Unit 3 Functions And Linear Equations Homework 1 Answers

Unit 3: Functions and Linear Equations – Homework 1 Answers: A Comprehensive Guide

This comprehensive guide provides detailed answers and explanations for common homework problems found in Unit 3, focusing on functions and linear equations. We'll tackle various problem types, from identifying functions to graphing linear equations and solving systems of equations. Remember, understanding the why behind the solution is just as important as the what. This guide aims to build your conceptual understanding alongside providing the solutions.

Section 1: Identifying and Evaluating Functions

This section focuses on understanding the definition of a function and evaluating functions given various inputs.

What is a Function?

A function is a relationship between a set of inputs (domain) and a set of possible outputs (range) where each input is related to exactly one output. This "exactly one output" rule is crucial. If a single input maps to multiple outputs, it's not a function.

Methods to Identify Functions:

• Vertical Line Test: If you can draw a vertical line anywhere on the graph and it intersects the graph at more than one point, it's not a function.
• Mapping Diagram: Check if each element in the domain is connected to only one element in the range.
• Equation: Solve for the output variable (usually 'y'). If, for every input 'x', you get only one output 'y', it's a function.

Example Problems and Solutions:

Problem 1: Determine if the following relation is a function: {(1, 2), (2, 4), (3, 6), (4, 8)}.

Solution: This is a function. Each input (x-value) is paired with exactly one output (y-value).

Problem 2: Determine if the following relation is a function: {(1, 2), (2, 4), (3, 6), (2, 8)}.

Solution: This is not a function. The input value 2 is associated with two different output values (4 and 8).

Problem 3: Evaluate the function f(x) = 2x + 3 for x = 5.

Solution: Substitute x = 5 into the function: f(5) = 2(5) + 3 = 13.

Problem 4: Given the function g(x) = x² - 4, find g(-2).

Solution: Substitute x = -2 into the function: g(-2) = (-2)² - 4 = 0.

Problem 5: Is the graph of x = 4 a function?

Solution: No. A vertical line at x = 4 would intersect the graph at infinitely many points.

Section 2: Linear Equations and Their Graphs

This section dives into the properties of linear equations and how to represent them graphically.

Slope-Intercept Form (y = mx + b)

• m: Represents the slope (rise/run) of the line. It indicates the steepness and direction of the line. A positive slope means the line goes uphill from left to right; a negative slope means it goes downhill.
• b: Represents the y-intercept, the point where the line crosses the y-axis (when x = 0).

Point-Slope Form (y - y₁ = m(x - x₁))

This form is useful when you know the slope (m) and one point (x₁, y₁) on the line.

Standard Form (Ax + By = C)

This form is useful for certain manipulations and finding intercepts.

Example Problems and Solutions:

Problem 1: Find the slope and y-intercept of the equation y = 3x - 2.

Solution: The slope (m) is 3, and the y-intercept (b) is -2.

Problem 2: Graph the line y = 2x + 1.

Solution: The y-intercept is 1. From this point, use the slope (2/1) to find another point: go up 2 units and right 1 unit. Connect the points to draw the line.

Problem 3: Write the equation of a line with a slope of -1/2 and passing through the point (4, 1).

Solution: Use the point-slope form: y - 1 = (-1/2)(x - 4). This can be simplified to y = (-1/2)x + 3.

Problem 4: Find the x- and y-intercepts of the equation 2x + 4y = 8.

Solution: To find the x-intercept, set y = 0 and solve for x: 2x = 8, x = 4. To find the y-intercept, set x = 0 and solve for y: 4y = 8, y = 2.

Problem 5: Convert the equation 3x - 6y = 12 into slope-intercept form.

Solution: Solve for y: -6y = -3x + 12, y = (1/2)x - 2.

Section 3: Systems of Linear Equations

This section covers solving systems of linear equations using various methods.

Methods for Solving Systems of Linear Equations:

• Graphing: Graph both equations and find the point of intersection. This method is useful for visualizing the solution.
• Substitution: Solve one equation for one variable and substitute it into the other equation.
• Elimination (Addition): Multiply equations by constants to eliminate one variable when adding the equations together.

Example Problems and Solutions:

Problem 1: Solve the system of equations using graphing: y = x + 2 y = -x + 4

Solution: Graph both lines. The point of intersection is (1, 3).

Problem 2: Solve the system using substitution: x + y = 5 x = y + 1

Solution: Substitute 'y + 1' for 'x' in the first equation: (y + 1) + y = 5. Solve for y: 2y = 4, y = 2. Substitute y = 2 back into either equation to find x = 3. The solution is (3, 2).

Problem 3: Solve the system using elimination: 2x + y = 7 x - y = 2

Solution: Add the two equations together to eliminate y: 3x = 9, x = 3. Substitute x = 3 into either equation to find y = 1. The solution is (3, 1).

Problem 4: A system of equations has no solution. What does this mean graphically?

Solution: Graphically, this means the lines are parallel and never intersect.

Problem 5: A system of equations has infinitely many solutions. What does this mean graphically?

Solution: Graphically, this means the lines are coincident (they are the same line).

Section 4: Applications of Linear Equations and Functions

This section shows how linear equations and functions are applied to real-world scenarios.

Word Problems and Modeling:

Many real-world problems can be modeled using linear equations. These often involve rates, costs, distances, and other proportional relationships.

Example Problems and Solutions:

Problem 1: A phone company charges a monthly fee of $20 plus $0.10 per minute of usage. Write a linear equation to represent the monthly cost (C) as a function of minutes used (m).

Solution: C = 0.10m + 20

Problem 2: A car travels at a constant speed of 60 mph. Write a linear equation that represents the distance (d) traveled as a function of time (t) in hours.

Solution: d = 60t

Problem 3: Two trains leave the same station at the same time, traveling in opposite directions. One train travels at 50 mph, and the other at 70 mph. How far apart are they after 3 hours?

Solution: Train 1 travels 50 mph * 3 hrs = 150 miles. Train 2 travels 70 mph * 3 hrs = 210 miles. The total distance apart is 150 + 210 = 360 miles.

Problem 4: A company's profit (P) is given by the equation P = 5x - 1000, where x is the number of units sold. How many units must be sold to break even (P = 0)?

Solution: Set P = 0: 0 = 5x - 1000. Solve for x: 5x = 1000, x = 200 units.

Problem 5: A rental car company charges $30 per day plus $0.25 per mile. If you rent a car for 3 days and the total cost was $135, how many miles did you drive?

Solution: Let 'm' be the number of miles. The equation is: 30(3) + 0.25m = 135. Solve for m: 90 + 0.25m = 135, 0.25m = 45, m = 180 miles.

This comprehensive guide provides a solid foundation for understanding and solving problems related to functions and linear equations. Remember to practice regularly to solidify your understanding and build confidence in tackling more complex problems. By understanding the underlying concepts and applying the various solution methods, you'll be well-equipped to succeed in your studies. Remember to always check your work and revisit concepts that you find challenging. Good luck!