College Algebra: Core Concepts and Practice for Test 1

Core Concepts in College Algebra

Domain of Functions

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

• Square Root Functions: The expression inside the square root must be greater than or equal to zero.

• Rational Functions: The denominator cannot be zero.

Example: For $f(x) = \sqrt{x-5}$, the domain is $x \geq 5$.

Distance Formula

The distance formula calculates the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane.

• Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Example: The distance between $(2,3)$ and $(6,11)$ is $\sqrt{(6-2)^2 + (11-3)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$.

Midpoint Formula

The midpoint of a segment connecting $(x_1, y_1)$ and $(x_2, y_2)$ is the point exactly halfway between them.

• Formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Example: The midpoint of $(-4,5)$ and $(2,-1)$ is $\left( \frac{-4+2}{2}, \frac{5+(-1)}{2} \right) = (-1, 2)$.

Slope of a Line

The slope measures the steepness of a line connecting two points.

• Formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$

• Parallel lines have the same slope.

• Perpendicular lines have slopes that are negative reciprocals: $m_1 \cdot m_2 = -1$.

Example: The slope between $(-2,5)$ and $(4,-1)$ is $m = \frac{-1-5}{4-(-2)} = \frac{-6}{6} = -1$.

Equations of Lines

There are two common forms for the equation of a line:

• Slope-Intercept Form: $y = mx + b$

• Point-Slope Form: $y - y_1 = m(x - x_1)$

Example: A line with slope $3$passing through$(1,4)$:$y - 4 = 3(x - 1)$or$y = 3x + 1$.

Average Rate of Change

The average rate of change of a function $f(x)$ from $x = a$ to $x = b$ is the change in $f(x)$ divided by the change in $x$.

• Formula: $\frac{f(b) - f(a)}{b - a}$

Example: For $f(x) = x^2$ from $x=1$ to $x=3$: $\frac{3^2 - 1^2}{3-1} = \frac{9-1}{2} = 4$.

Difference Quotient

The difference quotient is used to compute the average rate of change over an interval of length $h$ and is foundational for calculus.

• Formula: $\frac{f(x+h) - f(x)}{h}$

Circle: Standard Form

The standard form of a circle with center $(h, k)$ and radius $r$ is:

• Equation: $(x - h)^2 + (y - k)^2 = r^2$

Example: $(x - 2)^2 + (y + 1)^2 = 9$ has center $(2, -1)$ and radius $3$.

Absolute Value Equations

Solving equations of the form $|x - a| = b$ involves considering two cases:

• $x - a = b$

• $x - a = -b$

Example: $|x - 3| = 7$ gives $x = 10$ or $x = -4$.

Inequalities and Multiplying/Dividing by Negatives

When solving inequalities, if you multiply or divide both sides by a negative number, reverse the direction of the inequality sign.

Study and Review Strategies

Crash Review Plan (30 Minutes)

• Minutes 1–5: Memorize distance, midpoint, and slope formulas.

• Minutes 6–10: Practice writing equations of lines (parallel and perpendicular).

• Minutes 11–15: Solve two domain problems (square root and rational functions).

• Minutes 16–20: Solve two absolute value problems.

• Minutes 21–25: Solve one circle equation problem.

• Minutes 26–30: Review mistakes and rewrite formulas.

Practice Problems and Solutions

Practice Problems

1. Find the domain of $f(x) = \sqrt{x - 5}$.

2. Find the distance between $(2,3)$ and $(6,11)$.

3. Find the midpoint of $(-4,5)$ and $(2,-1)$.

4. Write the equation of a line with slope $3$passing through$(1,4)$.

5. Solve $|x - 3| = 7$.

6. Find the center and radius of $(x - 2)^2 + (y + 1)^2 = 9$.

Answer Key

1. Domain: $x \geq 5$

2. Distance: $\sqrt{80} = 4\sqrt{5}$

3. Midpoint: $(-1, 2)$

4. Equation: $y - 4 = 3(x - 1)$ or $y = 3x + 1$

5. $x = 10$ or $x = -4$

6. Center: $(2, -1)$, Radius: $3$

Mini Mock Quiz

1. Find the slope between $(-2,5)$ and $(4,-1)$.

2. Write the equation of a line parallel to $y = 2x - 3$ through $(0,5)$.

3. Solve $2|x + 4| = 10$.

4. Find the average rate of change of $f(x) = x^2$ from $1$to $3$.

Summary Table: Key Formulas and Concepts

Concept	Formula/Rule	Example
Distance	$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$	$(2,3)$ and $(6,11)$: $4\sqrt{5}$
Midpoint	$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$	$(-4,5)$ and $(2,-1)$: $(-1,2)$
Slope	$m = \frac{y_2 - y_1}{x_2 - x_1}$	$(-2,5)$ and $(4,-1)$: $-1$
Line Equation	$y = mx + b$ or $y - y_1 = m(x - x_1)$	Slope $3$, point$(1,4)$:$y = 3x + 1$
Circle	$(x - h)^2 + (y - k)^2 = r^2$	Center $(2,-1)$, radius $3$
Absolute Value	$|x - a| = b \rightarrow x = a \pm b$	$|x - 3| = 7 \rightarrow x = 10$ or $x = -4$
Average Rate of Change	$\frac{f(b) - f(a)}{b - a}$	$f(x) = x^2$, $1 \rightarrow 3$: $4$