Module 2 Practice Problems
Practice Makes Perfect!
These 15 problems cover all topics from Module 2: Linear and Quadratic Functions. Work through them to reinforce your learning!
- Take your time - Understanding is more important than speed
- Try before revealing - Attempt each problem before looking at the solution
- Learn from mistakes - Wrong answers are learning opportunities!
- Show your work - Write out steps to build problem-solving skills
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Part 1: Slope and Equations of Lines
1 Finding Slope from Two Points Easy
Find the slope of the line passing through the points (-3, 7) and (5, -1).
Solution:
Use the slope formula: m = (y₂ - y₁) / (x₂ - x₁)
Step 1: Identify the coordinates
Point 1: (x₁, y₁) = (-3, 7)
Point 2: (x₂, y₂) = (5, -1)
Step 2: Substitute into the formula
m = (-1 - 7) / (5 - (-3))
m = (-8) / (8)
m = -1
Answer: The slope is -1
Interpretation: For every 1 unit increase in x, y decreases by 1 unit.
2 Point-Slope Form Medium
Write the equation of a line with slope -2 that passes through the point (4, 3). Express your answer in slope-intercept form (y = mx + b).
Solution:
Step 1: Start with point-slope form
y - y₁ = m(x - x₁)
Step 2: Substitute m = -2 and point (4, 3)
y - 3 = -2(x - 4)
Step 3: Distribute
y - 3 = -2x + 8
Step 4: Solve for y (slope-intercept form)
y = -2x + 8 + 3
y = -2x + 11
Answer: y = -2x + 11
Check: When x = 4, y = -2(4) + 11 = -8 + 11 = 3
3 Finding Equation from Two Points Medium
Find the equation of the line passing through points (1, 4) and (3, 10). Write your answer in slope-intercept form.
Solution:
Step 1: Find the slope
m = (10 - 4) / (3 - 1) = 6 / 2 = 3
Step 2: Use point-slope form with either point (using (1, 4))
y - 4 = 3(x - 1)
Step 3: Distribute
y - 4 = 3x - 3
Step 4: Solve for y
y = 3x - 3 + 4
y = 3x + 1
Answer: y = 3x + 1
Verify with both points: (1, 4): y = 3(1) + 1 = 4 | (3, 10): y = 3(3) + 1 = 10
4 Parallel and Perpendicular Lines Hard
Consider the line y = (2/3)x + 5.
- What is the slope of a line parallel to this line?
- What is the slope of a line perpendicular to this line?
- Write the equation of the line perpendicular to the given line that passes through the point (6, 0).
Solution:
(a) Parallel line slope:
Parallel lines have the same slope.
Answer: m = 2/3
(b) Perpendicular line slope:
Perpendicular lines have slopes that are negative reciprocals.
Original slope: m = 2/3
Perpendicular slope: m⊥ = -1/(2/3) = -3/2
Answer: m = -3/2
(c) Perpendicular line through (6, 0):
Use point-slope form with m = -3/2 and point (6, 0)
y - 0 = (-3/2)(x - 6)
y = (-3/2)x + 9
Answer: y = (-3/2)x + 9
Key Concept: Product of perpendicular slopes = -1. Check: (2/3) × (-3/2) = -1
Part 2: Modeling with Linear Functions
5 Cost Function Easy
A cell phone plan costs $30 per month plus $0.10 per text message sent. Write a linear equation for the total monthly cost C as a function of the number of text messages t sent.
Solution:
Step 1: Identify the components
Fixed cost (y-intercept): $30 per month
Variable cost (slope): $0.10 per text message
Step 2: Write the equation
C(t) = 0.10t + 30
Answer: C(t) = 0.10t + 30
Example: If you send 100 texts: C(100) = 0.10(100) + 30 = $40
6 Temperature Conversion Medium
The relationship between Fahrenheit (F) and Celsius (C) is linear. Water freezes at 32°F (0°C) and boils at 212°F (100°C).
- Find the equation that converts Celsius to Fahrenheit: F = mC + b
- Use your equation to convert 25°C to Fahrenheit.
Solution:
(a) Finding the equation:
Step 1: Identify two points: (0, 32) and (100, 212) where C is x and F is y
Step 2: Find slope m = (212 - 32) / (100 - 0) = 180 / 100 = 9/5
Step 3: Use point (0, 32) to find b
F = (9/5)C + b
32 = (9/5)(0) + b
b = 32
Answer: F = (9/5)C + 32
(b) Convert 25°C to Fahrenheit:
F = (9/5)(25) + 32
F = 45 + 32
F = 77
Answer: 25°C = 77°F
7 Depreciation Medium
A car purchased for $25,000 depreciates linearly. After 4 years, it is worth $17,000.
- Write a linear equation for the value V of the car after t years.
- What will the car be worth after 7 years?
- When will the car be worth $5,000?
Solution:
(a) Linear equation:
Points: (0, 25000) and (4, 17000)
Slope: m = (17000 - 25000) / (4 - 0) = -8000 / 4 = -2000
Using V = mt + b with b = 25000:
Answer: V(t) = -2000t + 25000
(b) Value after 7 years:
V(7) = -2000(7) + 25000
V(7) = -14000 + 25000
V(7) = 11000
Answer: $11,000
(c) When will it be worth $5,000?
5000 = -2000t + 25000
-20000 = -2000t
t = 10
Answer: After 10 years
Note: The car loses $2,000 in value each year (slope = -2000).
Part 3: Graphing and Identifying Quadratic Features
8 Identifying Vertex from Vertex Form Easy
Identify the vertex and whether the parabola opens upward or downward for each function:
- f(x) = (x - 3)² + 7
- g(x) = -2(x + 1)² - 4
Solution:
Vertex form: f(x) = a(x - h)² + k, where vertex is (h, k)
(a) f(x) = (x - 3)² + 7
Here: a = 1, h = 3, k = 7
Vertex: (3, 7)
Opens: Upward (because a = 1 > 0)
(b) g(x) = -2(x + 1)² - 4
Rewrite: g(x) = -2(x - (-1))² + (-4)
Here: a = -2, h = -1, k = -4
Vertex: (-1, -4)
Opens: Downward (because a = -2 < 0)
Key Rule: If a > 0, parabola opens upward. If a < 0, opens downward.
9 Converting to Vertex Form Hard
Convert f(x) = x² - 6x + 13 to vertex form by completing the square. Then identify the vertex.
Solution:
Step 1: Start with f(x) = x² - 6x + 13
Step 2: Complete the square for x² - 6x
Take half of the coefficient of x: -6/2 = -3
Square it: (-3)² = 9
Step 3: Add and subtract 9
f(x) = x² - 6x + 9 - 9 + 13
f(x) = (x² - 6x + 9) + 4
f(x) = (x - 3)² + 4
Answer: f(x) = (x - 3)² + 4
Vertex: (3, 4)
Verify: Expand (x - 3)² + 4 = x² - 6x + 9 + 4 = x² - 6x + 13
10 Finding Axis of Symmetry and Intercepts Medium
For the function f(x) = x² - 4x - 5:
- Find the axis of symmetry
- Find the vertex
- Find the y-intercept
- Find the x-intercepts
Solution:
Given: f(x) = x² - 4x - 5, where a = 1, b = -4, c = -5
(a) Axis of symmetry:
x = -b/(2a) = -(-4)/(2×1) = 4/2 = 2
Answer: x = 2
(b) Vertex:
x-coordinate: 2 (from axis of symmetry)
y-coordinate: f(2) = (2)² - 4(2) - 5 = 4 - 8 - 5 = -9
Answer: Vertex = (2, -9)
(c) Y-intercept:
Set x = 0: f(0) = 0² - 4(0) - 5 = -5
Answer: (0, -5)
(d) X-intercepts:
Set f(x) = 0: x² - 4x - 5 = 0
Factor: (x - 5)(x + 1) = 0
x = 5 or x = -1
Answer: (5, 0) and (-1, 0)
11 Projectile Motion Medium
A ball is thrown upward from a height of 5 feet with an initial velocity of 48 feet per second. Its height h (in feet) after t seconds is given by:
h(t) = -16t² + 48t + 5
- What is the maximum height the ball reaches?
- When does the ball reach its maximum height?
Solution:
Given: h(t) = -16t² + 48t + 5, where a = -16, b = 48, c = 5
(b) When does it reach maximum height?
Maximum occurs at the vertex. Time: t = -b/(2a)
t = -48/(2×(-16)) = -48/(-32) = 1.5
Answer: At t = 1.5 seconds
(a) Maximum height:
h(1.5) = -16(1.5)² + 48(1.5) + 5
h(1.5) = -16(2.25) + 72 + 5
h(1.5) = -36 + 72 + 5
h(1.5) = 41
Answer: Maximum height = 41 feet
Note: The parabola opens downward (a = -16 < 0), so the vertex represents a maximum.
Part 4: Solving Quadratic Equations
12 Solving by Factoring Easy
Solve by factoring: x² + 7x + 12 = 0
Solution:
Step 1: Factor the quadratic
We need two numbers that multiply to 12 and add to 7
Numbers: 3 and 4 (because 3 × 4 = 12 and 3 + 4 = 7)
x² + 7x + 12 = (x + 3)(x + 4)
Step 2: Set each factor equal to zero
(x + 3)(x + 4) = 0
x + 3 = 0 or x + 4 = 0
x = -3 or x = -4
Answer: x = -3 or x = -4
Check x = -3: (-3)² + 7(-3) + 12 = 9 - 21 + 12 = 0
Check x = -4: (-4)² + 7(-4) + 12 = 16 - 28 + 12 = 0
13 Solving by Quadratic Formula Medium
Solve using the quadratic formula: 2x² + 5x - 3 = 0
Solution:
Quadratic formula: x = [-b ± √(b² - 4ac)] / (2a)
Step 1: Identify a, b, c
2x² + 5x - 3 = 0
a = 2, b = 5, c = -3
Step 2: Calculate the discriminant
b² - 4ac = (5)² - 4(2)(-3) = 25 + 24 = 49
Step 3: Apply the formula
x = [-5 ± √49] / (2×2)
x = [-5 ± 7] / 4
Step 4: Solve for both values
x = (-5 + 7) / 4 = 2/4 = 1/2
x = (-5 - 7) / 4 = -12/4 = -3
Answer: x = 1/2 or x = -3
14 Completing the Square Hard
Solve by completing the square: x² + 8x + 7 = 0
Solution:
Step 1: Move constant to the right side
x² + 8x = -7
Step 2: Complete the square on the left
Take half of 8: 8/2 = 4
Square it: 4² = 16
Add 16 to both sides:
x² + 8x + 16 = -7 + 16
x² + 8x + 16 = 9
Step 3: Factor the left side as a perfect square
(x + 4)² = 9
Step 4: Take square root of both sides
x + 4 = ±3
Step 5: Solve for x
x + 4 = 3 or x + 4 = -3
x = -1 or x = -7
Answer: x = -1 or x = -7
15 Application Problem Hard
A rectangular garden has a length that is 4 feet more than twice its width. If the area of the garden is 70 square feet, find the dimensions of the garden.
Solution:
Step 1: Define variables
Let w = width of the garden (in feet)
Then length = 2w + 4
Step 2: Write equation using Area = length × width
70 = w(2w + 4)
70 = 2w² + 4w
2w² + 4w - 70 = 0
Step 3: Simplify by dividing by 2
w² + 2w - 35 = 0
Step 4: Factor
(w + 7)(w - 5) = 0
w = -7 or w = 5
Step 5: Choose the positive solution
Width cannot be negative, so w = 5 feet
Length = 2(5) + 4 = 14 feet
Answer: Width = 5 feet, Length = 14 feet
Check: Area = 5 × 14 = 70 square feet
Great Work!
You've completed all 15 practice problems! These problems covered:
- Finding slopes and equations of lines
- Modeling real-world situations with linear functions
- Graphing and identifying features of quadratic functions
- Solving quadratic equations by factoring, quadratic formula, and completing the square
Next step: Test your knowledge with the Module 2 Quiz!