Unit 10 Function Transformations and Review Study Guide

Identification of Basic Function Graphs
- Function (1) $y = |x|$ : Represented by a V-shaped graph centered at the origin.
- Function (2) $y = \sqrt{x}$ : Represented by a curve starting at the origin and moving into the first quadrant.
- Function (3) $y = x^2$ : Represented by a U-shaped parabola.
- Function (4) $y = 2x$ : Represented by a straight line passing through the origin with a slope of 2.
Exponential Growth Characteristics
- The function $y = b^x$ where b > 1 represents exponential growth.
- The graph must be increasing, remain above the x-axis, and have a y-intercept at $(0, 1)$ .

Division by Zero Limitations
- In the function $f(x) = \frac{1}{x - 12}$ , the function cannot be evaluated when the denominator equals zero.
- Calculation: $x - 12 = 0 \implies x = 12$ .
- Therefore, the function cannot be evaluated at $x = 12$ .
Evaluating Transformed Functions from Graphs
- Given a graph of $f(x)$ and a second definition $g(x) = f(x) - 9$ , the value of $g(6)$ is found by determining $f(6)$ and subtracting 9.
- Procedure:
  - Identify point on graph: At $x = 6$ , the height of $f(x)$ is 6 ( $f(6) = 6$ ).
  - Substitute into formula: $g(6) = 6 - 9$ .
  - Final Value: $g(6) = -3$ .

Shifting Absolute Value Functions
- The function $y = |x| - 4$ has a turning point at $(0, -4)$ .
- This represents a vertical shift downwards of 4 units from the parent function $y = |x|$ .
Shifting Quadratic Functions
- Formula: $y = (x - 4)^2 + 10$ .
- The coordinates of the turning point are $(h, k)$ , which in this case are $(4, 10)$ .
- The sign inside the parentheses is negative, reflecting a shift to the right.

Horizontal vs. Vertical Translation
- A shift of a graph 3 units to the left is represented by the formula $y = |x + 3|$ .
- To determine the value of $k$ in $g(x) = f(x + k)$ , observe the horizontal movement from the parent function. If the graph of $g(x)$ is shifted 5 units to the right of $f(x)$ , then $k = -5$ .
Composite Transformations and Formula Substitutions
- If $h(x) = 5x - 7$ and $g(x) = h(x - 3)$ , the new formula is derived by substitution:
  - $g(x) = 5(x - 3) - 7$
  - Distribute: $g(x) = 5x - 15 - 7$
  - Simplify: $g(x) = 5x - 22$
Point-Specific Transformations
- If a function $f(x)$ passes through the point $(-3, 11)$ and there exists a function $g(x) = f(x + 2)$ , the graph of $g(x)$ is a horizontal shift of $f(x)$ by 2 units to the left.
- New x-coordinate calculation: $-3 - 2 = -5$ .
- New point: $(-5, 11)$ .
Reflections over the x-axis
- A reflection of the function $f(x) = -3x + 9$ across the x-axis is modeled by the equation $y = -f(x)$ .
- Calculation: $y = -(-3x + 9) \implies y = 3x - 9$ .
Vertical Scaling (Dilation)
- Given $f(x) = x^2 - 8$ and its transformation $g(x) = k \cdot f(x)$ .
- Observation from graph: The vertex of $f(x)$ is at $(0, -8)$ and the vertex of $g(x)$ is at $(0, -4)$ .
- Solve for $k$ : $-4 = k(-8) \implies k = 0.5$ or $k = \frac{1}{2}$ .

Algorithm for Finding Roots
1. Given $f(x) = x^2 - 2x - 15$ .
2. Define $h(x) = f(x) + 7 \implies h(x) = x^2 - 2x - 15 + 7 = x^2 - 2x - 8$ .
3. Factor the quadratic: $(x - 4)(x + 2)$ .
4. Solve for zero: $x - 4 = 0$ or $x + 2 = 0$ .
5. Roots: $x = 4$ and $x = -2$ . The set of zeros is $\{-2, 4\}$ .

Case Study: $f(x) = x^2 - 6x + 3$
- Transformation: If $g(x) = f(x) + 5$ , the formula becomes $g(x) = x^2 - 6x + 8$ .
- Solutions: The equation $g(x) = 0$ has 2 solutions because it crosses the x-axis twice.
  - Factoring $x^2 - 6x + 8 = 0 \implies (x - 4)(x - 2) = 0$ , resulting in $x = 2$ and $x = 4$ .
- Range: The vertex of $f(x)$ was at $(-3, -6)$ . Adding 5 shifting it upward puts the vertex at $(-3, -1)$ .
  - Range of $g(x)$ : $y \ge -1$ .
Defining Parabolas in Vertex Form
- Formula: $y = a(x + h)^2 + k$ .
- Example with turning point $(4, 4)$ and y-intercept of 16:
  - Identify constants: $h = -4$ and $k = 4$ , so $y = a(x - 4)^2 + 4$ .
  - Solve for $a$ using $(0, 16)$ : $16 = a(0 - 4)^2 + 4$
  - $16 = 16a + 4 \implies 12 = 16a \implies a = 0.75$ or $a = \frac{12}{16}$ .
  - Final Equation: $y = 0.75(x - 4)^2 + 4$ .

Toy Rocket Trajectory Analysis
- A toy rocket is fired from a height of $5\,ft$ above the ground.
- Parameters:
  - Maximum height ( $k$ ): $85\,ft$ .
  - Horizontal distance at max height ( $h$ ): $40\,ft$ .
  - Starting point (y-intercept): $(0, 5)$ .
- Solving for Scale Factor $a$ :
  - Use Vertex Form: $y = a(x - 40)^2 + 85$ .
  - Substitute intercept: $5 = a(0 - 40)^2 + 85 \implies 5 = 1600a + 85$ .
  - Simplify: $-80 = 1600a \implies a = -\frac{80}{1600} = -0.05$ .
  - Model Equation: $y = -0.05(x - 40)^2 + 85$ .
- Interval Determination:
  - To find over what distance the rocket is at or above $60\,ft$ :
  - Set inequality: $-0.05(x - 40)^2 + 85 \ge 60$ .
  - Subtract 85: $-0.05(x - 40)^2 \ge -25$ .
  - Divide by $-0.05$ (flip sign): $(x - 40)^2 \le 500$ .
  - Solve for $x$ : $x - 40 \le \pm \sqrt{500} \approx \pm 22.36$ .
  - Result: $x \approx 17.6\,ft$ to $62.4\,ft$ .