Learn Without Walls

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!

Your Progress

0 / 15

Solutions revealed: 0 / 15

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.

  1. What is the slope of a line parallel to this line?
  2. What is the slope of a line perpendicular to this line?
  3. 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).

  1. Find the equation that converts Celsius to Fahrenheit: F = mC + b
  2. 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.

  1. Write a linear equation for the value V of the car after t years.
  2. What will the car be worth after 7 years?
  3. 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:

  1. f(x) = (x - 3)² + 7
  2. 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:

  1. Find the axis of symmetry
  2. Find the vertex
  3. Find the y-intercept
  4. 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

  1. What is the maximum height the ball reaches?
  2. 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:

Next step: Test your knowledge with the Module 2 Quiz!