SAT Math Lesson Plan 19: Functions and Their Properties

Lesson Summary

In this lesson, we explored essential concepts related to functions, including function notation, domain/range, and transformations. Understanding these concepts is crucial as they are frequently tested on the SAT, specifically in the Heart of Algebra and Passport to Advanced Math sections. In fact, approximately 20% of the questions on the SAT Math section involve functions, making mastery of these topics essential for success. By the end of this lesson, you will be able to interpret function notation, determine the domain and range, and apply transformations to functions both algebraically and graphically.

This 45-minute lesson includes warm-up activities, a detailed Teach section, a Review with additional practice problems, and a quiz to reinforce the concepts learned. By the end of the lesson, students will be confident in their ability to analyze and solve function-related questions on the SAT.

Lesson Objectives

Understand function notation and evaluate functions for specific values of $x$ , e.g., $f(x) = 2x + 3$ .
Identify and determine the domain and range of functions.
Apply transformations (shifts, reflections, stretches/compressions) to functions.
Solve rational equations involving functions and their properties.
Find the zeros (roots) of a function by solving for $x$ when $f(x) = 0$ .

Common Core Standards

CCSS.MATH.CONTENT.HSA.SSE.A.1: Interpret expressions that represent a quantity in terms of its context.
CCSS.MATH.CONTENT.HSF.IF.C.7: Analyze functions using different representations.
CCSS.MATH.CONTENT.HSF.IF.C.8: Write a function in different forms to interpret various function properties.
CCSS.MATH.CONTENT.HSF.BF.B.3: Identify and analyze transformations of functions.

Prerequisite Skills

Basic understanding of algebraic operations (addition, subtraction, multiplication, division).
Familiarity with solving linear equations and graphing linear functions.
Ability to work with exponents and radicals.
Previous exposure to function notation and graphing of basic functions (e.g., linear, quadratic).

Key Vocabulary

Function: A relation between a set of inputs and a set of possible outputs where each input is related to exactly one output.
Function Notation: A way of writing functions in the form $f(x)$ where $f$ is the function and $x$ is the variable.
Evaluate a Function: To find the value of a function for a given value of $x$ by substituting that value into the function.
Domain: The set of all possible input values (usually $x$ -values) for which a function is defined.
Range: The set of all possible output values (usually $y$ -values) that a function can produce.
Transformation: A change in the position, size, or orientation of a graph, such as shifts, reflections, or stretches/compressions.
Vertical Shift: A transformation that moves the graph of a function up or down.
Horizontal Shift: A transformation that moves the graph of a function left or right.
Reflection: A transformation that flips the graph of a function across a line (e.g., the x-axis or y-axis).
Stretch/Compression: A transformation that either stretches or compresses the graph of a function vertically or horizontally.
Zero of a Function (Root): The value of $x$ where the function $f(x) = 0$ , which corresponds to the points where the graph intersects the x-axis.
Asymptote: A line that a graph approaches but never actually reaches. Asymptotes can be vertical, horizontal, or slant (oblique) and are important for understanding the behavior of rational functions.

Warm Up

Choose from one or more of these activities to activate prior knowledge and prepare students for this lesson.

Activity 1: Function Notation

Example: If $f(x) = 2x + 3$ , find $f(-1), f(0), f(4)$ .

For f(−1):
- Substitute $x = -1$ : $f(-1) = 2(-1) + 3 = -2 + 3 = 1$
For f(0):
- Substitute $x = 0$ : $f(0) = 2(0) + 3 = 0 + 3 = 3$
For f(4):
- Substitute $x = 4$ : $f(4) = 2(4) + 3 = 8 + 3 = 11$

Self-Study Tip: When given a function notation problem, simply substitute the provided value for $x$ and simplify.

Activity 2: Domain and Range

Example: Determine the domain and range of $g(x) = \sqrt{x - 2}$ .

Domain: The expression inside the square root, $x - 2$ , must be greater than or equal to zero. So, $x \geq 2$ . Therefore, the domain is $[2, \infty)$ .
Range: Since a square root cannot be negative, the range is $[0, \infty)$ .

Self-Study Tip: For square roots, ensure the radicand is non-negative to find the domain. The range is based on the behavior of the square root function.

Teach

In this section, we will explore important concepts related to functions, including function notation, domain/range, and transformations. These concepts are foundational for solving SAT Math problems related to functions and their graphs. You will learn how to evaluate functions at specific values, determine their domain and range, and understand how transformations (such as shifts and reflections) affect the graph of a function. Additionally, you will be able to apply these concepts to both algebraic equations and graph interpretations.

Key Concepts Covered:

Function Notation: Understanding how to evaluate a function at different values of $x$ , e.g., $f(x) = 2x + 3$ .
Domain and Range: Finding the domain and range of functions such as $g(x) = \sqrt{x - 2}$ , including identifying restrictions on $x$ .
Transformations: Learning how horizontal and vertical shifts, reflections, and stretches/compressions affect the graph of a function.
Solving Rational Equations: Using function notation to solve rational equations involving function variables.
Identifying Graphical Behavior: Interpreting function transformations from both algebraic and graphical perspectives.

We will break down these concepts through detailed examples and guided exercises to help you develop a deeper understanding of each. Let's begin by reviewing function notation.

Example 1: Function Notation

Problem: Find $f(-2)$ for $f(x) = x^2 - 3x + 5$ .

Step 1: Substitute $x = -2$ into the function

$f(-2) = (-2)^2 - 3(-2) + 5$ $f(-2) = 4 + 6 + 5$ $f(-2) = 15$

Final Answer: $f(-2) = 15$

Example 2: Identifying Domain and Range

Problem: Find the domain and range of the function $f(x) = \frac{1}{x-3}$ .

Step 1: Find the domain

The domain of a function includes all possible input values (x-values) for which the function is defined. For rational functions, the function is undefined where the denominator is zero.

Set the denominator equal to zero and solve for $x$ :

$x - 3 = 0$ $x = 3$

Since the function is undefined at $x = 3$ , the domain is all real numbers except $x = 3$ .

Domain: $(-\infty, 3) \cup (3, \infty)$

Step 2: Find the range

The range is the set of possible output values (y-values). For rational functions where the denominator cannot be zero, the range will often depend on the horizontal asymptote. In this case, the function has a vertical asymptote at $x = 3$ , but no horizontal asymptote as the function tends toward infinity as $x$ approaches 3 from either direction.

Range: $(-\infty, 0) \cup (0, \infty)$ (since the function never crosses the horizontal line $y = 0$ )

Step 3: Conclusion

The domain of the function is $(-\infty, 3) \cup (3, \infty)$ and the range is $(-\infty, 0) \cup (0, \infty)$ .

Self-Study Tip: When finding the domain and range, check for values that would cause the function to be undefined (such as division by zero) and consider the behavior of the graph, such as asymptotes, to help determine the range.

Example 3: Transformations of Functions

Problem: Describe the transformations that occur when you transform $f(x) = x^2$ to $g(x) = (x + 3)^2 - 4$ .

Step 1: Compare the transformations

The equation $g(x) = (x + 3)^2 - 4$ represents a transformation of the parent function $f(x) = x^2$ .
The term $(x + 3)$ shifts the graph 3 units to the left.
The term $-4$ shifts the graph 4 units downward.

Step 2: Visualize the transformation

Parent Function:

$f(x) = x^2$

Transformed Function:

$g(x) = (x + 3)^2 - 4$

Self-Study Tip: Use the Desmos Graphing Calculator to experiment with graphing transformations of various functions to better understand how shifts and reflections affect the graphs.

Example 4: Zeros of a Function

Problem: Find the zeros of the polynomial function $f(x) = x^2 - 5x + 6$ .

Step 1: Set the function equal to zero

To find the zeros, we set $f(x) = 0$ : $0 = x^2 - 5x + 6$

Step 2: Factor the quadratic

We need to find two numbers whose product is 6 and whose sum is -5. The numbers are -2 and -3: $0 = (x - 2)(x - 3)$

Step 3: Solve for $x$

Set each factor equal to zero: $x - 2 = 0 \quad \text{or} \quad x - 3 = 0$ Solving these gives: $x = 2 \quad \text{or} \quad x = 3$

Final Answer: The zeros of the function are $x = 2$ and $x = 3$ , which means the graph of the function intersects the x-axis at these points. These are the zeros of the function or roots.

Self-Study Tip: To find the zeros of a function, set the function equal to zero and solve for $x$ . If the function is a quadratic, you can often factor it to find the zeros.

Example 5: Asymptotes

Problem: Find the vertical and horizontal asymptotes of the rational function:

$f(x) = \frac{3x^2 - 5x + 2}{x^2 - 4}$

Step 1: Find the vertical asymptote

The vertical asymptote occurs where the denominator is zero, and the numerator is non-zero at the same value. Set the denominator equal to zero:

$x^2 - 4 = 0$ Solve for $x$ : $x^2 = 4$ $x = \pm 2$

Thus, there are vertical asymptotes at $x = 2$ and $x = -2$ .

Step 2: Find the horizontal asymptote

For rational functions, the horizontal asymptote depends on the degrees of the numerator and denominator.

If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $y = 0$ .
If the degrees are equal, the horizontal asymptote is at $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$ .

In this case, both the numerator and denominator have degree 2 (the highest power of $x$ is 2 in both), so the horizontal asymptote is determined by the ratio of the leading coefficients:

$y = \frac{3}{1} = 3$

Thus, the horizontal asymptote is at $y = 3$ .