 9/2 or x = 7/2

• x = − 5/2 or x = 7/2

• x = 9/2 or x = − 7/2

Solve the equation algebraically.

|4−3𝑥|/ 3 = 5

• x = − 19/3 or x = 11/3

• x = − 1/3 or x = 3

• x = 1/3 or x = −3

• x = 19/3 or x = − 11/3

Solve the inequality algebraically.

|x − 2| > 3 is

• x < −1 or x > 5

• x < −1 or x > 5

• − 1 < x < 5

• − 1 < x < 5

The solution for the inequality |2x + 3| > 19 is

• 
• 
• 
• 

Solve the inequality algebraically.

|3𝑥 + 6| ≤ 30

• -12 ≤ x ≤ 8

• -8 ≤ x ≤ 12

• x ≤ -8 or x ≥ 12

• x ≤ -12 or x ≥ 8

Solve the inequality algebraically.

|4𝑥 + 8| ≥ 16

• -6 ≤ x ≤ 2

• -2 ≤ x ≤ 6

• x ≤ -6 or x ≥ 2

• x ≤ -2 or x ≥ 6