What We Review

**Introduction**

### What Should You Know Going into AP® Precalculus?

You might be wondering, is AP® Precalculus hard? Well, AP® Precalculus is designed to prepare students for advanced mathematics courses, particularly calculus. Therefore, some might say that AP® Pre calc is hard. The course covers various topics that build on your previous mathematical knowledge. So, we can make sure to review topics from your mathematical background that are considered prerequisites. These prerequisites are listed in the AP® Precalculus CED (college and exam description). If you practice these foundational skills, you’ll be ready for the various topics in the AP® Precalculus syllabus and, ultimately, ready to tackle the AP® Precalculus exam.

### Proficiency vs. Familiarity: Understanding the Difference

In preparation for AP® Precalculus, you need proficiency in some skills and concepts. This means you should be able to execute these tasks accurately and efficiently. On the other hand, other areas require familiarity, where a basic understanding and recognition are sufficient. This guide will help you differentiate between these levels and ensure you are well-prepared for the course.

### Goals of This Guide

This guide aims to outline the key AP® Precalculus prerequisites that are mentioned in the AP® Precalculus CED. We will provide a clear understanding of the skills and concepts you need to master or be familiar with. By the end, you will know exactly what areas to focus on to succeed in your AP® Precalculus course. Additionally, we will provide examples and links to more practice on our website, Albert.io.

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**Proficiency with Linear and Quadratic Functions**

The first bullet point in the AP® Precalculus prerequisites to cover is linear and quadratic functions. To succeed in AP® Precalculus, you need a solid understanding of linear and quadratic functions. This includes proficiency in algebraic manipulation, solving equations, and solving inequalities.

### Algebraic Manipulation

Above all, you should be comfortable manipulating algebraic expressions involving linear and quadratic functions. This includes operations like combining like terms, distributing, and factoring.

#### Key Skills:

Combining Like Terms: Simplify expressions by adding or subtracting terms with the same variable raised to the same power.

Distributive Property: Apply a(b + c) = ab + ac to expand or factor expressions.

Factoring: Break down expressions into products of simpler factors.

Example: Simplify 3x^2 - 2x + 4 - (x^2 - 3x + 5).

#### Solution:

3x^2 - 2x + 4 - (x^2 - 3x + 5) = 3x^2 - 2x + 4 - x^2 + 3x - 5

Firstly, combine like terms:

= 2x^2 + x - 1

As can be seen, we first distribute the negative sign across the terms inside the parentheses. Then, combine like terms by adding or subtracting the coefficients of terms with the same power on the variables.

### Solving Equations

Secondly, being able to solve both linear and quadratic equations is crucial. This involves isolating the variable using inverse operations, factoring, or using the quadratic formula.

#### Key Skills:

Solving Linear Equations: Find the value of x in equations like 2x + 3 = 7.

Solving Quadratic Equations: Use factoring, completing the square, or the quadratic formula.

Example: Solve 2x^2 - 3x - 2 = 0 using the quadratic formula.

#### Solution:

Most important, recall the quadratic formula.

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute a = 2, b = -3, and c = -2:

x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{4} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}

Thus, we get two solutions:

x = 2 \quad \text{or} \quad x = -\frac{1}{2}

To sum up, we used the quadratic formula to find the values of the variable. Then, we substituted the coefficients into the formula, simplified the expression under the square root, and solved for the variable.

### Solving Inequalities

Finally, you should be able to solve inequalities and represent their solutions graphically or on a number line. This includes understanding both linear and quadratic inequalities.

#### Key Skills:

Linear Inequalities: Solve and graph solutions on a number line.

Quadratic Inequalities: Solve by finding critical points and testing intervals.

Example: Solve and graph x^2 - 4x + 3 > 0.

#### Solution:

Most important, factor the quadratic expression:

(x - 1)(x - 3) > 0

The critical points are x = 1 and x = 3. Subsequently, test the intervals around these points to find where the inequality holds:

x < 1 \quad \text{or} \quad x > 3

In any case, this means the solution set includes values of the variable that are less than 1 or greater than 3.

In conclusion, we factored the quadratic expression to find the critical points. Then, tested the intervals determined by these points to identify where the inequality is true. Finally, the solution is represented on a number line, showing the intervals where the quadratic expression is positive.

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**Proficiency with Solving Right Triangle Problems with Trigonometry**

Secondly, the next topic listed in the AP® Precalculus prerequisites is having a solid foundation in trigonometry, especially when solving problems involving right triangles. At the same time, this includes understanding basic trigonometric ratios and being able to apply them to solve right triangle problems.

### Basic Trigonometric Ratios

Basically, you need to be familiar with the primary trigonometric ratios: sine, cosine, and tangent. These ratios are used to relate the angles of a right triangle to the lengths of its sides.

#### Key Ratios:

Sine: \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}

Cosine: \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}

Tangent: \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}

Example: In a right triangle, if the angle \theta is 30 degrees, the opposite side is 3, and the hypotenuse is 6, find \sin(\theta), \cos(\theta), and \tan(\theta).

#### Solution:

Firstly, calculate the sine:

\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{6} = \frac{1}{2}

Then, calculate the cosine:

\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} (using the Pythagorean identity for a 30-60-90 triangle)

Finally, calculate the tangent:

\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

As shown above, we used the definitions of the trigonometric ratios to find the sine, cosine, and tangent of the given angle.

### Solving Right Triangle Problems

Additionally, being able to apply trigonometric ratios to solve problems involving right triangles is crucial. This includes finding missing sides or angles.

#### Key Skills:

Using Trigonometric Ratios: Apply sine, cosine, and tangent to find missing sides or angles in right triangles.

Pythagorean Theorem: Use a^2 + b^2 = c^2 to find the lengths of sides in right triangles.

Example: In a right triangle, the hypotenuse is 10, and one of the angles is 30 degrees. Find the lengths of the other two sides.

#### Solution:

Firstly, use the sine ratio to find the length of the side opposite the 30-degree angle:

\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}

\frac{1}{2} = \frac{\text{opposite}}{10}

Then, multiply both sides by 10 to solve for the opposite side:

\text{opposite} = 10 \times \frac{1}{2} = 5

Following, use the cosine ratio to find the length of the adjacent side:

\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}}

\frac{\sqrt{3}}{2} = \frac{\text{adjacent}}{10}

Finally, multiply both sides by 10 to solve for the adjacent side:

\text{adjacent} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}

Therefore, the lengths of the other two sides are 5 and 5\sqrt{3}.

In essence, we used the sine and cosine ratios to find the lengths of the sides in the right triangle.

**Proficiency with Solving Systems of Equations in Two and Three Variables**

Thirdly, the AP® Precalculus prerequisites list mentions proficiency in solving systems of equations as vital. You need to be able to solve systems of equations in two and three variables. This involves using various methods such as substitution, elimination, and matrix operations.

### Solving Systems in Two Variables

Undoubtedly, you should be able to solve systems of linear equations in two variables. This can be done using methods like substitution and elimination.

#### Key Methods:

Substitution Method: Solve one equation for one variable and substitute this expression into the other equation.

Elimination Method: Add or subtract equations to eliminate one of the variables, making it easier to solve for the remaining variable.

Example: Solve the system of equations:

2x + y = 10

3x - y = 5

#### Solution:

Firstly, add the two equations to eliminate one of the variables:

2x + y + 3x - y = 10 + 5

5x = 15

Then, solve for the other variable:

x = \frac{15}{5} = 3

Lastly, substitute that value into the first equation to find the other variable:

2(3) + y = 10

6 + y = 10

y = 10 - 6 = 4

Therefore, the solution to the system has been found because we solved for both variables.

To sum up, we used the elimination method to eliminate one variable, making it easier to solve for the other variable. Finally, we substituted the value back into one of the original equations to find the original variable.

### Solving Systems in Three Variables

Next, solving systems of equations in three variables typically requires using methods such as substitution, elimination, or matrix operations.

#### Key Methods:

Substitution Method: Solve one equation for one variable, then substitute this expression into the other equations.

Elimination Method: Use elimination to reduce the system to two equations in two variables, then solve.

Matrix Operations: Use matrices and row reduction techniques to solve the system.

Example: Solve the system of equations:

x + y + z = 6

2x - y + 3z = 14

-x + 2y - z = -2

#### Solution:

Firstly, use the elimination method to eliminate the third variable. To rephrase it, add the first and second equations:

(x + y + z) + (2x - y + 3z) = 6 + 14

3x + 4z = 20

Then, add the first and third equations:

(x + y + z) + (-x + 2y - z) = 6 - 2

3y = 4

Thirdly, solve for the second variable:

y = \frac{4}{3}

Then, substitute this expression into one of the original equations to find the other two missing variables. In particular, using the first equation:

x + \frac{4}{3} + z = 6

x + z = 6 - \frac{4}{3} = \frac{14}{3}

Using the equation 3x + 4z = 20 :

x = \frac{2}{3}

Finally, substitute the first two variables into one of the equations to solve for the third variable:

x + \frac{4}{3} + z = 6

z = 4

Thus, the solution to the system has been found. In coordinate form, it i s (x, y, z) = (\frac{2}{3}, \frac{4}{3}, 4).

Here, we used elimination to reduce the system to two equations in two variables, solved for one variable, and then used substitution to find the remaining variables.

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**Familiarity with Piecewise-Defined Functions**

So far, we have only looked at topics that the list of AP® Precalculus prerequisites recommends proficiency. From this point on, the AP® Precalculus prerequisites list mentions that you should only be familiar with the rest of the topics. Firstly, you should understand piecewise-defined functions. These functions are defined by different expressions depending on the input value. You need to be able to interpret, evaluate, and graph these functions.

Definition and Examples

A piecewise-defined function is a function that is defined by multiple sub-functions, each of which applies to a specific interval of the domain.

#### Example: Consider the piecewise-defined function:

x + 2 \text{ if } x < 0

x^2 \text{ if } 0 \leq x < 2

3x - 1 \text{ if } x \geq 2

In short, this function has three different expressions based on the value of the variable.

### Evaluating Piecewise-Defined Functions:

Unlike a normal function, to evaluate a piecewise-defined function, determine which piece of the function to use based on the input value.

#### Example: Evaluate f(x) for x = -1, x = 1, and x = 3 for the given function f(x).

Solution:

1. For x = -1:

Since this value is less than zero, use the first piece:.

f(-1) = -1 + 2 = 1

2. For x = 1:

Since this is between zero and two, use the second piece:

f(1) = 1^2 = 1

3. For x = 3:

Since this value is greater than two, use the third piece:

f(3) = 3(3) - 1 = 9 - 1 = 8

In summary, we evaluated the function by selecting the appropriate piece based on the value of the variable.

### Graphical Representation

To graph a piecewise-defined function, graph each piece of the function over its specified interval. Pay attention to whether the endpoints of the intervals are included (closed dots) or excluded (open dots).

Example: Graph the piecewise-defined function:

g(x) = 2x + 1 \text{ if } x < 1

x^2 \text{ if } 1 \leq x \leq 3

-2x + 5 \text{ if } x > 3

#### Solution:

1. For x < 1:

Firstly, graph the linear equation with a slope of 2 and y-intercept of 1. Most importantly, stop when the x values reach 1.

2. For 1 \leq x \leq 3:

Secondly, graph the parabola between the x values of 1 and 3.

3. For x > 3:

Finally, graph the linear equation with a slope of -2 and a y-intercept of 5. However, only do so for x values greater than 3.

Once you’ve done this, use an open dot for any interval endpoints that are less than or greater than. Then, use a closed dot for any interval endpoints that are less than or equal to or greater than or equal to.

In detail, we graphed each piece of the function within its specified interval, ensuring we accurately represented the endpoints.

**Familiarity with Exponential Functions and Rules for Exponents**

According to the list of AP® Precalculus prerequisites, you should be familiar with exponential functions. These functions are crucial for modeling growth and decay in various contexts. You need to understand their properties and be able to apply the rules for exponents.

### Definition and Properties

An exponential function is a function of the form f(x) = a \cdot b^x, where a is a constant, b is the base, and x is the exponent.

#### Key Properties:

Base Greater Than 1: For b > 1, the function models exponential growth.

Base Between 0 and 1: For 0 < b < 1, the function models exponential decay.

Horizontal Asymptote: The line y = 0 is a horizontal asymptote for exponential functions.

Example: Consider the exponential function f(x) = 2 \cdot 3^x. Describe its properties.

#### Solution:

The base b = 3 > 1, so it models exponential growth.

As x increases, f(x) grows rapidly.

The horizontal asymptote is y = 0.

### Rules for Exponents

Moreover, to work with exponential functions effectively, you need to be familiar with the rules for exponents.

#### Key Rules:

Product of Powers: a^m \cdot a^n = a^{m+n}

Quotient of Powers: \frac{a^m}{a^n} = a^{m-n}

Power of a Power: latex^n = a^{mn}[/latex]

Negative Exponent: a^{-n} = \frac{1}{a^n}

Zero Exponent: a^0 = 1 (for a \neq 0)

Example: Simplify the expression \frac{2^5 \cdot 2^{-3}}{2^2}.

#### Solution:

First, apply the product of powers rule:

2^5 \cdot 2^{-3} = 2^{5-3} = 2^2

Next, apply the quotient of powers rule:

\frac{2^2}{2^2} = 2^{2-2} = 2^0 = 1

Here, we use the product of powers and quotient of powers rules to simplify the expression.

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### Graphing Exponential Functions

Graphing exponential functions involves plotting points and understanding the shape of the graph based on the base.

Example: Graph the exponential function f(x) = 2 \cdot 3^x.

#### Solution:

Plot Points: Calculate and plot points for several values of x.

Firstly, for x = -2: f(-2) = 2 \cdot 3^{-2} = \frac{2}{9}

Secondly, for x = -1: f(-1) = 2 \cdot 3^{-1} = \frac{2}{3}

Thirdly, for x = 0: f(0) = 2 \cdot 3^0 = 2

Furthermore, x = 1: f(1) = 2 \cdot 3^1 = 6

Finally, x = 2: f(2) = 2 \cdot 3^2 = 18

Draw the Curve: Connect the points smoothly, showing the rapid growth as x increases.

Horizontal Asymptote: Indicate the horizontal asymptote y = 0 on the graph.

By plotting these points and connecting them, we can visualize the exponential growth of the function.

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**Familiarity with Radicals (Square Roots and Cube Roots)**

Following exponents, the AP® Precalculus prerequisites list mentions that you should be familiar with radicals, including square roots and cube roots. In any case, understanding how to simplify and manipulate radical expressions is essential.

Square Roots and Cube Roots

Radicals are expressions that include a root, such as a square root or cube root. The square root of a number a is a number b such that b^2 = a. Similarly, the cube root of a is a number b such that b^3 = a.

#### Key Concepts:

Square Root: \sqrt{a} is a number that, when squared, gives a.

Cube Root: \sqrt[3]{a} is a number that, when cubed, gives a.

Example: Simplify \sqrt{36} and \sqrt[3]{27}.

#### Solution:

\sqrt{36} = 6 because 6^2 = 36.

\sqrt[3]{27} = 3 because 3^3 = 27.

In essence, we identified the given numbers’ square root and cube root.

Simplifying Radical Expressions

Simplifying radicals involves expressing the radical in its simplest form. This can include factoring out perfect squares or cubes.

Example: Simplify \sqrt{50}.

#### Solution:

First, factor 50 into its prime factors: 50 = 2 \cdot 5^2.

Then, simplify by taking the square root of the perfect square:

\sqrt{50} = \sqrt{2 \cdot 5^2} = 5\sqrt{2}

Here, we factor 50 and simplify the square root by taking out the perfect square.

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Operations with Radicals

To work with radical expressions, you need to be able to add, subtract, multiply, and divide them.

#### Key Operations:

Addition/Subtraction: Combine like radicals (same radicand).

Multiplication: Multiply the coefficients and radicands separately.

Division: Divide the coefficients and radicands separately.

Example: Simplify latex(3\sqrt{6})[/latex].

#### Solution:

First, multiply the coefficients:

2 \cdot 3 = 6

Next, multiply the radicands:

\sqrt{3} \cdot \sqrt{6} = \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}

Combine the results:

(3\sqrt{6}) = 6 \cdot 3\sqrt{2} = 18\sqrt{2}

Here, we multiply the coefficients and radicands separately then simplify the resulting radical.

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**Familiarity with Complex Numbers**

Furthermore, the AP® Precalculus prerequisites list mentions familiarity with complex numbers. Complex numbers extend the real number system and are essential in various advanced mathematical contexts.

### Definition and Operations

A complex number is a number of the form a + bi, where a and b are real numbers, and i is the imaginary unit with the property that i^2 = -1.

#### Key Concepts:

Real Part: a in a + bi.

Imaginary Part: b in a + bi.

Imaginary Unit: i, where i^2 = -1.

Example: Consider the complex number 3 + 4i.

The real part is 3.

The imaginary part is 4i.

### Operations with Complex Numbers

In addition, you need to know how to perform basic operations with complex numbers, including addition, subtraction, multiplication, and division.

#### Addition and Subtraction:

In general, to add or subtract complex numbers, combine the real parts and the imaginary parts separately.

Example: Add (3 + 2i) and (1 + 4i).

Solution:

(3 + 2i) + (1 + 4i) = (3 + 1) + (2i + 4i) = 4 + 6i

Here, we add the real parts 3 and 1, and the imaginary parts 2i and 4i.

#### Multiplication:

Specifically, to multiply complex numbers, use the distributive property (FOIL method for binomials).

Example: Multiply (2 + 3i) and (1 - 4i).

Solution:

(2 + 3i)(1 - 4i) = 2(1) + 2(-4i) + 3i(1) + 3i(-4i)

= 2 - 8i + 3i - 12i^2

Since i^2 = -1, -12i^2 = 12:

= 2 - 8i + 3i + 12 = 14 - 5i

In order to multiply the real parts and the imaginary parts, we simplified using the fact that squaring the imaginary unit equals -1.

#### Division:

To divide complex numbers, multiply the numerator and the denominator by the conjugate of the denominator and then simplify.

Example: Divide \frac{3 + 4i}{1 - 2i}.

Solution:

Multiply the numerator and the denominator by the conjugate of the denominator (1 + 2i):

\dfrac{(3 + 4i)(1 + 2i)}{(1 - 2i)(1 + 2i)}

Secondly, simplify the numerator using the distributive property:

(3 + 4i)(1 + 2i) = 3(1) + 3(2i) + 4i(1) + 4i(2i)

= 3 + 6i + 4i + 8i^2 = 3 + 10i + 8(-1) = 3 + 10i - 8 = -5 + 10i

Then, simplify the denominator using the difference of squares formula:

(1 - 2i)(1 + 2i) = 1 - (2i)^2 = 1 - 4(-1) = 1 + 4 = 5

Finally, combine the results:

\frac{-5 + 10i}{5} = -1 + 2i

Here, we multiply the numerator and the denominator by the conjugate of the denominator, simplify both the numerator and the denominator, and then divide to get the final result.

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**Familiarity with Multiple Representations of Functions**

Lastly, as indicated in the AP® Precalculus prerequisites list, it is crucial to understand and communicate functions in various forms. You need to be familiar with graphical, numerical, analytical, and verbal representations of functions.

### Graphical Representation

Graphical representation involves plotting the function on a coordinate plane. This visual representation helps to understand the behavior and properties of the function, such as intercepts, maxima and minima, and asymptotes.

Example: Graph the function f(x) = x^2 - 4x + 3.

Solution:

Find the intercepts:

Y-intercept: Set x = 0:

f(0) = 0^2 - 4(0) + 3 = 3

X-intercepts: Set f(x) = 0:

x^2 - 4x + 3 = 0

Factoring the quadratic:

(x - 1)(x - 3) = 0

So, x = 1 and x = 3

Then, plot the points on the coordinate plane.

Finally, draw the parabola passing through these points, opening upwards.

Here, we graph the function by finding and plotting the intercepts and then sketching the curve.

### Numerical Representation

Numerical representation involves using tables of values to represent the function. This is useful for understanding how the function behaves at specific points.

Example: Create a table of values for f(x) = x^2 - 4x + 3.

Solution:

Here, we calculate the function’s value at several points and organize them in a table.

### Analytical Representation

Analytical representation involves expressing the function in a symbolic form, such as an equation or an expression. This form allows for algebraic manipulation and deeper analysis.

Example: Consider the function f(x) = x^2 - 4x + 3.

We can analyze the function by factoring:

f(x) = (x - 1)(x - 3)

Here, the analytical form of the function allows us to easily find the x-intercepts and understand the behavior of the function.

### Verbal Representation

Verbal representation involves describing the function and its properties in words. This form is useful for communicating the function’s behavior and characteristics.

Example: Describe the function f(x) = x^2 - 4x + 3 verbally.

Solution:

The function f(x) = x^2 - 4x + 3 is a quadratic function that opens upwards. It has x-intercepts at x = 1 and x = 3, and a y-intercept at y = 3. The vertex of the parabola occurs at x = 2, and the minimum value of the function is f(2) = -1.

Here, we describe the function’s key properties and behavior in words.

**Conclusion**

Understanding the AP® Precalculus prerequisites is essential for success in the course. By mastering the foundational skills and concepts outlined in this guide, you will be well-prepared for the AP® Precalculus curriculum and, ultimately, the AP® Precalculus exam. Keep practicing, stay curious, and don’t hesitate to seek additional resources and support as you embark on this exciting mathematical journey.

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