Algebra 2 Unit 1 Review Answer Key

Algebra 2 Unit 1 sets the foundation for more advanced mathematical concepts, and mastering its key topics is crucial for success in the course. A comprehensive review, accompanied by an answer key, allows students to identify their strengths and weaknesses, reinforcing their understanding of the material. This article will delve into the primary concepts covered in Algebra 2 Unit 1, providing examples, explanations, and, most importantly, solutions that will solidify your grasp of these fundamental algebraic principles.

Linear Equations and Inequalities

At the heart of Algebra 2 Unit 1 lies the understanding and manipulation of linear equations and inequalities. These serve as building blocks for more complex mathematical models and are applicable in various real-world scenarios.

Solving Linear Equations:

The core objective is to isolate the variable on one side of the equation. This involves applying inverse operations to both sides to maintain equality.

• Example: Solve for x: 3x + 5 = 14

  • Subtract 5 from both sides: 3x = 9
  • Divide both sides by 3: x = 3

Solving Linear Inequalities:

Similar to solving equations, but with an additional rule: multiplying or dividing by a negative number flips the inequality sign.

• Example: Solve for x: -2x + 4 > 10

  • Subtract 4 from both sides: -2x > 6
  • Divide both sides by -2 (and flip the inequality): x < -3

Answer Key Examples:

1. Solve: 5x - 7 = 18

   • Solution: x = 5

2. Solve: -4x + 9 ≤ 1

   • Solution: x ≥ 2

3. Solve: 2(x + 3) = 10

   • Solution: x = 2

4. Solve: 3x - 5 > 4x + 2

   • Solution: x < -7

5. Solve: -2(x - 1) ≤ 6

   • Solution: x ≥ -2

Functions and Their Graphs

Functions are a fundamental concept in algebra, representing a relationship between inputs and outputs. Understanding how to represent them graphically is crucial.

Defining Functions:

A function is a relation where each input (x-value) has exactly one output (y-value).

Graphing Linear Functions:

Linear functions can be represented in the form y = mx + b, where m is the slope and b is the y-intercept.

• The slope indicates the steepness and direction of the line.
• The y-intercept is the point where the line crosses the y-axis.

Graphing Absolute Value Functions:

Absolute value functions take the form y = |x|. The graph is V-shaped, with the vertex at the point where the expression inside the absolute value is equal to zero.

• Example: y = |x - 2| + 1. The vertex is at (2, 1).

Answer Key Examples:

1. Graph the function: y = 2x - 1

   • Solution: A straight line with a slope of 2 and a y-intercept of -1.

2. Graph the function: y = -x + 3

   • Solution: A straight line with a slope of -1 and a y-intercept of 3.

3. Graph the function: y = |x + 1|

   • Solution: A V-shaped graph with a vertex at (-1, 0).

4. Graph the function: y = |x| - 2

   • Solution: A V-shaped graph with a vertex at (0, -2).

5. Determine if the following is a function: {(1, 2), (2, 3), (3, 4), (1, 5)}

   • Solution: No, because the input 1 has two different outputs (2 and 5).

Systems of Linear Equations

Solving systems of linear equations involves finding the values of the variables that satisfy all equations simultaneously. Common methods include substitution and elimination.

Substitution Method:

Solve one equation for one variable, then substitute that expression into the other equation.

• Example: Solve the system:

  • y = x + 1

  • 2x + y = 5

    • Substitute (x + 1) for y in the second equation: 2x + (x + 1) = 5
    • Simplify and solve for x: 3x + 1 = 5 => 3x = 4 => x = 4/3
    • Substitute x = 4/3 back into the first equation to find y: y = 4/3 + 1 => y = 7/3
    • Solution: (x, y) = (4/3, 7/3)

Elimination Method:

Multiply one or both equations by a constant so that the coefficients of one variable are opposites. Then, add the equations together to eliminate that variable.

• Example: Solve the system:

  • 2x + y = 7

  • x - y = 2

    • Add the two equations: 3x = 9
    • Solve for x: x = 3
    • Substitute x = 3 back into either equation to find y: 2(3) + y = 7 => y = 1
    • Solution: (x, y) = (3, 1)

Answer Key Examples:

1. Solve the system:

   • y = 3x - 2

   • y = -x + 6

     • Solution: (x, y) = (2, 4)
2. Solve the system:

   • x + y = 5

   • x - y = 1

     • Solution: (x, y) = (3, 2)
3. Solve the system:

   • 2x + y = 8

   • x - y = 1

     • Solution: (x, y) = (3, 2)
4. Solve the system:

   • 3x + 2y = 7

   • x - y = -1

     • Solution: (x, y) = (1, 2)
5. Solve the system:

   • 4x - 3y = 6

   • 2x + y = 8

     • Solution: (x, y) = (3, 2)

Absolute Value Equations and Inequalities

Absolute value represents the distance from zero, so solving absolute value equations and inequalities requires considering both positive and negative cases.

Solving Absolute Value Equations:

Set up two equations: one where the expression inside the absolute value is equal to the given value, and another where it's equal to the negative of the given value.

• Example: Solve |x - 3| = 5

  • Case 1: x - 3 = 5 => x = 8
  • Case 2: x - 3 = -5 => x = -2
  • Solutions: x = 8, x = -2

Solving Absolute Value Inequalities:

• For |x| < a, solve -a < x < a

• For |x| > a, solve x < -a or x > a

• Example: Solve |2x + 1| ≤ 3

  • Solve: -3 ≤ 2x + 1 ≤ 3
  • Subtract 1 from all parts: -4 ≤ 2x ≤ 2
  • Divide all parts by 2: -2 ≤ x ≤ 1

• Example: Solve |x - 2| > 4

  • Solve: x - 2 < -4 or x - 2 > 4
  • Add 2 to both parts: x < -2 or x > 6

Answer Key Examples:

1. Solve: |x + 2| = 4

   • Solution: x = 2, x = -6

2. Solve: |2x - 1| = 5

   • Solution: x = 3, x = -2

3. Solve: |x - 3| < 2

   • Solution: 1 < x < 5

4. Solve: |3x + 2| ≥ 4

   • Solution: x ≤ -2 or x ≥ 2/3

5. Solve: |x + 1| > 3

   • Solution: x < -4 or x > 2

Transformations of Functions

Understanding how functions transform, including translations, reflections, stretches, and compressions, is essential for visualizing and manipulating graphs.

Vertical Translations:

Adding a constant c to a function shifts the graph vertically.

• y = f(x) + c shifts the graph up by c units if c > 0, and down by |c| units if c < 0.

Horizontal Translations:

Replacing x with (x - c) shifts the graph horizontally.

• y = f(x - c) shifts the graph right by c units if c > 0, and left by |c| units if c < 0.

Reflections:

• y = -f(x) reflects the graph across the x-axis.
• y = f(-x) reflects the graph across the y-axis.

Stretches and Compressions:

• y = a f(x) vertically stretches the graph by a factor of a if a > 1, and compresses it if 0 < a < 1.
• y = f(bx) horizontally compresses the graph by a factor of b if b > 1, and stretches it if 0 < b < 1.

Answer Key Examples:

1. Describe the transformation of y = x² to y = x² + 3

   • Solution: Vertical translation up by 3 units.

2. Describe the transformation of y = |x| to y = |x - 2|

   • Solution: Horizontal translation right by 2 units.

3. Describe the transformation of y = √x to y = -√x

   • Solution: Reflection across the x-axis.

4. Describe the transformation of y = x to y = 2x

   • Solution: Vertical stretch by a factor of 2.

5. Describe the transformation of y = x² to y = (x/2)²

   • Solution: Horizontal stretch by a factor of 2.

Piecewise Functions

Piecewise functions are defined by different equations over different intervals of their domain.

Evaluating Piecewise Functions:

To evaluate a piecewise function, determine which interval the input value belongs to, and then use the corresponding equation to find the output.

• Example:
  • f(x) = { x + 1, if x < 0; x², if x ≥ 0 }
  • Find f(-2): Since -2 < 0, use f(x) = x + 1 => f(-2) = -2 + 1 = -1
  • Find f(3): Since 3 ≥ 0, use f(x) = x² => f(3) = 3² = 9

Graphing Piecewise Functions:

Graph each piece of the function over its specified interval. Be careful to use open or closed circles at the endpoints to indicate whether the endpoint is included or not.

Answer Key Examples:

1. Evaluate the piecewise function:

   • f(x) = { 2x, if x ≤ 1; x + 3, if x > 1 }

   • Find f(0)

     • Solution: 0 ≤ 1, so f(0) = 2(0) = 0

2. Evaluate the piecewise function:

   • f(x) = { -x, if x < 2; x - 4, if x ≥ 2 }

   • Find f(5)

     • Solution: 5 ≥ 2, so f(5) = 5 - 4 = 1

3. Graph the piecewise function:

   • f(x) = { 1, if x < 0; x + 1, if x ≥ 0 }

     • Solution: A horizontal line at y = 1 for x < 0, and a line with slope 1 and y-intercept 1 for x ≥ 0.

4. Graph the piecewise function:

   • f(x) = { x², if x ≤ 1; 2, if x > 1 }

     • Solution: A parabola y = x² for x ≤ 1, and a horizontal line at y = 2 for x > 1.

5. Write a piecewise function for the following graph: (Assume a horizontal line at y=3 for x<0 and a line with slope 1 and y-intercept -1 for x≥0)

   • Solution: f(x) = { 3, if x < 0; x - 1, if x ≥ 0 }

Linear Programming

Linear programming involves optimizing (maximizing or minimizing) a linear objective function subject to a set of linear constraints.

Steps in Linear Programming:

1. Define variables: Identify the decision variables.
2. Write the objective function: Express the quantity to be optimized as a linear function of the variables.
3. Write the constraints: Express the limitations as a set of linear inequalities.
4. Graph the feasible region: Graph the constraints and identify the region that satisfies all constraints simultaneously.
5. Find the vertices: Determine the coordinates of the vertices (corner points) of the feasible region.
6. Evaluate the objective function at each vertex: Substitute the coordinates of each vertex into the objective function and determine the maximum or minimum value.

Example:

Maximize P = 3x + 2y subject to:

• x ≥ 0

• y ≥ 0

• x + y ≤ 4

• 2x + y ≤ 6

  • Vertices of the feasible region: (0, 0), (3, 0), (2, 2), (0, 4)
  • Evaluate P at each vertex:
    • (0, 0): P = 3(0) + 2(0) = 0
    • (3, 0): P = 3(3) + 2(0) = 9
    • (2, 2): P = 3(2) + 2(2) = 10
    • (0, 4): P = 3(0) + 2(4) = 8
  • Maximum value: P = 10 at (2, 2)

Answer Key Examples:

1. Maximize Z = 4x + 5y subject to:

   • x ≥ 0

   • y ≥ 0

   • x + y ≤ 6

   • 2x + y ≤ 10

     • Solution: Maximum Z = 30 at (0, 6)

2. Minimize C = 2x + 3y subject to:

   • x ≥ 0

   • y ≥ 0

   • x + y ≥ 5

   • 2x + y ≥ 8

     • Solution: Minimum C = 12 at (3, 2)

3. A factory makes two products, A and B. Each product requires time on two machines: Machine 1 and Machine 2. Product A requires 1 hour on Machine 1 and 2 hours on Machine 2. Product B requires 3 hours on Machine 1 and 1 hour on Machine 2. Machine 1 can be used for at most 9 hours and Machine 2 for at most 8 hours. If the profit on each product A is $2 and on B is $3, how many of each product should be made to maximize profit?

   • Solution: Let x = number of product A and y = number of product B. Maximize P = 2x + 3y subject to x + 3y ≤ 9 and 2x + y ≤ 8. The optimal solution is to make 3 of product A and 2 of product B for a maximum profit of $12.

4. A diet is to contain at least 2400 units of vitamins, 1800 units of minerals, and 1200 calories. Two foods are available: Food A costs $0.20 per unit and contains 40 units of vitamins, 30 units of minerals, and 10 calories. Food B costs $0.30 per unit and contains 20 units of vitamins, 30 units of minerals, and 40 calories. How many units of each food should be purchased to minimize the cost?

   • Solution: Let x = units of Food A and y = units of Food B. Minimize C = 0.20x + 0.30y subject to 40x + 20y ≥ 2400, 30x + 30y ≥ 1800, and 10x + 40y ≥ 1200. The optimal solution is to purchase 48 units of Food A and 24 units of Food B.

In Summary

Algebra 2 Unit 1 lays a robust foundation for advanced mathematical exploration. Mastering linear equations and inequalities, functions and their transformations, systems of equations, absolute value concepts, piecewise functions, and linear programming is essential for success in subsequent units. Diligent practice with the examples and answer keys provided will not only enhance comprehension but also build confidence in tackling complex algebraic challenges. Remember, consistent effort and a clear understanding of the underlying principles are the keys to unlocking the full potential of Algebra 2.