Algebra 2 Study Guide: Exponential Functions, Transformations, and Symmetry

Scenario Context: The population of a city is measured over time, starting from the year $1950$ . * Data Points: * The population in $1950$ (where $t = 0$ ) was $250,000$ . * The population in $1951$ (where $t = 1$ ) was $252,500$ . * Finding the Growth Rate: * To determine the growth factor, the population of the second year is divided by the population of the first year: * $\frac{252,500}{250,000} = 1.01$ * The Population Function: * The function $P(t)$ models the population in terms of $t$ , the number of years since $1950$ . * $P(t) = 250,000(1.01)^t$

Scenario Context: A substance undergoes decay with a specific half-life. * Parameters: * Half-life ( $h$ ): Approximately $14$ years. * Initial Quantity: $20$ \ units. * Function Formulation: * The function $F(t)$ models the remaining amount of the substance where $t$ is measured in years. * $F(t) = 20\left(\frac{1}{2}\right)^{\frac{t}{14}}$ * Amount Remaining Over Time (Table): * At $t = 0$ years: $F(0) = 20\left(\frac{1}{2}\right)^{\frac{0}{14}} = 20$ * At $t = 14$ years: $F(14) = 20\left(\frac{1}{2}\right)^{\frac{14}{14}} = 20\left(\frac{1}{2}\right) = 10$ * At $t = 28$ years: $F(28) = 20\left(\frac{1}{2}\right)^{\frac{28}{14}} = 20\left(\frac{1}{2}\right)^2 = 20\left(\frac{1}{4}\right) = 5$

Problem 3 Analysis: Given a base graph defined by the function $y = f(x)$ . * Transformation Case (a): * Function: $g(x) = f(x) + 2$ * Effect: This represents a vertical shift. The entire graph of $f(x)$ is moved up by $2$ units on the y-axis. * Transformation Case (b): * Function: $h(x) = f(x + 2)$ * Effect: This represents a horizontal shift. The entire graph of $f(x)$ is moved to the left by $2$ units on the x-axis. * Comparison of $g$ and $h$ to $f$ : * $g(x)$ changes the output (y-values), while $h(x)$ changes the input (x-values) before the function is applied.

Core Definitions: * Even Functions: * Identity: A function is even if $f(x) = f(-x)$ . * Graphic Property: Reflecting the graph around the y-axis does not change the appearance of the graph. * Odd Functions: * Identity: A function is odd if $f(-x) = -f(x)$ . * Graphic Property: Reflecting the graph around both the y-axis and the x-axis does not change the appearance of the graph.

Example (a): $f(x) = 3x^2 + 3$ * Classification: Even. * Algebraic Proof: * Substitute $-x$ for $x$ : $f(-x) = 3(-x)^2 + 3$ * Simplify: $f(-x) = 3x^2 + 3$ * Conclusion: Since $f(-x) = f(x)$ , the function is Even.
Example (b): $f(x) = x^3 - 4x$ * Classification: Odd. * Algebraic Proof: * Substitute $-x$ for $x$ : $f(-x) = (-x)^3 - 4(-x)$ * Simplify: $f(-x) = -x^3 + 4x$ * Factor out $-1$ : $f(-x) = -(x^3 - 4x)$ * Conclusion: Since $f(-x) = -f(x)$ , the function is Odd.
Example (c): $f(x) = \frac{1}{x^2 + 1}$ * Classification: Even. * Explanation: The denominator contains an $x^2$ term, which ensures that substituting $-x$ for $x$ results in the same value ( $(-x)^2 = x^2$ ). Reflection around the y-axis does not change the graph.
Example (d): $f(x) = x^2 + x - 3$ * Classification: Neither. * Explanation: While the $x^2$ term is even, the $x$ term (linear) is odd. The combination of both even and odd powers of $x$ (without a structure that balances them) results in a function that is neither even nor odd.

Problem 7: Completing a table for an even function where $f(x) = f(-x)$ . * Input-Output Values: * If $x = 4$ , $f(x) = 2$ . Therefore, at $x = -4$ , $f(x) = 2$ . * If $x = 3$ , $f(x) = 1$ . Therefore, at $x = -3$ , $f(x) = 1$ . * If $x = 2$ , $f(x) = 8$ . Therefore, at $x = -2$ , $f(x) = 8$ . * If $x = 1$ , $f(x) = 10$ . Therefore, at $x = -1$ , $f(x) = 10$ . * At the center, $x = 0$ , $f(x) = -1$ . * Completed Data Points Table: * $x = -4, f(x) = 2$ * $x = -3, f(x) = 1$ * $x = -2, f(x) = 8$ * $x = -1, f(x) = 10$ * $x = 0, f(x) = -1$ * $x = 1, f(x) = 10$ * $x = 2, f(x) = 8$ * $x = 3, f(x) = 1$ * $x = 4, f(x) = 2$