Module 2 Study Guide
Linear & Quadratic Functions
College Algebra Learning Platform - Safaa Dabagh
1. Linear Functions
Linear Function: A function of the form f(x) = mx + b, where m is the slope and b is the y-intercept. The graph is a straight line.
Slope
Slope Formula:
m = (y₂ - y₁) / (x₂ - x₁)
Given two points (x₁, y₁) and (x₂, y₂)
m = (y₂ - y₁) / (x₂ - x₁)
Given two points (x₁, y₁) and (x₂, y₂)
Slope (m): Represents the rate of change of y with respect to x. It measures the steepness and direction of a line.
- Positive slope: Line rises from left to right
- Negative slope: Line falls from left to right
- Zero slope: Horizontal line
- Undefined slope: Vertical line
- Positive slope: Line rises from left to right
- Negative slope: Line falls from left to right
- Zero slope: Horizontal line
- Undefined slope: Vertical line
Forms of Linear Equations
| Form | Equation | When to Use |
|---|---|---|
| Slope-Intercept | y = mx + b | When you know slope and y-intercept |
| Point-Slope | y - y₁ = m(x - x₁) | When you know slope and one point |
| Standard Form | Ax + By = C | General form, useful for intercepts |
Parallel Lines: Same slope (m₁ = m₂)
Perpendicular Lines: Slopes are negative reciprocals (m₁ × m₂ = -1)
Perpendicular Lines: Slopes are negative reciprocals (m₁ × m₂ = -1)
Example: Find the equation of the line with slope 2 passing through (3, 5).
Solution:
1. Use point-slope form: y - 5 = 2(x - 3)
2. Distribute: y - 5 = 2x - 6
3. Solve for y: y = 2x - 1
Solution:
1. Use point-slope form: y - 5 = 2(x - 3)
2. Distribute: y - 5 = 2x - 6
3. Solve for y: y = 2x - 1
2. Modeling with Linear Functions
Real-World Applications
In real-world problems:
- Slope (m) = rate of change (cost per unit, speed, etc.)
- Y-intercept (b) = starting value (initial cost, starting point, etc.)
- Slope (m) = rate of change (cost per unit, speed, etc.)
- Y-intercept (b) = starting value (initial cost, starting point, etc.)
Example: A taxi charges $3.50 plus $0.75 per mile.
Solution:
- Fixed cost (b) = $3.50
- Variable cost (m) = $0.75 per mile
- Equation: C(x) = 0.75x + 3.50
where C is total cost and x is miles driven
Solution:
- Fixed cost (b) = $3.50
- Variable cost (m) = $0.75 per mile
- Equation: C(x) = 0.75x + 3.50
where C is total cost and x is miles driven
Common Applications
- Cost Functions: Total cost = (unit cost)(quantity) + fixed cost
- Revenue Functions: Revenue = (price)(quantity sold)
- Depreciation: Value decreases linearly over time
- Temperature Conversion: F = (9/5)C + 32
3. Quadratic Functions
Quadratic Function: A function of the form f(x) = ax² + bx + c, where a ≠ 0. The graph is a parabola.
Forms of Quadratic Equations
| Form | Equation | Key Information |
|---|---|---|
| Standard Form | f(x) = ax² + bx + c | Shows y-intercept (c) |
| Vertex Form | f(x) = a(x - h)² + k | Shows vertex (h, k) |
| Factored Form | f(x) = a(x - r₁)(x - r₂) | Shows x-intercepts (r₁, r₂) |
Key Features of Parabolas
Direction:
- If a > 0: Parabola opens upward (minimum)
- If a < 0: Parabola opens downward (maximum)
Vertex (from standard form):
x-coordinate: h = -b/(2a)
y-coordinate: k = f(h)
Axis of Symmetry:
x = -b/(2a)
- If a > 0: Parabola opens upward (minimum)
- If a < 0: Parabola opens downward (maximum)
Vertex (from standard form):
x-coordinate: h = -b/(2a)
y-coordinate: k = f(h)
Axis of Symmetry:
x = -b/(2a)
The vertex represents:
- Maximum value if a < 0 (opens down)
- Minimum value if a > 0 (opens up)
- Maximum value if a < 0 (opens down)
- Minimum value if a > 0 (opens up)
Finding Intercepts
| Intercept | How to Find |
|---|---|
| Y-intercept | Set x = 0, solve for y. Result: (0, c) |
| X-intercepts | Set y = 0, solve ax² + bx + c = 0 |
Example: Find the vertex of f(x) = x² - 6x + 5
Solution:
1. x-coordinate: x = -(-6)/(2×1) = 6/2 = 3
2. y-coordinate: f(3) = (3)² - 6(3) + 5 = 9 - 18 + 5 = -4
3. Vertex: (3, -4)
Solution:
1. x-coordinate: x = -(-6)/(2×1) = 6/2 = 3
2. y-coordinate: f(3) = (3)² - 6(3) + 5 = 9 - 18 + 5 = -4
3. Vertex: (3, -4)
4. Solving Quadratic Equations
Method 1: Factoring
Zero Product Property: If ab = 0, then a = 0 or b = 0
Example: Solve x² + 5x + 6 = 0
Solution:
1. Factor: (x + 2)(x + 3) = 0
2. Set each factor to zero: x + 2 = 0 or x + 3 = 0
3. Solve: x = -2 or x = -3
Solution:
1. Factor: (x + 2)(x + 3) = 0
2. Set each factor to zero: x + 2 = 0 or x + 3 = 0
3. Solve: x = -2 or x = -3
Method 2: Quadratic Formula
Quadratic Formula:
For ax² + bx + c = 0:
x = [-b ± √(b² - 4ac)] / (2a)
Discriminant: b² - 4ac
- If b² - 4ac > 0: Two real solutions
- If b² - 4ac = 0: One real solution (repeated)
- If b² - 4ac < 0: No real solutions (two complex)
For ax² + bx + c = 0:
x = [-b ± √(b² - 4ac)] / (2a)
Discriminant: b² - 4ac
- If b² - 4ac > 0: Two real solutions
- If b² - 4ac = 0: One real solution (repeated)
- If b² - 4ac < 0: No real solutions (two complex)
Example: Solve 2x² + 3x - 2 = 0
Solution:
1. Identify: a = 2, b = 3, c = -2
2. Discriminant: b² - 4ac = 9 - 4(2)(-2) = 9 + 16 = 25
3. Apply formula: x = [-3 ± √25] / (2×2) = [-3 ± 5] / 4
4. Solutions: x = (-3 + 5)/4 = 1/2 or x = (-3 - 5)/4 = -2
Solution:
1. Identify: a = 2, b = 3, c = -2
2. Discriminant: b² - 4ac = 9 - 4(2)(-2) = 9 + 16 = 25
3. Apply formula: x = [-3 ± √25] / (2×2) = [-3 ± 5] / 4
4. Solutions: x = (-3 + 5)/4 = 1/2 or x = (-3 - 5)/4 = -2
Method 3: Completing the Square
Steps:
1. Move constant to right side
2. Take half of b coefficient, square it
3. Add to both sides
4. Factor left side as perfect square
5. Take square root of both sides
6. Solve for x
1. Move constant to right side
2. Take half of b coefficient, square it
3. Add to both sides
4. Factor left side as perfect square
5. Take square root of both sides
6. Solve for x
Example: Solve x² + 6x + 5 = 0 by completing the square
Solution:
1. x² + 6x = -5
2. Half of 6 is 3, squared is 9
3. x² + 6x + 9 = -5 + 9
4. (x + 3)² = 4
5. x + 3 = ±2
6. x = -3 + 2 = -1 or x = -3 - 2 = -5
Solution:
1. x² + 6x = -5
2. Half of 6 is 3, squared is 9
3. x² + 6x + 9 = -5 + 9
4. (x + 3)² = 4
5. x + 3 = ±2
6. x = -3 + 2 = -1 or x = -3 - 2 = -5
5. Transformations of Quadratic Functions
Starting with f(x) = x²:
f(x) = a(x - h)² + k
- |a| > 1: Vertical stretch (narrower)
- 0 < |a| < 1: Vertical compression (wider)
- a < 0: Reflection over x-axis
- h: Horizontal shift (right if h > 0, left if h < 0)
- k: Vertical shift (up if k > 0, down if k < 0)
f(x) = a(x - h)² + k
- |a| > 1: Vertical stretch (narrower)
- 0 < |a| < 1: Vertical compression (wider)
- a < 0: Reflection over x-axis
- h: Horizontal shift (right if h > 0, left if h < 0)
- k: Vertical shift (up if k > 0, down if k < 0)
Quick Reference: Key Formulas
Linear Functions
- Slope: m = (y₂ - y₁) / (x₂ - x₁)
- Slope-intercept: y = mx + b
- Point-slope: y - y₁ = m(x - x₁)
- Perpendicular slopes: m₁ × m₂ = -1
Quadratic Functions
- Standard form: f(x) = ax² + bx + c
- Vertex form: f(x) = a(x - h)² + k
- Vertex x-coordinate: x = -b/(2a)
- Quadratic formula: x = [-b ± √(b² - 4ac)] / (2a)
- Discriminant: b² - 4ac
Important Reminders:
1. Parallel lines have equal slopes
2. Perpendicular lines have negative reciprocal slopes
3. The vertex is the maximum or minimum point of a parabola
4. The axis of symmetry passes through the vertex
5. Always check your solutions by substituting back into the original equation
1. Parallel lines have equal slopes
2. Perpendicular lines have negative reciprocal slopes
3. The vertex is the maximum or minimum point of a parabola
4. The axis of symmetry passes through the vertex
5. Always check your solutions by substituting back into the original equation
Module 2: Linear & Quadratic Functions
College Algebra Learning Platform - Safaa Dabagh - sdabagh.github.io
© 2026