Comprehensive Introduction to Algebra Two Study Guide

Overview of Parent Functions

Definition of a Function: A relationship between $x$ and $y$ where each $x$ -value is associated with exactly one $y$ -value.
Definition of a Parent Function: The simplest form of a function in a family of functions.

The Nine Fundamental Parent Functions

Linear Function:
- Equation: $y = x$
- Graph characteristics: A straight diagonal line passing through the origin $(0, 0)$ with a slope of $1$ .
Absolute Value Function:
- Equation: $y = |x|$
- Graph characteristics: A V-shaped graph with a vertex at $(0, 0)$ , symmetric across the $y$ -axis.
Quadratic Function:
- Equation: $y = x^2$
- Graph characteristics: A U-shaped parabola opening upward with a vertex at $(0, 0)$ .
Cubic Function:
- Equation: $y = x^3$
- Graph characteristics: An S-shaped curve passing through the origin, extending from quadrant III to quadrant I.
Square Root Function:
- Equation: $y = \frac{\sqrt{x}}{1}$ or $y = \sqrt{x}$
- Graph characteristics: Starts at $(0, 0)$ and curves gradually to the right in quadrant I.
Cube Root Function:
- Equation: $y = \sqrt[3]{x}$
- Graph characteristics: An sideways S-curve passing through the origin and extending infinitely in both directions.
Rational Function:
- Equation: $y = \frac{1}{x}$
- Graph characteristics: A hyperbola with vertical asymptote at $x = 0$ and horizontal asymptote at $y = 0$ .
Exponential Function:
- Equation: $y = a^x$
- Graph characteristics: A curve approaching the horizontal asymptote $y = 0$ to the left and increasing rapidly to the right.
Logarithmic Function:
- Equation: $y = \log(x)$
- Graph characteristics: Has a vertical asymptote at $x = 0$ , passes through $(1, 0)$ , and increases slowly to the right.

Parent function identification practice grid

Practice: Parent Function Identification

Problem 1: S-shaped curve centered at origin $\rightarrow$ Cube Root ( $y = \sqrt[3]{x}$ ).
Problem 2: Inverted S-curve centered at origin $\rightarrow$ Cubic ( $y = x^3$ ).
Problem 3: U-shaped curve opening upwards $\rightarrow$ Quadratic ( $y = x^2$ ).
Problem 4: Inverted U-shaped curve opening downwards $\rightarrow$ Quadratic ( $y = x^2$ ).
Problem 5: Increasing curve with horizontal asymptote $\rightarrow$ Exponential ( $y = a^x$ ).
Problem 6: Curve starting at origin moving right $\rightarrow$ Square Root ( $y = \sqrt{x}$ ).
Problem 7: U-shaped parabola shifted down $\rightarrow$ Quadratic ( $y = x^2$ ).
Problem 8: Two separate hyperbolic branches $\rightarrow$ Rational ( $y = \frac{1}{x}$ ).
Problem 9: S-curve shifted vertically $\rightarrow$ Cubic ( $y = x^3$ ).

Key Features of Functions

Core Vocabulary and Concepts:
- Domain and Range
- Function Evaluation at a point or value
- $X$ -intercepts and $Y$ -intercepts
- Location and value of absolute/relative maxima and minima
- Intervals of increasing, decreasing, or constant behavior

Set-Builder Notation vs. Interval Notation

Set-Builder Notation: Uses a variable and a logical rule to describe elements of a set.
- Example: $\{y \mid y \text{ is an integer and } y \ge 10\}$ represents "The set of all numbers $y$ such that $y$ is an integer and is greater than or equal to $10$ ".
Interval Notation: Describes a continuous set of real numbers using minimum and maximum values.
- Use parentheses () when the boundary value is not included in the set ( $<$ or $>$ ).
- Use square brackets [] when the boundary value is included ( $\le$ or $\ge$ ).
- Use $\pm \infty$ when there is no lower or upper bound.
- Conversion Examples:
- $(-5, 0]$ is equivalent to $\{y \mid -5 < y \le 0\}$
- $[-3, \infty)$ is equivalent to $\{y \mid -3 \le y\}$
- $(-\infty, 7)$ is equivalent to $\{y \mid y < 7\}$

Domain and Range Analysis

Domain: The set of all possible $x$ inputs (read graph from left to right).
- Example Analysis: The graph starts at $x = -3$ (open circle) and ends at $x = 1$ (closed dot) $\rightarrow$ Domain: $(-3, 1]$ .
Range: The set of all possible $y$ outputs (read graph from bottom to top).
- Example Analysis: Lowest point is $y = -4$ (included) and highest point is $y = 0$ (included) $\rightarrow$ Range: $[-4, 0]$ .

Evaluating Functions Graphically

Method: To evaluate $f(a)$ , find the point on the curve where $x = a$ and read the corresponding $y$ -value ( $f(x) = y$ ).
Algebraic Practice: Domain and range of $y = x^2 - 1$ :
- Domain: $(-\infty, \infty)$
- Range: $[-1, \infty)$
Graphical Evaluation Practice 1:
- Given graph evaluation:
- $f(0) = 3$
- $f(-2) = 4$
- $f(2) = -1$
- Finding $x$ when $f(x) = 2.5$ : $x = -3$ and $x = 0.1$
Graphical Evaluation Practice 2:
- Is it a function? Yes
- Domain: $[-3, 6]$
- Range: $[-2, 3]$
- $f(-1) = 3$
- $f(0) = 3$

Intercepts and Extrema

$X$ --Intercept(s): Coordinates $(x, 0)$ where the graph crosses or touches the $x$ -axis.
- Example Graph Intercepts: $(3, 0)$ and $(9, 0)$
$Y$ --Intercept: Coordinates $(0, y)$ where the graph crosses the $y$ -axis.
- Example Graph Intercept: $(0, 5)$
Real-World Application (Gas Consumption):
- Equation: $y = 16 - 0.05x$ , where $y$ is gallons of gas remaining and $x$ is miles traveled.
- $x$ -intercept: $(320, 0)$ $\rightarrow$ The car traveled $320\text{ miles}$ before running completely out of gas.
- $y$ -intercept: $(0, 16)$ $\rightarrow$ The car's tank can hold $16\text{ gallons}$ of gas at the start.
Extrema (Maxima and Minima):
- Local/Relative Maximum: Peak point relative to nearby values. Example: $\{2.5, 3.5\}$
- Global/Absolute Maximum: The highest overall $y$ -value on the entire graph. Example: $3.5$
- Local/Relative Minimum: Valley point relative to nearby values. Example: $1.5$
- Global/Absolute Minimum: The lowest overall $y$ -value across the entire graph. Example: None (extends infinitely downwards).

Roller coaster path graph and increasing decreasing intervals

Behavior Intervals (Increasing, Decreasing, Constant)

Rule: Reading from left to right, state the $x$ -intervals over which the curve goes up (increasing), goes down (decreasing), or stays flat (constant). Always use interval notation with parentheses for open intervals.
Roller Coaster Application:
- Highest point: $25\text{ inches}$ at $0\text{ inches}$ horizontal displacement.
- Number of hills: $2\text{ hills}$ , located at the local maxima.
- Relative minima: At $0\text{ inches}$ in height.
- Climbing (Increasing) Intervals: $(20, 38) \cup (55, 70) \cup (85, 95]$
- Falling (Decreasing) Intervals: $[0, 20) \cup (38, 55) \cup (70, 85)$
- Constant Interval: None

End Behavior of Functions

End behavior describes the trend of the graph as $x$ approaches positive infinity ( $x \rightarrow \infty$ ) or negative infinity ( $x \rightarrow -\infty$ ).
End Behavior Summary Table:
- Left Behavior Rising: $x \rightarrow -\infty, f(x) \rightarrow \infty$
- Left Behavior Falling: $x \rightarrow -\infty, f(x) \rightarrow -\infty$
- Right Behavior Rising: $x \rightarrow \infty, f(x) \rightarrow \infty$
- Right Behavior Falling: $x \rightarrow \infty, f(x) \rightarrow -\infty$
End Behavior Combinations:
- Rising on Both Sides: As $x \rightarrow -\infty, f(x) \rightarrow \infty$ and as $x \rightarrow \infty, f(x) \rightarrow \infty$
- Falling on Both Sides: As $x \rightarrow -\infty, f(x) \rightarrow -\infty$ and as $x \rightarrow \infty, f(x) \rightarrow -\infty$
- Opposite Behavior: Left rising ( $x \rightarrow -\infty, f(x) \rightarrow \infty$ ) and right falling ( $x \rightarrow \infty, f(x) \rightarrow -\infty$ )

Function Transformations

Transformation Formula Structure

General Form: $y = a f(x - h) + k$
- Vertical Stretch / Compression / Reflection ( $a$ ):
- If $a$ is negative: Reflection (flip) over the $x$ -axis.
- If $|a| > 1$ : Vertical stretch by a factor of $|a|$
- If $0 < |a| < 1$ : Vertical compression by a factor of $|a|$
- Horizontal Shift ( $h$ ):
- $x - h$ : Translate right by $h$ units.
- $x + h$ : Translate left by $h$ units.
- Vertical Shift ( $k$ ):
- $+k$ : Translate up by $k$ units.
- $-k$ : Translate down by $k$ units.

Writing Transformed Equations

Parent: $y = |x|$ ; Transformations: left $5$ , up $3$
- New Equation: $y = |x + 5| + 3$
Parent: $y = \sqrt[3]{x}$ ; Transformations: compression by $\frac{1}{2}$ , right $2$
- New Equation: $y = \frac{1}{2} \sqrt[3]{x - 2}$
Parent: $y = x^2$ ; Transformations: stretch of $3$ , down $5$
- New Equation: $y = 3x^2 - 5$
Parent: $y = \sqrt[3]{x}$ ; Transformations: reflection in $x$ -axis
- New Equation: $y = -\sqrt[3]{x}$
Parent: $y = x^3$ ; Transformations: right $2$ , down $3$
- New Equation: $y = (x - 2)^3 - 3$
Parent: $y = |x|$ ; Transformations: reflection in $x$ -axis, left $1$
- New Equation: $y = -|x + 1|$
Parent: $y = \sqrt{x}$ ; Transformations: stretch of $5$ , left $8$
- New Equation: $y = 5\sqrt{x + 8}$
Parent: $y = x^3$ ; Transformations: reflection in $x$ -axis, stretch of $2$
- New Equation: $y = -2x^3$
Parent: $y = \sqrt{x}$ ; Transformations: compression by $\frac{2}{3}$ , right $7$
- New Equation: $y = \frac{2}{3}\sqrt{x - 7}$
Parent: $y = x^2$ ; Transformations: reflection in $x$ -axis, stretch of $3$ , left $2$ , down $10$
- New Equation: $y = -3(x + 2)^2 - 10$

Identifying Parent Functions and Transformations

Graph Analysis Examples:
- Main Parent: $y = x^2$
- Shifted graph 1: Left $3$ , Down $5$
- Shifted graph 2: Vertical Compression, Left $2$ , Up $1$
- Shifted graph 3: Right $1$ , Up $1$
- Shifted graph 4: Right $2$ , Down $5$
Algebraic Practice Problems:
1. $f(x) = \frac{1}{2}\sqrt{x + 4}$
- Parent: $y = \sqrt{x}$
- Transformations: Vertical compression by $\frac{1}{2}$ , Left $4$
1. $f(x) = 2x^2 - 3$
- Parent: $y = x^2$
- Transformations: Vertical stretch by $2$ , Down $3$
1. $f(x) = -|x + 6|$
- Parent: $y = |x|$
- Transformations: Reflection over $x$ -axis, Left $6$
1. $f(x) = (x + 1)^3 + 5$
- Parent: $y = x^3$
- Transformations: Left $1$ , Up $5$
1. $f(x) = 5\sqrt{x - 7} - 2$
- Parent: $y = \sqrt{x}$
- Transformations: Vertical stretch by $5$ , Right $7$ , Down $2$
1. $f(x) = \frac{1}{3}x^2 + 8$
- Parent: $y = x^2$
- Transformations: Vertical compression by $\frac{1}{3}$ , Up $8$
1. $f(x) = -4\sqrt[3]{x} + 9$
- Parent: $y = \sqrt[3]{x}$
- Transformations: Reflection over $x$ -axis, Vertical stretch by $4$ , Up $9$
1. $f(x) = 3|x + 2|$
- Parent: $y = |x|$
- Transformations: Vertical stretch by $3$ , Left $2$
1. $f(x) = -(x - 5)^3 + 4$
- Parent: $y = x^3$
- Transformations: Reflection over $x$ -axis, Right $5$ , Up $4$
1. $f(x) = -\frac{1}{4}\sqrt{x} - 7$
  - Parent: $y = \sqrt{x}$
  - Transformations: Reflection over $x$ -axis, Vertical compression by $\frac{1}{4}$ , Down $7$
2. $f(x) = -x^3 + 2$
  - Parent: $y = x^3$
  - Transformations: Reflection over $x$ -axis, Up $2$
3. $f(x) = \sqrt[3]{x - 1} + 3$
  - Parent: $y = \sqrt[3]{x}$
  - Transformations: Right $1$ , Up $3$

Piecewise-Defined Functions and Step Functions

Piecewise functions definition and example

Definitions

Continuous Function: A function with no breaks, holes, or jumps. It can be graphed completely without picking up the pencil.
Discontinuous Function: A function containing breaks, gaps, holes, jump discontinuities, or vertical asymptotes.
Piecewise Function: A function constructed from pieces of different functions defined over distinct sub-intervals.

Comprehensive Piecewise Graph Analysis

Example Analysis:
- Continuous? No
- Domain: $(-\infty, \infty)$
- Range: $(-\infty, \infty)$
- $x$ -intercepts: $(-3, 0)$ and $(1, 0)$
- $y$ -intercept: $(0, 3)$
- Maximum: None; Minimum: None
- Increasing Interval: $(-\infty, 1) \cup [1, \infty)$
- Decreasing Interval: None
- Constant Interval: None
- End Behavior: As $x \rightarrow -\infty, y \rightarrow -\infty$ ; As $x \rightarrow \infty, y \rightarrow \infty$

Piecewise Practice Problems

Graph 1:
- Continuous? No
- Domain: $(-\infty, \infty)$
- Range: $[-4] \cup [0, \infty)$
- $x$ -intercept: $(0, 0)$ ; $y$ -intercept: $(0, 0)$
- Minimum: $-4$
- Right End Behavior: As $x \rightarrow \infty, y \rightarrow \infty$
- Increasing Interval: $(0, \infty)$
- Decreasing Interval: $[-2, 0)$
- Constant Interval: $(-\infty, -2)$
- Function values: $f(-2) = 4$ , $f(0) = 0$ , $f(x) = 10 \implies x = 3$
Graph 2:
- Continuous? No
- Domain: $(-\infty, \infty)$
- Range: $(-\infty, \infty)$
- $y$ -intercept: $(0, 0)$
- Left End Behavior: As $x \rightarrow -\infty, y \rightarrow -\infty$
- Increasing Interval: $(-\infty, 2]$
- Decreasing Interval: None
- Constant Interval: $(2, \infty)$
- Function values: $f(-2) = -10$ , $f(0) = 0$ , $f(x) = 8 \implies x = 2$
Graph 3:
- Continuous? Yes
- Domain: $(-\infty, \infty)$
- Range: $(-\infty, \infty)$
- $x$ -intercept: $(-2, 0)$ ; $y$ -intercept: $(0, 1)$
- Left End Behavior: As $x \rightarrow -\infty, y \rightarrow -\infty$
- Right End Behavior: As $x \rightarrow \infty, y \rightarrow \infty$
- Increasing Interval: $(-\infty, -1) \cup (1, \infty)$
- Decreasing Interval: None
- Constant Interval: $(-1, 1)$
- Function values: $f(-2) = 0$ , $f(0) = 1$ , $f(x) = 0 \implies x = -2$

Step function graph and piecewise extension

Extensions: Step Functions and Piecewise Rewriting

Rewriting Absolute Value Functions as Piecewise Functions:
- Find the boundary (zero of the absolute value expression), then test sign regions.
- a. $f(x) = |-5x - 10| = |-5(x + 2)|$ $f(x) = \begin{cases} -5x - 10 & x < -2 \\ 5x + 10 & x \ge -2 \end{cases}$
- b. $g(x) = |6x + 18| = |6(x + 3)|$ $g(x) = \begin{cases} -6x - 18 & x < -3 \\ 6x + 18 & x \ge -3 \end{cases}$
Step Functions:
- Visual appearance: Looks like a series of stair steps.
- Structural comparison: Composed of pieces of horizontal line segments.
- Continuous? No.
- Graph Analysis (Parking/Time cost example):
- Domain: $(0, 12]$
- Range: $\{10, 15, 20, 25\}$
- Maximum Value: $25$
- Minimum Value: $10$

Absolute Value Functions and Equations

Graphing Absolute Value Functions

Basic Function f(x) = |x|**:\n - Table: (-1, 1) $,$ (0, 0) $,$ (1, 1)\n - Piecewise Form:\n f(x) = \begin{cases} -x & x < 0 \ x & x \ge 0 \end{cases}\n - Key Values: f(4) = 4 $,$ f(6) = 6 $,$ f(5) = 5 $,$ f(2) = 2\n - Domain: (-\infty, \infty) $; Range:$ [0, \infty)\n - Positive Interval: (-\infty, 0) \cup (0, \infty); Negative Interval: **None**\n - Increasing: (0, \infty) $; Decreasing:$ (-\infty, 0)\n - Average Rate of Change (AROC) on [-2, 0]:\n \text{AROC} = \text{slope} = \frac{f(0) - f(-2)}{0 - (-2)} = \frac{0 - 2}{0 + 2} = \frac{-2}{2} = -1\n\n- **Shifted Function f(x) = |x - 3|:
- Table: $(2, 1)$ , $(3, 0)$ , $(4, 1)$
- Values: $f(6) = 3$ , $f(-2) = 5$ , $f(-3, 9) = 6$ , $f(-1, 7) = 4$
- Domain: $(-\infty, \infty)$ ; Range: $[0, \infty)$
- End Behavior: As $x \rightarrow -\infty, f(x) \rightarrow \infty$ ; As $x \rightarrow \infty, f(x) \rightarrow \infty$

Additional Absolute Value Graphs

f(x) = -|x| + 2**:\n - Table: (-1, 1) $,$ (0, 2) $,$ (1, 1)\n - Values: f(-5) = -3 $,$ f(4) = -2 $,$ f(-6, 6) = -4 $,$ f(-3, 3) = -1\n - Domain: (-\infty, \infty) $; Range:$ (-\infty, 2]\n\n- **f(x) = 2|x|:
- Table: $(-1, 2)$ , $(0, 0)$ , $(1, 2)$
- Values: $f(2) = 4$ , $f(-3) = 6$ , $f(-1, 1) = 2$ , $f(-2, 2) = 4$
- Domain: $(-\infty, \infty)$ ; Range: $[0, \infty)$
Practice 1: f(x) = -|x + 4|**:\n - Table: (-5, -1) $,$ (-4, 0) $,$ (-3, -1)\n - Values: f(0) = -4 $,$ f(-2) = -2 $,$ f(-7, -1) = -3 $,$ f(-9, 1) = -5\n - Domain: (-\infty, \infty) $; Range:$ (-\infty, 0]\n\n- **Practice 2: f(x) = |x - 1| + 2:
- Table: $(0, 3)$ , $(1, 2)$ , $(2, 3)$
- Values: $f(-1) = 4$ , $f(3) = 4$ , $f(-2, 4) = 5$ , $f(-3, 5) = 6$
- Domain: $(-\infty, \infty)$ ; Range: $[2, \infty)$

Solving linear and absolute value equations

Solving Linear Equations Review

Steps: Simplify by combining like terms, identify variable to isolate, apply order of operations in reverse.
Review Problem 1: $-\frac{3}{2}x - 6 = 9 \implies -\frac{3}{2}x = 15 \implies x = 15 \times \left(-\frac{2}{3}\right) = -10$
Review Problem 2: $8x - 2 = -10 + 4x \implies 4x - 2 = -10 \implies 4x = -8 \implies x = -2$

Procedure for Solving Absolute Value Equations

Isolate the absolute value expression on one side of the equation.
Split into two linear equations:
- Set expression equal to the positive number.
- Set expression equal to the negative number.
Solve both independent linear equations.
Check all solutions in the original equation to identify any extraneous solutions.

Demonstration Example: $4|x| + 1 = 13 \implies 4|x| = 12 \implies |x| = 3$ Split: $-x = 3 \implies x = -3$ and $x = 3$
Worked Examples:
1. $3 - 2|x + 1| = -7$ $-2|x + 1| = -10 \implies |x + 1| = 5$ Case A: $-(x + 1) = 5 \implies x + 1 = -5 \implies x = -6$ Case B: $x + 1 = 5 \implies x = 4$
2. $\frac{|3x + 1|}{2} = 8$ $|3x + 1| = 16$ Case A: $-(3x + 1) = 16 \implies 3x + 1 = -16 \implies 3x = -17 \implies x = -\frac{17}{3}$ Case B: $3x + 1 = 16 \implies 3x = 15 \implies x = 5$

Number of solutions for absolute value equations

Number of Solutions Rules

$|\text{expression}| = \text{positive value} \implies 2 \text{ solutions}$
$|\text{expression}| = 0 \implies 1 \text{ solution}$
$|\text{expression}| = \text{negative value} \implies 0 \text{ solutions (No Solution)}$
Determining Number of Solutions:
- $7 + |2x - 5| = 4 \implies |2x - 5| = -3 \implies$ No Solution ( $0$ solutions).
- $|1 - 4x| + 8 = 13 \implies |1 - 4x| = 5 \implies$ 2 Solutions.
- $|3x + 6| - 6 = -6 \implies |3x + 6| = 0 \implies$ 1 Solution.
Equation Practice:
1. $|3x| = 9 \implies -3x = 9 \implies x = -3$ ; $3x = 9 \implies x = 3$
2. $|-3x| = 9 \implies -(-3x) = 9 \implies 3x = 9 \implies x = 3$ ; $-3x = 9 \implies x = -3$
3. $\left|\frac{x}{5}\right| = 1 \implies -\frac{x}{5} = 1 \implies x = -5$ ; $\frac{x}{5} = 1 \implies x = 5$
4. $|5x - 4| = 16$ $-(5x - 4) = 16 \implies 5x - 4 = -16 \implies 5x = -12 \implies x = -\frac{12}{5}$ $5x - 4 = 16 \implies 5x = 20 \implies x = 4$
5. $|-2x - 1| = 11$ $-(-2x - 1) = 11 \implies -2x - 1 = -11 \implies -2x = -10 \implies x = 5$ $-2x - 1 = 11 \implies -2x = 12 \implies x = -6$

Advanced Absolute Value Practice and Extensions

Practice Problems:
1. $|x + 8| - 5 = 2 \implies |x + 8| = 7 \implies x = -15$ and $x = -1$
2. $|5x| + 5 = 45 \implies |5x| = 40 \implies x = -8$ and $x = 8$
3. $3|-8x| + 8 = 80 \implies 3|-8x| = 72 \implies |-8x| = 24 \implies x = 3$ and $x = -3$
4. $5 - 8|-2x| = -75 \implies -8|-2x| = -80 \implies |-2x| = 10 \implies x = 5$ and $x = -5$
5. $-5|3 + 4x| = -115 \implies |3 + 4x| = 23 \implies x = -\frac{13}{2}$ and $x = 5$
6. $\frac{|7x + 4|}{8} = 3 \implies |7x + 4| = 24 \implies x = -4$ and $x = \frac{20}{7}$
Extensions:
1. Equal Absolute Values: $|x| = |2x - 1|$
  - Case A: $x = 2x - 1 \implies -x = -1 \implies x = 1$
  - Case B: $-x = 2x - 1 \implies -3x = -1 \implies x = \frac{1}{3}$
2. Sum of Absolute Values: $|2x - 4| + |x + 2| = 9$
  - Step 1: Set each term to zero to locate critical interval boundaries: $2x - 4 = 0 \implies x = 2$ ; $x + 2 = 0 \implies x = -2$
  - Step 2: Perform interval analysis over 3 cases:
  - Interval 1 ( $x < -2$ ): $-(2x - 4) - (x + 2) = 9 \implies -2x + 4 - x - 2 = 9 \implies -3x + 2 = 9 \implies -3x = 7 \implies x = -\frac{7}{3}$ (Valid: $-\frac{7}{3} < -2$ ).
  - Interval 2 ( $-2 \le x < 2$ ): $-(2x - 4) + (x + 2) = 9 \implies -2x + 4 + x + 2 = 9 \implies -x + 6 = 9 \implies x = -3$ (Invalid: $-3$ is outside $[-2, 2)$ ).
  - Interval 3 ( $x \ge 2$ ): $(2x - 4) + (x + 2) = 9 \implies 3x - 2 = 9 \implies 3x = 11 \implies x = \frac{11}{3} = 3\frac{2}{3}$ (Valid: $3\frac{2}{3} \ge 2$ ).

Absolute Value Inequalities

Steps to Solve Absolute Value Inequalities

Isolate the absolute value expression.
Split into two separate inequalities:
- First inequality keeps original expression and direction.
- Second inequality flips the inequality symbol and flips the sign of the number.
Graph solutions on a number line:
- Use closed/solid dot $\bullet$ for $\le$ or $\ge$
- Use open dot $\circ$ for $<$ or $>$

Introductory Example: $|x + 5| < 9$
- Case 1: $-(x + 5) < 9 \implies x + 5 > -9 \implies x > -14$
- Case 2: $x + 5 < 9 \implies x < 4$
- Solution: $-14 < x < 4$ or $(-14, 4)$
Inequality Practice Problems:
1. $\left|\frac{x}{4}\right| \le 3 \implies -\frac{x}{4} \le 3 \implies x \ge -12$ and $\frac{x}{4} \le 3 \implies x \le 12$
- Solution: $[-12, 12]$
1. $|x| - 3 > -2 \implies |x| > 1 \implies -x > 1 \implies x < -1$ and $x > 1$
- Solution: $(-\infty, -1) \cup (1, \infty)$
1. $2|x - 1| \le 14 \implies |x - 1| \le 7 \implies -(x - 1) \le 7 \implies x - 1 \ge -7 \implies x \ge -6$ and $x - 1 \le 7 \implies x \le 8$
- Solution: $[-6, 8]$
1. $\frac{|x - 4|}{-5} \le 2 \implies |x - 4| \ge -10$
- Special Case: Absolute value output is always $\ge 0$ , which is always greater than $-10$ . Solution: All Real Numbers ( $\mathbb{R}$ ).

Absolute value inequalities practice and extension

Inequality Practice and Misconception Analysis

Practice:
1. $2|x| > 3 \implies |x| > \frac{3}{2} \implies x < -\frac{3}{2}$ or $x > \frac{3}{2}$
2. $|2x| \le 8 \implies -2x \le 8 \implies x \ge -4$ and $2x \le 8 \implies x \le 4 \implies [-4, 4]$
3. $2|x - 4| \le 14 \implies |x - 4| \le 7 \implies x \ge -3$ and $x \le 11 \implies [-3, 11]$
4. $|2x - 3| > -7 \implies$ Absolute value is always $\ge 0 > -7$ , so $x \in (-\infty, \infty)$ .
5. $\frac{1}{3}|2x - 3| < 5 \implies |2x - 3| < 15 \implies x > -6$ and $x < 9 \implies (-6, 9)$
6. $|6x + 7| \le 5 \implies 6x + 7 \ge -5 \implies 6x \ge -12 \implies x \ge -2$ and $6x + 7 \le 5 \implies 6x \le -2 \implies x \le -\frac{1}{3} \implies \left[-2, -\frac{1}{3}\right]$
Extension Critique Problem:
- Problem: Two classmates solved $|2x - 3| \le 1 - x$ . Janice found $\frac{4}{3} \le x \le 2$ , while Paola claims there is no solution.
- Mathematical Explanation: For an absolute value inequality $|A| \le B$ , the right-hand side $B$ must be non-negative ( $1 - x \ge 0 \implies x \le 1$ ). Janice's solution range $\frac{4}{3} \le x \le 2$ requires $x \ge \frac{4}{3} > 1$ , which directly violates the constraint $x \le 1$ . Therefore, Paola is correct; there is no solution.

Solving Equations and Inequalities by Graphing

Principles of Graphical Solution

Statement $f(x) = g(x)$ : Asks where outputs are equal $\rightarrow$ Report $x$ -coordinates of intersection points.
Statement $f(x) < g(x)$ : Asks where $f$ is below $g$ $\rightarrow$ Report $x$ -interval(s).
Statement $f(x) > g(x)$ : Asks where $f$ is above $g$ $\rightarrow$ Report $x$ -interval(s).

Standard Graphing Method Steps

Rewrite: Set $y_1 = \text{left-hand side}$ and $y_2 = \text{right-hand side}$ .
Graph: Plot both functions on the same coordinate plane.
Interpret: Locate intersection points for equations, or region bounds where $y_1$ is above/below $y_2$ for inequalities.
Verify: Check key points algebraically or with technology.

Graphical Examples:
1. Solve $x^2 - 3 = 2x + 1$ by graphing:
- Intersection coordinates from graphing utility: $(-1.236, -1.472)$ and $(3.236, 7.472)$ .
- Solution $x$ -values: $x \approx -1.236$ and $x \approx 3.236$
1. Solve $x^2 - x - 6 \le 0$ by graphing:
- Parabola is on or below the $x$ -axis between $x = -2$ and $x = 3$ .
- Solution interval: $[-2, 3]$

Solving equations by graphing practice page

Graphical Practice Problems

Problem 1 & 2: Solving $-2|x - 1| + 3 = \frac{2}{5}x - \frac{11}{5}$
- Graphing shows intersection points at $x = -2$ and $x = 3$ .
- Correct Option: C ( $x = -2, x = 3$
Problem 3 & 4 (Art Museum Membership Application):
- Nonmember cost: Parking $\$15$ + Admission $\$25$ = $\$40$ per visit. Equation: $y = 40x$
- Member cost: Membership $\$110$ + Parking $\$10$ = $\$110 + 10x$
- Inequality for Member Cost < Nonmember Cost: $110 + 10x < 40x$
- Algebraic Solution: $110 < 30x \implies x > \frac{110}{30} = 3.67$
- Fewest Whole Visits Required: 4 visits
Problem 5: Evaluation of $x = -8$ in $\left|\frac{1}{2}x - 6\right| < 12$
- Solving inequality: $-\left(\frac{1}{2}x - 6\right) < 12 \implies \frac{1}{2}x - 6 > -12 \implies \frac{1}{2}x > -6 \implies x > -12$ $\frac{1}{2}x - 6 < 12 \implies \frac{1}{2}x < 18 \implies x < 36$
- Full solution interval: $(-12, 36)$
- Since $-8 \in (-12, 36)$ , the answer is Yes.