Comprehensive Study Notes on Data Visualization, Coordinate Geometry, and Circle Equations

Single-Variable Data Analysis

• One-Variable Data Concept:

  • One-variable data consists of a single list of numerical values representing a single characteristic or measurement (for example, points scored by blocks in the $2026$ NBA Finals).

• Measures of Central Tendency:

  • Mean: The arithmetic average calculated by summing all data points in a set and dividing by the total number of values $n$.
  • Median: The middle value of an ordered dataset.
    • When $n$ is odd, the median is the single central number.
    • When $n$ is even, the median is calculated as the mean of the two central numbers.
    • Interpretation: The median divides the distribution into two equal halves, meaning $50\%$ of the data points lie at or below the median, and $50\%$ lie at or above it.

• Example 1: Sleep Duration Dataset:

  • Dataset: $5$, $5$, $4$, $10$, $9$, $7$.
  • Sample size: $n = 6$.
  • Mean Calculation:
    • $\text{Mean} = \frac{5 + 5 + 4 + 10 + 9 + 7}{6} = \frac{40}{6} \approx 6.67$
  • Median Calculation:
    • Order the data from smallest to largest: $4$, $5$, $5$, $7$, $9$, $10$.
    • Since $n = 6$ is even, locate the two middle entries: $5$ and $7$
    • $\text{Median} = \frac{5 + 7}{2} = 6$

• Example 2: Decimal Sample Statistics:

  • Dataset: $3.61$, $4.3$, $4.12$, $4.92$, $8.82$, $9.86$
  • Total sum: $35.63$
  • $\text{Mean} = \frac{35.63}{6} \approx 5.94$
  • Related radical expression evaluation: $\frac{\sqrt{70}}{6} \approx 1.39$

• Example 3: Minnesota Low Temperatures:

  • Recorded low temperatures across six nights (in $^\circ\text{F}$): $-12$, $-4$, $-5$, $21$, $8$, $9$
  • Number Line Visualization: Data values are plotted along a single horizontal number line spanning from $-30$ to $30$
  • Extrema:
    • $\text{Maximum Value} = 21\,^\circ\text{F}$
    • $\text{Minimum Value} = -12\,^\circ\text{F}$
  • Mean Calculation:
    • $\text{Sum} = (-12) + (-4) + (-5) + 21 + 8 + 9 = -24 + 45 = 21$
    • $\text{Mean} = \frac{21}{6} = 3.5\,^\circ\text{F}$

Relations, Functions, and Two-Variable Data

• Fundamental Definitions:

  • Relation: A mapping or rule that pairs input values $x$ with output values $y$, represented as a set of ordered pairs $(x, y)$.
  • Domain: The set of all possible input values ($x$-values). Repeating values are listed only once.
  • Range: The set of all possible output values ($y$-values). Repeating values are listed only once.

• Example 1: Falling Objects Experiment:

  • Experiment tracking height fallen $v$ (in feet) and time $y$ (in seconds).
  • Set of ordered pairs $(v, y)$: $\{(20, 1.2), (40, 1.5), (40, 1.6)\}$
  • $\text{Domain} = \{20, 40\}$
  • $\text{Range} = \{1.2, 1.5, 1.6\}$

• Example 2: Discrete Relation Set $S$:

  • Relation definition: $S = \{(5, 10), (5, -5), (-10, 5), (10, 15), (-13, -10)\}$
  • $\text{Domain} = \{-13, -10, 5, 10\}$
  • $\text{Range} = \{-10, -5, 5, 10, 15\}$
  • Coordinate Extrema:
    • $X_{\text{max}} = 5$
    • $X_{\text{min}} = -13$
    • $Y_{\text{max}} = 15$
    • $Y_{\text{min}} = -10$
  • Graph Axis Scale Determination:
    • To display all elements of relation $S$ effectively, an appropriate coordinate scale spans $x$ from $-20$ to $20$ and $y$ from $-20$ to $20$.
    • The scatterplot graphs ordered pairs including $(15, 10)$, $(15, 20)$, and related data coordinates.

Coordinate Geometry: Midpoint and Distance Formulas

• The Midpoint Formula:

  • The midpoint $M$ of a line segment connecting endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is found by taking the mean of the $x$-coordinates and the mean of the $y$-coordinates:
  • $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
  • Example: Find the midpoint of $(6, -7)$ and $(-4, 6)$.
    • $M = \left(\frac{6 + (-4)}{2}, \frac{-7 + 6}{2}\right) = \left(\frac{2}{2}, \frac{-1}{2}\right) = \left(1, -\frac{1}{2}\right)$

• Finding an Unknown Endpoint from a Midpoint:

  • Given Midpoint $M = (-2, 6)$ and Endpoint 1 at $(2, 4)$, determine Endpoint 2 $(x_2, y_2)$.
  • $x$-coordinate derivation:
    • $\frac{2 + x_2}{2} = -2$
    • $2 + x_2 = -4 \implies x_2 = -6$
  • $y$-coordinate derivation:
    • $\frac{4 + y_2}{2} = 6$
    • $4 + y_2 = 12 \implies y_2 = 8$
  • Endpoint 2: $(-6, 8)$

• The Distance Formula:

  • Derived from the Pythagorean theorem $a^2 + b^2 = d^2$, where $a = x_2 - x_1$ and $b = y_2 - y_1$:
  • $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
  • Example: Calculate the distance between $(6, -7)$ and $(-4, 6)$.
    • $d = \sqrt{(-4 - 6)^2 + (6 - (-7))^2}$
    • $d = \sqrt{(-10)^2 + 13^2} = \sqrt{100 + 169} = \sqrt{269}$

• Applied Distance Problem 1: Two-Car Kinematics:

  • Scenario: At noon ($12:00\,\text{PM}$), Car A is located $80\,\text{miles}$ north at $(0, 80)$ traveling south at $20\,\text{mph}$. Car B is located at $(0, 0)$ traveling east at $60\,\text{mph}$.
  • Positions at $1:30\,\text{PM}$ ($1.5\,\text{hours}$ elapsed):
    • Car B travels east: $1.5\,\text{hours} \times 60\,\text{mph} = 90\,\text{miles}$. Position is $(90, 0)$.
    • Car A travels south: $1.5\,\text{hours} \times 20\,\text{mph} = 30\,\text{miles}$. Position drops from $(0, 80)$ to $(0, 50)$.
  • Distance between Car A and Car B at $1:30\,\text{PM}$:
    • $d = \sqrt{(90 - 0)^2 + (0 - 50)^2} = \sqrt{90^2 + (-50)^2} = \sqrt{8100 + 2500} = \sqrt{10600}$

• Applied Distance Problem 2: Map Distance:

  • Calculate distance between coordinates $(-40, 0)$ and $(0, 50)$.
  • $d = \sqrt{(0 - (-40))^2 + (50 - 0)^2} = \sqrt{40^2 + 50^2} = \sqrt{1600 + 2500} = \sqrt{4100} = 10\sqrt{41} \approx 64.03\,\text{miles}$

Equations of Circles in the Coordinate Plane

• Standard Form Equation of a Circle:

  • $(x - h)^2 + (y - k)^2 = r^2$
  • Center: $(h, k)$
  • Radius: $r$

• Example 1: Center and Radius Identification:

  • Given equation: $(x - 4)^2 + (y + 3)^2 = 16$
  • Rewrite parameters: $x - h = x - 4 \implies h = 4$, and $y - k = y - (-3) \implies k = -3$
  • Radius: $r^2 = 16 \implies r = 4$
  • Center: $(4, -3)$, Radius: $r = 4$

• Example 2: Completing the Square for Standard Form:

  • Given equation: $x^2 - 12x + y^2 + 10y = -25$
  • Group variables: $(x^2 - 12x) + (y^2 + 10y) = -25$
  • Complete square for $x$ and $y$:
    • For $x$: $\left(\frac{-12}{2}\right)^2 = 36$
    • For $y$: $\left(\frac{10}{2}\right)^2 = 25$
  • Add terms to both sides:
    • $(x^2 - 12x + 36) + (y^2 + 10y + 25) = -25 + 36 + 25$
    • $(x - 6)^2 + (y + 5)^2 = 36$
  • Resulting parameters: Center at $(6, -5)$, Radius $r = \sqrt{36} = 6$

• Example 3: Completing the Square with Leading Coefficients:

  • Given equation: $4x^2 + 4x + 4y^2 - 16y - 19 = 0$
  • Divide all terms by $4$:
    • $x^2 + x + y^2 - 4y - \frac{19}{4} = 0$
    • $x^2 + x + y^2 - 4y = \frac{19}{4}$
  • Complete the square for $x$ and $y$:
    • For $x$: $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$
    • For $y$: $\left(\frac{-4}{2}\right)^2 = 4$
  • Add constants to both sides:
    • $\left(x^2 + x + \frac{1}{4}\right) + (y^2 - 4y + 4) = \frac{19}{4} + \frac{1}{4} + 4$
    • Simplify right side: $\frac{19}{4} + \frac{1}{4} = \frac{20}{4} = 5$; $5 + 4 = 9$
  • Standard form equation:
    • $\left(x + \frac{1}{2}\right)^2 + (y - 2)^2 = 9$
  • Resulting parameters: Center at $\left(-\frac{1}{2}, 2\right)$ (also noted as $(-1, 2)$), Radius $r = \sqrt{9} = 3$

Plotting and Scale Guidelines for Discrete Coordinates

• Domain and Range Analysis of Discrete Points:

  • Points set: $(1, 1)$, $(3, 0)$, $(-5, -5)$, $(8, -2)$, $(0, 3)$
  • $\text{Domain} = \{1, 3, -5, 8, 0\}$
  • $\text{Range} = \{1, 0, -5, -2, 3\}$

• Coordinate Extrema:

  • $x$-values: Maximum $x = 8$, Minimum $x = -5$
  • $y$-values: Maximum $y = 3$, Minimum $y = -5$

• Axis Setup Guidelines:

  • $x$--axis range: Span from $-6$ to $10$, with tick marks every $1$ or $2$ units.
  • $y$--axis range: Span from $-6$ to $4$, with tick marks every $1$ or $2$ units.