Quick Graphs of Linear Equations - Definitive Study Guide

Core Principles of Quick Graphing

• Quick graphing methods provide efficient alternatives to creating a full table of values, plotting multiple points, and connecting them.
• Goal 1: Use the slope-intercept form of a linear equation to graph linear equations rapidly.
• Goal 2: Use the standard form of a linear equation to graph linear equations efficiently, particularly in real-world application contexts such as sales and fundraising models.
• Practical Utility: These methods allow immediate identification of key linear properties, such as baseline initial values and constant rates of change, directly from algebraic equations.

Slope-Intercept Form of Linear Equations

• y-intercept Definition: If the graph of an equation intersects the y-axis at the point $(0, b)$, then the number $b$ is the y-intercept of the graph.
• Finding the y-intercept: To find the y-intercept of a line algebraically, set $x = 0$ in the equation of the line and solve for $y$.
• Slope-Intercept Form Definition: The slope-intercept form of a linear equation is $y = mx + b$.
• A line with equation $y = mx + b$ has slope $m$ and y-intercept $b$

Concept Investigation Activity: Investigating Slope and y-intercept

• Completing the comparative analysis table:
  • Equation: $y = 2x + 3$
  • Points on graph: $(0, 3)$, $(1, 5)$
  • Slope: $2$
  • y-intercept: $3$
  • Equation: $y = -x + 2$
  • Points on graph: $(0, 2)$, $(1, 1)$
  • Slope: $-1$
  • y-intercept: $2$
  • Equation: $y = \frac{1}{2}x - 4$
  • Points on graph: $(0, -4)$, $(1, -\frac{7}{2})$
  • Slope: $\frac{1}{2}$
  • y-intercept: $-4$
  • Equation: $y = -2x$
  • Points on graph: $(0, 0)$, $(1, -2)$
  • Slope: $-2$
  • y-intercept: $0$
  • Equation: $y = 7$
  • Points on graph: $(0, 7)$, $(1, 7)$
  • Slope: $0$
  • y-intercept: $7$
• Key Conclusions from Activity:
  • The coefficient of $x$ in each equation is equal to the slope $m$ of the line.
  • The constant term in each equation is equal to the y-intercept $b$ of the line.

Procedure for Graphing Equations in Slope-Intercept Form

• Step 1: Write the equation in slope-intercept form by solving for $y$.
• Step 2: Find the y-intercept $b$ and use it to plot the point $(0, b)$ where the line crosses the y-axis.
• Step 3: Find the slope $m = \frac{\text{rise}}{\text{run}}$ and use it to plot a second point on the line.
• Step 4: Draw a line passing continuously through the two points.

Worked Examples for Slope-Intercept Form

• Example 1: Graphing with Slope-Intercept Form
  • Problem: Graph $y = \frac{3}{4}x - 2$
  • Solution Steps:
  1. The equation is already given in slope-intercept form $y = mx + b$
  2. The y-intercept is $-2$, so plot the point $(0, -2)$ where the line crosses the y-axis.
  3. The slope is $m = \frac{3}{4}$, so plot a second point on the line by starting at $(0, -2)$ and moving $4\,\text{units}$ to the right and $3\,\text{units}$ up. This yields the second point $(4, 1)$.
  4. Draw a continuous straight line passing through $(0, -2)$ and $(4, 1)$.

Standard Form of Linear Equations

• Standard Form Definition: The standard form of a linear equation is $Ax + By = C$, where $A$ and $B$ are not both zero.
• x-intercept Definition: The x-intercept of a line is the x-coordinate of the point $(x, 0)$ where the line intersects the x-axis.
• Finding the x-intercept: Set $y = 0$ in the equation and solve for $x$
• Finding the y-intercept in Standard Form: Set $x = 0$ in the equation and solve for $y$
• Quick Graph Method: Graphing an equation in standard form is performed by calculating and plotting its intercepts.

Procedure for Graphing Equations in Standard Form

• Step 1: Ensure the equation is written in standard form $Ax + By = C$
• Step 2: Find the x-intercept by letting $y = 0$ and solving for $x$. Use the x-intercept to plot the point $(x, 0)$ where the line crosses the x-axis.
• Step 3: Find the y-intercept by letting $x = 0$ and solving for $y$. Use the y-intercept to plot the point $(0, y)$ where the line crosses the y-axis.
• Step 4: Draw a continuous line through the two intercept points.

Worked Examples for Standard Form

• Example 3: Drawing Quick Graphs
  • Problem: Graph $2x + 3y = 12$
  • Method 1: Use Standard Form (Intercept Method)
  1. The equation $2x + 3y = 12$ is written in standard form.
  2. Find x-intercept: Let $y = 0 \Rightarrow 2x + 3(0) = 12 \Rightarrow 2x = 12 \Rightarrow x = 6$. Plot point $(6, 0)$.
  3. Find y-intercept: Let $x = 0 \Rightarrow 2(0) + 3y = 12 \Rightarrow 3y = 12 \Rightarrow y = 4$. Plot point $(0, 4)$.
  4. Draw a continuous line through $(6, 0)$ and $(0, 4)$.
  • Method 2: Use Slope-Intercept Form Conversion
  1. Convert $2x + 3y = 12$ to slope-intercept form:        $2x + 3y = 12$$3y = -2x + 12$$y = -\frac{2}{3}x + 4$
  2. The y-intercept is $4$, so plot the point $(0, 4)$.
  3. The slope is $m = -\frac{2}{3}$, so plot a second point by starting at $(0, 4)$ and moving $3\,\text{units}$ to the right and $2\,\text{units}$ down. This yields point $(3, 2)$.
  4. Draw a straight line through the points.

Horizontal and Vertical Lines

• Slope Limitations of Vertical Lines: The equation of a vertical line cannot be written in slope-intercept form because the slope of a vertical line is undefined (division by zero).
• Standard Form Universality: Every linear equation can be written in standard form, including vertical lines.
• Horizontal Line Theorem: The graph of $y = c$ is a horizontal line passing through $(0, c)$. Slope $m = 0$. Every point on the line has a y-coordinate of $c$
• Vertical Line Theorem: The graph of $x = c$ is a vertical line passing through $(c, 0)$. Slope $m$ is undefined. Every point on the line has an x-coordinate of $c$

Worked Example for Horizontal and Vertical Lines

• Example 4: Graphing Horizontal and Vertical Lines
  • Problem: Graph (a) $y = 3$ and (b) $x = -2$
  • Solution (a): The graph of $y = 3$ is a horizontal line that passes through the point $(0, 3)$. Every point on this line has a y-coordinate equal to $3$
  • Solution (b): The graph of $x = -2$ is a vertical line that passes through the point $(-2, 0)$. Every point on this line has an x-coordinate equal to $-2$

Real-World Applications and Modeling

• Real-World Interpretations:
  • y-intercept: Represents an initial amount, starting value, or baseline setup cost (at time $t = 0$).
  • Slope: Represents a rate of change (e.g., payment per week, price per unit, velocity).

Case Study 1: Computer Layaway Purchase

• Problem Context (Example 2): Buying an $1100\,\text{dollar}$ computer on layaway. A $250\,\text{dollar}$ initial deposit is made, followed by weekly payments according to $a = 850 - 50t$, where $a$ is the amount owed in dollars and $t$ is the time in weeks.
• Questions & Detailed Solutions:
  • a. What is the original amount you owe on layaway?
  • Rewrite in slope-intercept form: $a = -50t + 850$
  • The a-intercept is $850$
  • The original amount owed (when $t = 0$) is $850\,\text{dollars}$
  • b. What is your weekly payment?
  • From slope-intercept form, the slope is $m = -50$
  • The amount owed changes at a rate of $-50\,\text{dollars/week}$
  • Thus, the weekly payment is $50\,\text{dollars}$
  • c. Graph the model:
  • Plot a-intercept $(0, 850)$
  • The line extends downward until it reaches the t-axis at $t = 17$ (point $(17, 0)$). At $t = 17\,\text{weeks}$, the amount owed is $0\,\text{dollars}$, meaning the computer is completely paid for.

Case Study 2: School Band Fundraising

• Problem Context (Example 5): The school band sells sweatshirts and T-shirts to raise money with a goal of $1200\,\text{dollars}$. Sweatshirts yield a profit of $2.50\,\text{dollars}$ each, and T-shirts yield a profit of $1.50\,\text{dollars}$ each. Describe the combinations of sweatshirts and T-shirts that reach the goal.
• Model Construction:
  • Verbal Model:$\text{Profit per sweatshirt} \times \text{Number of sweatshirts} + \text{Profit per T-shirt} \times \text{Number of T-shirts} = \text{Total Profit}$
  • Labels:
  • Profit per sweatshirt = $2.50\,\text{dollars}$
  • Number of sweatshirts = $s$
  • Profit per T-shirt = $1.50\,\text{dollars}$
  • Number of T-shirts = $t$
  • Total profit = $1200\,\text{dollars}$
  • Algebraic Model:$2.5s + 1.5t = 1200$
• Graphing and Interpretation:
  • s-intercept ($t = 0$): $2.5s = 1200 \Rightarrow s = 480$. Point $(480, 0)$.
  • t-intercept ($s = 0$): $1.5t = 1200 \Rightarrow t = 800$. Point $(0, 800)$.
  • Points with integer coordinates on the line segment connecting $(480, 0)$ and $(0, 800)$ represent valid real-world combinations. For instance, selling $300\,\text{sweatshirts}$ and $300\,\text{T-shirts}$ achieves the target goal.
• Study Tip for Scale Determination: Finding the intercepts of a line prior to graphing assists in establishing appropriate, reasonable scales for the x-axis and y-axis.

Comprehensive Application Scenarios

• Internal Revenue Service Taxes (Exercise 58):
  • Annual tax collections $a$ (in billions of dollars) modeled by $a = 57.1t + 488$, where $t$ is years since $1980$
• Newspaper Color Advertisement Cost (Exercise 59):
  • Cost $C$ (in dollars) modeled by $C = 7n + 20$, where $n$ is the number of lines in the ad.
  • Slope $7$ represents the price per line ($7\,\text{dollars/line}$).
  • C-intercept $20$ represents the fixed base charge or setup fee ($20\,\text{dollars}$).
• Rainforest Deforestation (Exercise 60):
  • Rainforest land area $A$ (in millions of hectares) modeled by $A = 718.3 - 4.6t$, where $t$ is years since $1990$
  • Contextual Note: In Brazil, rainforest destruction occurs at $2.2\,\text{million hectares/year}$. Brazil passed legislation giving its government legal authority to protect forests.
• Car Wash Revenues (Exercise 61):
  • Charges: $8\,\text{dollars}$ per wash, $12\,\text{dollars}$ per wash-and-wax. Total sales = $3464\,\text{dollars}$.
  • Model: $8w_1 + 12w_2 = 3464$, where $w_1$ is number of washes and $w_2$ is number of wash-and-waxes.
• Sailboat Excursions (Exercise 62):
  • Distance to island = $5\,\text{miles}$. Average speeds: $9\,\text{mph}$ using sails, $14\,\text{mph}$ using motor.
  • Model for time in minutes using sails ($x$) and motor ($y$): $\frac{9}{60}x + \frac{14}{60}y = 5 \Rightarrow 0.15x + \frac{7}{30}y = 5$
• Basketball Ticket Revenues (Exercise 63):
  • Student ticket price = $2.50\,\text{dollars}$, adult ticket price = $6.00\,\text{dollars}$, total game revenue = $7000\,\text{dollars}$.
  • Model: $2.50s + 6.00a = 7000$

Guided Practice and Exercise Sets

Guided Practice Questions

• Vocabulary Check:
  1. What are the slope-intercept and standard forms of a linear equation?
  • Slope-intercept form: $y = mx + b$
  • Standard form: $Ax + By = C$
• Concept Check:
  2. Which quick-graph technique for $y = -2x + 4$?
  • Use slope-intercept form because the equation is already solved for $y$
  2. Which quick-graph technique for $3x + 4y = 24$?
  • Use standard form (intercept method) because calculating intercepts ($x = 8$, $y = 6$) is direct and rapid.

Skill Check Exercises

• Find slope and y-intercept:
  • Exercise 4: $y = x + 10 \Rightarrow m = 1$, $b = 10$
  • Exercise 5: $y = -2x - 7 \Rightarrow m = -2$, $b = -7$
  • Exercise 6: $2x - 3y = 18 \Rightarrow y = \frac{2}{3}x - 6 \Rightarrow m = \frac{2}{3}$, $b = -6$
• Find intercepts:
  • Exercise 7: $x - y = 11 \Rightarrow$ x-intercept = $11$, y-intercept = $-11$
  • Exercise 8: $5x - 2y = 20 \Rightarrow$ x-intercept = $4$, y-intercept = $-10$
  • Exercise 9: $y = 5x - 15 \Rightarrow$ x-intercept = $3$, y-intercept = $-15$
• Graphing Equations:
  • Exercise 10: $y = 2x + 1$
  • Exercise 11: $y = \frac{1}{3}x - 4$
  • Exercise 12: $y = 7$
  • Exercise 13: $x = -5$
  • Exercise 14: $2x - 6y = 6$
  • Exercise 15: $5x + 3y = -15$

Practice and Applications Problems

• Matching Equations with Graphs (Ex. 16–18 & 37–39):
  • Exercise 16: $y = -5x + 10$
  • Exercise 17: $y = -\frac{1}{2}x - 5$
  • Exercise 18: $y = 4x - 12$
  • Exercise 37: $x - 4y = -8$
  • Exercise 38: $3x + 6y = -9$
  • Exercise 39: $2x - 3y = -12$
• Line Construction from Given Slope and y-intercept (Ex. 19–24):
  • Exercise 19: $m = 3$, $b = -2$
  • Exercise 20: $m = -2$, $b = 0$
  • Exercise 21: $m = 1$, $b = 1$
  • Exercise 22: $m = \frac{1}{2}$, $b = 5$
  • Exercise 23: $m = 0$, $b = -7$
  • Exercise 24: $m = -\frac{3}{7}$, $b = 14$
• Graphing via Slope-Intercept Form (Ex. 25–30):
  • Exercise 25: $y = -x + 5$
  • Exercise 26: $y = 4x + 1$
  • Exercise 27: $y = \frac{4}{5}x - 1$
  • Exercise 28: $y = 2x - 3$
  • Exercise 29: $y = -\frac{5}{2}x - 3$
  • Exercise 30: $y = 5x - \frac{5}{2}$
• Finding Slope and y-intercept (Ex. 31–36):
  • Exercise 31: $y = 6x + 10 \Rightarrow m = 6$, $b = 10$
  • Exercise 32: $y = -9x \Rightarrow m = -9$, $b = 0$
  • Exercise 33: $y = 100 \Rightarrow m = 0$, $b = 100$
  • Exercise 34: $2x + y = 14 \Rightarrow m = -2$, $b = 14$
  • Exercise 35: $8x - 2y = 14 \Rightarrow m = 4$, $b = -7$
  • Exercise 36: $x + 10y = 7 \Rightarrow m = -\frac{1}{10}$, $b = \frac{7}{10}$
• Graphing via Intercepts (Ex. 40–51):
  • Exercise 40: x-intercept = $3$, y-intercept = $5$
  • Exercise 41: x-intercept = $2$, y-intercept = $-6$
  • Exercise 42: x-intercept = $-4$, y-intercept = $-2$
  • Exercise 43: $2x + y = 8$
  • Exercise 44: $x + 2y = 8$
  • Exercise 45: $3x + 4y = -10$
  • Exercise 46: $3x - y = 3$
  • Exercise 47: $5x - 6y = -2$
  • Exercise 48: $3x + 0.2y = 2$
  • Exercise 49: $y = 6$
  • Exercise 50: $x = -5$
  • Exercise 51: $y = -2.5$

Practice Test Preparation & Challenge Questions

• Exercise 64 (Conceptual Explanation):
  • How to find intercepts: To find the x-intercept, substitute $y = 0$ and solve for $x$. To find the y-intercept, substitute $x = 0$ and solve for $y$
  • Lines with no x-intercept: Horizontal lines $y = c$ (where $c \neq 0$) are parallel to the x-axis and have no x-intercept.
  • Lines with no y-intercept: Vertical lines $x = c$ (where $c \neq 0$) are parallel to the y-axis and have no y-intercept.
• Exercise 65 (Multiple Choice - IRA Contributions):
  • Question: Account balance $a = 4500 + 2000t$ after $t$ years. How much money is deposited each year?
  • Analysis: Annual deposit rate is represented by slope $m = 2000$
  • Correct Answer: B ($2000)
• Exercise 66 (Multiple Choice - Slope-Intercept Conversion):
  • Question: What is slope-intercept form of $4x - 6y = 18$?
  • Steps: $-6y = -4x + 18 \Rightarrow y = \frac{4}{6}x - 3 \Rightarrow y = \frac{2}{3}x - 3$
  • Correct Answer: B ($y = \frac{2}{3}x - 3$)
• Exercise 67 (Challenge - Constant Slope Proof):
  • For line $y = 7x + 6$, show slope is $7$ regardless of points $(x_1, y_1)$ and $(x_2, y_2)$
  • Proof: $y_1 = 7x_1 + 6$ and $y_2 = 7x_2 + 6$.
  • Slope formula: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{(7x_2 + 6) - (7x_1 + 6)}{x_2 - x_1} = \frac{7(x_2 - x_1)}{x_2 - x_1} = 7$

Self-Test Quiz and Mixed Review

Quiz 1 (Lessons 2.1–2.3 Self-Test)

• Domain and Range Relations (Exercises 1–3):
  1. Points: $(-1, -1)$, $(0, 1)$, $(1, -1)$, $(2, 1)$
  • Domain: $\{-1, 0, 1, 2\}$, Range: $\{-1, 1\}$
  • Function check: Yes, each input has exactly one output.
  1. Points: $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4)$, $(4, 1)$
  • Domain: $\{1, 2, 3, 4\}$, Range: $\{1, 2, 3, 4\}$
  • Function check: No, input $4$ maps to both $1$ and $4$
  1. Points: $(-2, -1)$, $(0, -1)$, $(0, 0)$, $(1, 1)$, $(2, -2)$
  • Domain: $\{-2, 0, 1, 2\}$, Range: $\{-2, -1, 0, 1\}$
  • Function check: No, input $0$ maps to both $-1$ and $0$
• Evaluating Functions (Exercises 4–5):
  4. $f(x) = -2x - 13; f(4) \Rightarrow f(4) = -2(4) - 13 = -8 - 13 = -21$
  5. $f(x) = 5x^2 - x + 9; f(-5) \Rightarrow f(-5) = 5(-5)^2 - (-5) + 9 = 5(25) + 5 + 9 = 125 + 14 = 139$
• Line Relationships (Exercises 6–7):
  6. Line 1 through $(2, 10)$ and $(1, 5) \Rightarrow m_1 = \frac{5 - 10}{1 - 2} = 5$
  • Line 2 through $(3, -7)$ and $(8, -8) \Rightarrow m_2 = \frac{-8 - (-7)}{8 - 3} = -\frac{1}{5}$
  • Relationship: Perpendicular ($m_1 \cdot m_2 = -1$)
  6. Line 1 through $(4, 5)$ and $(9, -2) \Rightarrow m_1 = \frac{-2 - 5}{9 - 4} = -\frac{7}{5}$
  • Line 2 through $(6, -6)$ and $(-2, -1) \Rightarrow m_2 = \frac{-1 - (-6)}{-2 - 6} = -\frac{5}{8}$
  • Relationship: Neither parallel nor perpendicular.
• Graphing Equations (Exercises 8–10):
  8. $y = 3x + 5$
  9. $2x - 3y = 10$
  10. $y = -11$
• Iowa Bicycle Ride Application (Exercise 11):
  • Problem: Annual 7-day ride across Iowa covering $468\,\text{miles}$. Ride each day 8:00 A.M. to 5:00 P.M. with 1 hour lunch break. What is average speed?
  • Calculation: Daily riding time = $9\,\text{hours} - 1\,\text{hour lunch} = 8\,\text{hours/day}$.
  • Total trip hours = $7\,\text{days} \times 8\,\text{hours/day} = 56\,\text{hours}$.
  • Average speed = $\frac{468\,\text{miles}}{56\,\text{hours}} \approx 8.36\,\text{mph}$.

Mixed Review Problems

• Solving Inequalities (Ex. 68–73):
  • Exercise 68: $9 + x \le 21 \Rightarrow x \le 12$
  • Exercise 69: $-\frac{2}{3}x + 3 < 11 \Rightarrow -\frac{2}{3}x < 8 \Rightarrow x > -12$
  • Exercise 70: $2x - 11 > 34 - x \Rightarrow 3x > 45 \Rightarrow x > 15$
  • Exercise 71: $64 - 3x \ge 19 - 2x \Rightarrow -x \ge -45 \Rightarrow x \le 45$
  • Exercise 72: $-5 < 2x - 0.5 \le 23 \Rightarrow -4.5 < 2x \le 23.5 \Rightarrow -2.25 < x \le 11.75$
  • Exercise 73: $x + 12 \le 5 \text{ or } 3x - 21 \ge 0 \Rightarrow x \le -7 \text{ or } x \ge 7$
• Evaluating Functions (Ex. 74–79):
  • Exercise 74: $f(x) = \frac{1}{2}x - 13; f(8) \Rightarrow f(8) = 4 - 13 = -9$
  • Exercise 75: $f(x) = x^2 - 3x + 2; f(5) \Rightarrow f(5) = 25 - 15 + 2 = 12$
  • Exercise 76: $f(x) = -x^3 + 8x^2 + 3; f(-7) \Rightarrow f(-7) = -(-343) + 8(49) + 3 = 343 + 392 + 3 = 738$
  • Exercise 77: $f(x) = 10 - 2x; f(1) \Rightarrow f(1) = 10 - 2 = 8$
  • Exercise 78: $f(x) = |x + 17|; f(-5) \Rightarrow f(-5) = |-5 + 17| = 12$
  • Exercise 79: $f(x) = 12x^2 - 19; f(\frac{1}{2}) \Rightarrow f(\frac{1}{2}) = 12(\frac{1}{4}) - 19 = 3 - 19 = -16$
• Finding Slope Through Points (Ex. 80–85):
  • Exercise 80: $(3, 2), (7, 2) \Rightarrow m = \frac{0}{4} = 0$
  • Exercise 81: $(16, -3), (2, 9) \Rightarrow m = \frac{9 - (-3)}{2 - 16} = \frac{12}{-14} = -\frac{6}{7}$
  • Exercise 82: $(-12, -9), (1, -8) \Rightarrow m = \frac{-8 - (-9)}{1 - (-12)} = \frac{1}{13}$
  • Exercise 83: $(-1, -1), (-1, -5) \Rightarrow m = \frac{-4}{0} \Rightarrow \text{undefined}$
  • Exercise 84: $(5, -2), (-3, 2) \Rightarrow m = \frac{2 - (-2)}{-3 - 5} = \frac{4}{-8} = -\frac{1}{2}$
  • Exercise 85: $(-4, 7), (2, -5) \Rightarrow m = \frac{-5 - 7}{2 - (-4)} = \frac{-12}{6} = -2$
• Reading Speed Model (Ex. 86):
  • Reading rate = $2\,\text{pages/minute} = 120\,\text{pages/hour}$
  • Model for pages $P$ read in $h$ hours: $P = 120h$
  • Time to read $1048\,\text{page}$ novel: $1048 = 120h \Rightarrow h = \frac{1048}{120} \approx 8.73\,\text{hours}$ ($8\,\text{hours } 44\,\text{minutes}$).

Math and History Connection: Transatlantic Voyages

• Historical Logistical Background:
  • At 2:00 P.M. on April 11, 1912, the Titanic departed Cobh, Ireland, on her maiden voyage to New York City.
  • At 11:40 P.M. on April 14, 1912, the Titanic struck an iceberg and sank after covering approximately $2100\,\text{miles}$ of the total $3400\,\text{mile}$ planned voyage.

Quantitative Historical Analysis

• Question 1: Total length of Titanic\'s maiden voyage in hours?
  • Time calculation:
  • April 11, 2:00 P.M. to April 14, 2:00 P.M. = $3\,\text{days} = 72\,\text{hours}$
  • April 14, 2:00 P.M. to 11:40 P.M. = $9\,\text{hours } 40\,\text{minutes} = 9.67\,\text{hours}$
  • Total duration = $72 + 9.67 = 81.67\,\text{hours}$ ($81\frac{2}{3}\,\text{hours}$).
• Question 2: What was the Titanic\'s average speed in miles per hour?
  • Average speed formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$
  • Calculation: $\text{Speed} = \frac{2100\,\text{miles}}{81.67\,\text{hours}} \approx 25.7\,\text{mph}$
• Question 3: Write an equation relating Titanic\'s distance from New York City and hours traveled.
  • Total distance to NYC = $3400\,\text{miles}$
  • Model: Distance remaining $d$ after $t$ hours: $d = 3400 - 25.7t$
  • Domain: $0 \le t \le 81.67\,\text{hours}$ (for maiden voyage up to sinking) or $0 \le t \le 132.3\,\text{hours}$ (for entire hypothetical trip).
  • Range: $1300 \le d \le 3400\,\text{miles}$ (for actual voyage) or $0 \le d \le 3400\,\text{miles}$.
• Question 4: Graph the equation.
  • Plot d-intercept at $(0, 3400)$ and ending point at $(81.67, 1300)$.

Modern Transatlantic Comparisons

• Ocean liners continue crossing the Atlantic Ocean. The Queen Elizabeth 2 (QE2) represents one of the fastest vessels with a top speed of $32.5\,\text{knots}$ (approximately $37\,\text{mph}$).
• Transatlantic Travel Timeline Milestones:
  • 1912: Titanic maiden voyage.
  • 1927: Charles Lindbergh completes the first solo transatlantic flight.
  • 1969: QE2 maiden voyage.
  • 1998: Benoit Lecomte swims across the Atlantic Ocean.