Comprehensive Study Guide: Library of Functions and Function Transformations

Library of Key Functions (Section 2.4)

Overview and Recognition Requirements

• Essential key functions (base graphs) must be memorized and recognized instantly without hesitation.

• Core competencies required:

  • Given a graph, immediately identify its corresponding function equation $f(x)$.

  • Given a function equation $f(x)$, rapidly construct an accurate graph.

  • Identify domain, range, x-intercepts, y-intercepts, and symmetry (even or odd function classification) for each base graph.

• Foundation importance: Rapid visual and algebraic recognition of basic functions forms the prerequisite foundation for all subsequent function transformations and calculus concepts.

Key Base Functions Summary

• Constant Function:

  • Equation: $f(x) = c$

  • Visual Shape: A straight horizontal line extending infinitely left and right.

• Identity Function:

  • Equation: $f(x) = x$

  • Visual Shape: An oblique (slanted) line passing through the origin $(0, 0)$ with a slope of $1$

• Square Function (Quadratic):

  • Equation: $f(x) = x^2$

  • Visual Shape: A U-shaped parabola opening upwards with its vertex at $(0, 0)$

• Cube Function:

  • Equation: $f(x) = x^3$

  • Visual Shape: An S-shaped curve passing through the origin, vertically stretched through Quadrants I and III.

• Square Root Function:

  • Equation: $f(x) = \sqrt{x}$

  • Visual Shape: Half of a horizontal parabola starting at $(0, 0)$ and extending into Quadrant I.

• Cube Root Function:

  • Equation: $f(x) = \sqrt[3]{x}$

  • Visual Shape: An S-shaped curve stretched out horizontally across Quadrants I and III.

• Reciprocal Function:

  • Equation: $f(x) = \frac{1}{x}$

  • Visual Shape: A rational curve formatted as two hyperbola branches situated in Quadrants I and III.

• Absolute Value Function:

  • Equation: $f(x) = |x|$

  • Visual Shape: A symmetric V-shaped graph with a right-angle vertex at $(0, 0)$

• Greatest Integer Function (Step Function):

  • Equation: $f(x) = \lfloor x \rfloor$

  • Visual Shape: A series of horizontal step segments with closed left endpoints and open right endpoints.

Point-by-Point Derivation of Base Graphs

Base graph shapes are formally constructed using a table of values to determine precise coordinate points.

• Square Function Table and Graph construction ($f(x) = x^2$):

  • For $x = -2$: $y = (-2)^2 = 4 \implies (-2, 4)$

  • For $x = -1$: $y = (-1)^2 = 1 \implies (-1, 1)$

  • For $x = 0$: $y = 0^2 = 0 \implies (0, 0)$

  • For $x = 1$: $y = 1^2 = 1 \implies (1, 1)$

  • For $x = 2$: $y = 2^2 = 4 \implies (2, 4)$

  • Graph properties: Smooth symmetric parabola passing through $(0, 0)$, $(1, 1)$, $(-1, 1)$, $(2, 4)$, and $(-2, 4)$.

• Reciprocal Function Table and Graph construction ($f(x) = \frac{1}{x}$):

  • For $x = -2$: $y = \frac{1}{-2} = -\frac{1}{2} \implies \left(-2, -\frac{1}{2}\right)$

  • For $x = -1$: $y = \frac{1}{-1} = -1 \implies (-1, -1)$

  • For $x = 0$: $y = \frac{1}{0}$ is undefined (y-axis is a vertical asymptote; graph never touches or crosses $x = 0$)

  • For $x = 1$: $y = \frac{1}{1} = 1 \implies (1, 1)$

  • For $x = 2$: $y = \frac{1}{2} \implies \left(2, \frac{1}{2}\right)$

  • Intercept analysis: Setting $0 = \frac{1}{x}$ yields no real solution. There are no x-intercepts and no y-intercepts.

  • Graph properties: Asymptotic curves approaching $x = 0$ vertically and $y = 0$ horizontally.

• Cube Root Function Table and Graph construction ($f(x) = \sqrt[3]{x}$):

  • For $x = -8$: $y = \sqrt[3]{-8} = -2 \implies (-8, -2)$

  • For $x = -2$: $y = \sqrt[3]{-2} \approx -1.25 \implies (-2, -1.25)$

  • For $x = -1$: $y = \sqrt[3]{-1} = -1 \implies (-1, -1)$

  • For $x = 0$: $y = \sqrt[3]{0} = 0 \implies (0, 0)$

  • For $x = 1$: $y = \sqrt[3]{1} = 1 \implies (1, 1)$

  • For $x = 2$: $y = \sqrt[3]{2} \approx 1.25 \implies (2, 1.25)$

  • For $x = 8$: $y = \sqrt[3]{8} = 2 \implies (8, 2)$

  • Graph properties: Slowly rising horizontal S-curve passing through $(0, 0)$, $(1, 1)$, $(-1, -1)$, $(8, 2)$, and $(-8, -2)$.

Piecewise-Defined Functions

• Definition: A piecewise function is a single function defined by different sub-equations applied across specific sub-domains of $x$.

• Real-world relevance: Critical for modeling systems that undergo threshold-based operational changes and used extensively throughout calculus.

• Evaluating Piecewise Functions:

  • Given function:     $f(x) = \begin{cases} -3x & \text{if } x < -1 \\ 0 & \text{if } x = -1 \\ 2x^2 + 1 & \text{if } x > -1 \end{cases}$

  • Example Evaluation 1: Find $f(-2)$

  • Identify domain condition: -2 < -1, so apply the top rule $f(x) = -3x$.

  • Calculation: $f(-2) = -3(-2) = 6$

  • Example Evaluation 2: Find $f(-1)$

  • Identify domain condition: $x = -1$, so apply the middle rule $f(x) = 0$.

  • Calculation: $f(-1) = 0$

  • Example Evaluation 3: Find $f(0)$

  • Identify domain condition: 0 > -1, so apply the bottom rule $f(x) = 2x^2 + 1$.

  • Calculation: $f(0) = 2(0)^2 + 1 = 1$

• Graphing Piecewise Functions Strategy:

  • Procedure: Graph each sub-equation lightly across the full plane, then erase the portions falling outside the designated domain boundaries. Pay close attention to strict inequalities (open circles) versus inclusive inequalities (closed solid dots).

• Piecewise Graphing Example 1:

  • Function:     $f(x) = \begin{cases} 2 - x &amp; \text{if } -3 \le x &lt; 1 \\ \sqrt{x} &amp; \text{if } x &gt; 1 \end{cases}$

  • Graphing Sub-piece 1 ($y = 2 - x$ on $[-3, 1)$):

  • Linear segment with y-intercept $2$ and slope $-1$

  • At $x = -3$: $y = 2 - (-3) = 5 \implies$ solid closed dot at $(-3, 5)$

  • At $x = 1$: $y = 2 - 1 = 1 \implies$ open circle at $(1, 1)$

  • Graphing Sub-piece 2 ($y = \sqrt{x}$ on $(1, \infty)$):

  • Square root curve starting after $x = 1$

  • At $x = 1$: $y = \sqrt{1} = 1 \implies$ open circle at $(1, 1)$

  • Plot points for x > 1: $(4, 2)$, $(9, 3)$ extending rightward.

• Piecewise Graphing Example 2:

  • Function:     $f(x) = \begin{cases} 2x + 5 &amp; \text{if } -3 \le x &lt; 0 \\ -3 &amp; \text{if } x = 0 \\ -5x &amp; \text{if } x &gt; 0 \end{cases}$

  • Graphing Sub-piece 1 ($y = 2x + 5$ on $[-3, 0)$):

  • Linear segment with slope $2$

  • At $x = -3$: $y = 2(-3) + 5 = -1 \implies$ solid closed dot at $(-3, -1)$

  • At $x = 0$: $y = 2(0) + 5 = 5 \implies$ open circle at $(0, 5)$

  • Graphing Sub-piece 2 ($y = -3$ at $x = 0$):

  • Single isolated coordinate point at $(0, -3)$

  • Graphing Sub-piece 3 ($y = -5x$ on $(0, \infty)$):

  • Steep linear line extending downward into Quadrant IV

  • At $x = 0$: $y = -5(0) = 0 \implies$ open circle at $(0, 0)$

  • At $x = 1$: $y = -5(1) = -5 \implies (1, -5)$

  • Continuity Analysis:

  • A function is continuous if its graph can be traced from left to right without lifting the pencil.

  • This function is discontinuous because it consists of separated, broken pieces.

• Constructing Piecewise Equations from Graphs:

  • Problem: Determine the algebraic equation for a given two-part graph.

  • Left Segment: Horizontal line segment at height $y = 1$ defined from $x = -1$ to $x = 0$ with closed endpoints at both ends.

  • Right Segment: Slanted line with y-intercept $0$ and slope $1$ ($y = x$) defined from $x = 0$ to $x = 2$ with an open end at $x = 0$ and closed endpoint at $x = 2$

  • Formulation:     $f(x) = \begin{cases} 1 &amp; \text{if } -1 \le x \le 0 \\ x &amp; \text{if } 0 &lt; x \le 2 \end{cases}$

Function Transformations (Section 2.5)

Comprehensive Rules of Transformation

Transformations modify base graphs through algebraic adjustments to the function equation:

• Vertical Shifts:

  • Equation modification: $y = f(x) + c$

  • Effect: Shifts graph UP by $c$ units (c > 0). Adds $c$ to all y-coordinates.

  • Equation modification: $y = f(x) - c$

  • Effect: Shifts graph DOWN by $c$ units (c > 0). Subtracts $c$ from all y-coordinates.

• Horizontal Shifts:

  • Equation modification: $y = f(x + c)$

  • Effect: Shifts graph LEFT by $c$ units (c > 0). Subtracts $c$ from all x-coordinates (opposite intuition).

  • Equation modification: $y = f(x - c)$

  • Effect: Shifts graph RIGHT by $c$ units (c > 0). Adds $c$ to all x-coordinates.

• Vertical Stretching and Compressing (Y-Axis Transformations):

  • Equation modification: $y = a f(x)$

  • Effect: Multiplies all y-coordinates by $a$

  • If a > 1: Vertical stretch by a factor of $a$

  • If 0 < a < 1: Vertical compression by a factor of $a$

• Horizontal Stretching and Compressing (X-Axis Transformations):

  • Equation modification: $y = f(a x)$

  • Effect: Multiplies all x-coordinates by $\frac{1}{a}$ (or divides x-coordinates by $a$)

  • If a > 1: Horizontal compression by a factor of $\frac{1}{a}$

  • If 0 < a < 1: Horizontal stretch by a factor of $\frac{1}{a}$

• Reflections:

  • Reflection over X-Axis: $y = -f(x)$

  • Negates all y-coordinates (minus symbol outside the main operation).

  • Reflection over Y-Axis: $y = f(-x)$

  • Negates all x-coordinates (minus symbol inside the operation attached directly to $x$).

Applying Single Transformations to Base Graph $f(x) = x^3$

Given the base function $f(x) = x^3$ with key points $(-2, -8)$, $(-1, -1)$, $(0, 0)$, $(1, 1)$, and $(2, 8)$:

• Shift Left 4 units:

  • Modify equation inside operation: $g(x) = (x + 4)^3$

• Shift Down 4 units:

  • Modify equation outside operation: $g(x) = x^3 - 4$

• Reflect across the X-Axis:

  • Place negative symbol outside: $g(x) = -x^3$

• Reflect across the Y-Axis:

  • Place negative symbol inside: $g(x) = (-x)^3$

• Horizontal Stretch by a factor of 4:

  • Multiply inside $x$ by reciprocal $\frac{1}{4}$: $g(x) = \left(\frac{1}{4}x\right)^3$

Multi-Step Graphing Transformations

• Recommended Order of Transformations:

  1. Reflections (across x-axis or y-axis)

  2. Compressions and Extensions (vertical or horizontal stretching/compressing)

  3. Shifts (horizontal and vertical movements)

• Multi-Step Transformation Example 1: Graph $g(x) = 3|x + 1| - 3$

  • Base Function: $f(x) = |x|$

  • Initial points: $(-2, 2)$, $(-1, 1)$, $(0, 0)$, $(1, 1)$, $(2, 2)$

  • Stage 1 (Vertical Stretch by factor of 3): $y = 3|x|$

  • Multiply all y-values by $3$

  • Intermediate points: $(-2, 6)$, $(-1, 3)$, $(0, 0)$, $(1, 3)$, $(2, 6)$

  • Stage 2 (Horizontal Shift Left 1 unit): $y = 3|x + 1|$

  • Subtract $1$ from all x-values

  • Intermediate points: $(-3, 6)$, $(-2, 3)$, $(-1, 0)$, $(0, 3)$, $(1, 6)$

  • Stage 3 (Vertical Shift Down 3 units): $g(x) = 3|x + 1| - 3$

  • Subtract $3$ from all y-values

  • Final coordinate points: $(-3, 3)$, $(-2, 0)$, $(-1, -3)$, $(0, 0)$, $(1, 3)$

• Multi-Step Transformation Example 2: Graph $h(x) = \frac{4}{x} + 2$

  • Base Function: $f(x) = \frac{1}{x}$

  • Initial points: $\left(-2, -\frac{1}{2}\right)$, $(-1, -1)$, $(1, 1)$, $\left(2, \frac{1}{2}\right)$

  • Stage 1 (Vertical Stretch by factor of 4): $y = \frac{4}{x}$

  • Multiply all y-values by $4$

  • Intermediate points: $(-2, -2)$, $(-1, -4)$, $(1, 4)$, $(2, 2)$

  • Stage 2 (Vertical Shift Up 2 units): $h(x) = \frac{4}{x} + 2$

  • Add $2$ to all y-values

  • Final coordinate points: $(-2, 0)$, $(-1, -2)$, $(1, 6)$, $(2, 4)$

  • Asymptotes: Vertical asymptote remains at $x = 0$; horizontal asymptote shifts from $y = 0$ up to $y = 2$

Coordinate Point Transformations

• Single Point Reflection Example:

  • Given point $(3, 6)$ on $y = f(x)$.

  • Find corresponding point on $y = f(-x)$.

  • Transformation rule: Inside negative reflects across the y-axis, multiplying x-coordinates by $-1$

  • Resulting Point: $(-3, 6)$

• Single Point Compression Example:

  • Given point $(4, 2)$ on $y = f(x)$.

  • Find corresponding point on $y = f(2x)$.

  • Transformation rule: Inside factor $2$ causes a horizontal compression by factor $\frac{1}{2}$, multiplying x-coordinates by $\frac{1}{2}$

  • Calculation: $x = 4 \times \frac{1}{2} = 2$

  • Resulting Point: $(2, 2)$

• X-Intercept Transformation Example:

  • Given x-intercepts at $(-8, 0)$ and $(1, 0)$ for generic function $y = f(x)$.

  • Problem A: Find x-intercepts for $y = f(x + 4)$

  • Rule: Shift left 4 units ($x - 4$)

  • Result: $(-8 - 4, 0) = (-12, 0)$ and $(1 - 4, 0) = (-3, 0)$

  • Problem B: Find x-intercepts for $y = f(x - 3)$

  • Rule: Shift right 3 units ($x + 3$)

  • Result: $(-8 + 3, 0) = (-5, 0)$ and $(1 + 3, 0) = (4, 0)$

  • Problem C: Find x-intercepts for $y = 2f(x)$

  • Rule: Vertical stretch by factor 2 (multiply y-values by 2)

  • Result: $y = 0 \times 2 = 0 \implies$ intercepts remain unchanged at $(-8, 0)$ and $(1, 0)$

  • Problem D: Find x-intercepts for $y = f(-x)$

  • Rule: Reflection across y-axis (multiply x-values by $-1$

  • Result: $(-(-8), 0) = (8, 0)$ and $(-(1), 0) = (-1, 0)$

Deriving Equation from Sequential Order Instructions

• Sequential Task: Determine final equation for base graph $y = \sqrt{x}$ following three ordered operations:

  1. Move UP 2 units:

  • Resulting expression: $y = \sqrt{x} + 2$

  1. Reflect across Y-Axis:

  • Substitute $-x$ for $x$

  • Resulting expression: $y = \sqrt{-x} + 2$

  1. Move LEFT 3 units:

  • Substitute $(x + 3)$ for $x$

  • Resulting expression: $y = \sqrt{-(x + 3)} + 2 = \sqrt{-x - 3} + 2$

    • Final Algebraic Equation: $y = \sqrt{-x - 3} + 2$