Unit 3 Homework 5 Vertex Form Of A Quadratic Equation

Unit 3 Homework 5: Mastering the Vertex Form of a Quadratic Equation

This comprehensive guide dives deep into the vertex form of a quadratic equation, a crucial concept in algebra. We'll explore its properties, how to convert between different forms, and tackle a variety of practice problems to solidify your understanding. By the end, you'll be confident in handling Unit 3 Homework 5 and beyond!

Understanding the Vertex Form: A Quadratic's Identity

The vertex form of a quadratic equation is a powerful tool for understanding a parabola's key features. It's written as:

y = a(x - h)² + k

Where:

• a: Determines the parabola's vertical stretch or compression and its direction (opens upwards if a > 0, downwards if a < 0). A larger absolute value of 'a' indicates a narrower parabola, while a smaller absolute value indicates a wider one.

• (h, k): Represents the coordinates of the vertex of the parabola. The vertex is the parabola's turning point – either its minimum (if a > 0) or maximum (if a < 0).

• x and y: Are the coordinates of any point on the parabola.

Understanding these components is paramount to interpreting and manipulating the equation.

Why is the Vertex Form Important?

The vertex form offers several advantages over the standard form (y = ax² + bx + c):

• Easy Vertex Identification: The vertex (h, k) is directly visible in the equation, eliminating the need for calculations like finding the x-coordinate using -b/2a.

• Quick Axis of Symmetry Determination: The axis of symmetry, a vertical line passing through the vertex, is simply x = h.

• Simple Transformations: The values of 'a', 'h', and 'k' clearly illustrate transformations (shifts, stretches, reflections) applied to the parent function y = x².

• Efficient Graphing: Knowing the vertex and the direction of opening allows for quick sketching of the parabola.

Converting to and from Vertex Form

While the vertex form is incredibly useful, you often encounter quadratic equations in other forms. Therefore, mastering conversions is essential.

1. Converting from Standard Form (y = ax² + bx + c) to Vertex Form

The key to this conversion lies in completing the square. Let's break down the process:

1. Factor out 'a' from the x² and x terms: This leaves the equation in the form y = a(x² + (b/a)x) + c.

2. Complete the Square: Take half of the coefficient of x ((b/a)/2 = b/2a), square it ((b/2a)² = b²/4a²), and add and subtract this value inside the parentheses. This doesn't change the equation's value because we're adding and subtracting the same quantity.

3. Rewrite as a Perfect Square: The expression inside the parentheses will now be a perfect square trinomial, which can be factored as (x + b/2a)².

4. Simplify: Distribute 'a' and combine constant terms to obtain the vertex form y = a(x - h)² + k.

Example:

Convert y = 2x² + 8x + 5 to vertex form.

1. Factor out 'a': y = 2(x² + 4x) + 5

2. Complete the square: Half of 4 is 2, and 2² = 4. So, we add and subtract 4 inside the parentheses: y = 2(x² + 4x + 4 - 4) + 5

3. Rewrite as a perfect square: y = 2((x + 2)² - 4) + 5

4. Simplify: y = 2(x + 2)² - 8 + 5 = 2(x + 2)² - 3

Therefore, the vertex form is y = 2(x + 2)² - 3. The vertex is (-2, -3).

2. Converting from Vertex Form to Standard Form

This conversion is simpler; it involves expanding the equation.

1. Expand the squared term: Use the formula (x - h)² = x² - 2hx + h².

2. Distribute 'a': Multiply 'a' by each term within the parentheses.

3. Combine like terms: Add the constant terms to obtain the standard form y = ax² + bx + c.

Example:

Convert y = 3(x - 1)² + 4 to standard form.

1. Expand the squared term: (x - 1)² = x² - 2x + 1

2. Distribute 'a': y = 3(x² - 2x + 1) + 4 = 3x² - 6x + 3 + 4

3. Combine like terms: y = 3x² - 6x + 7

Therefore, the standard form is y = 3x² - 6x + 7.

Practice Problems: Putting it All Together

Let's work through some problems to reinforce your understanding.

Problem 1: Find the vertex and axis of symmetry for the quadratic equation y = -1/2(x + 3)² + 5.

Solution:

The equation is already in vertex form. The vertex is (-3, 5), and the axis of symmetry is x = -3.

Problem 2: Convert y = x² - 6x + 11 to vertex form.

Solution:

1. Complete the square: y = (x² - 6x + 9 - 9) + 11

2. Rewrite as a perfect square: y = (x - 3)² - 9 + 11

3. Simplify: y = (x - 3)² + 2

The vertex form is y = (x - 3)² + 2. The vertex is (3, 2).

Problem 3: A parabola passes through the points (1, 4), (2, 7), and (3, 12). Is it possible to determine if this parabola can be represented by a quadratic equation in vertex form?

Solution:

While three points are not enough to uniquely define a parabola in vertex form (you need the vertex), it is possible to try and fit the points to a quadratic equation in standard form and then convert to vertex form. You would need to solve a system of three equations with three unknowns (a, b, and c in y = ax² + bx + c). If a solution exists, you can then convert that standard form equation to vertex form. If no solution exists using standard form, it means the points don't fit a parabolic curve.

Problem 4: A ball is thrown upwards and its height (in meters) above the ground after t seconds is given by the equation h(t) = -5t² + 20t + 1. Find the maximum height the ball reaches and the time it takes to reach that height.

Solution:

This is a quadratic equation representing a parabola that opens downwards (a = -5 < 0). The maximum height is the y-coordinate of the vertex. Convert to vertex form to find the vertex:

1. Factor out -5: h(t) = -5(t² - 4t) + 1

2. Complete the square: h(t) = -5(t² - 4t + 4 - 4) + 1

3. Simplify: h(t) = -5((t - 2)² - 4) + 1 = -5(t - 2)² + 20 + 1 = -5(t - 2)² + 21

The vertex is (2, 21). Therefore, the maximum height is 21 meters, reached after 2 seconds.

Advanced Applications and Further Exploration

The vertex form is not just a theoretical concept; it has numerous real-world applications in various fields, including:

• Physics: Modeling projectile motion, as seen in Problem 4.

• Engineering: Designing parabolic antennas and reflectors.

• Economics: Analyzing quadratic cost and revenue functions.

• Computer Graphics: Creating parabolic curves for animations and game design.

Further exploration of quadratic equations might involve investigating:

• Discriminant: Understanding how the discriminant (b² - 4ac) determines the number and nature of the roots (solutions) of a quadratic equation.

• Complex Numbers: Exploring the solutions of quadratic equations when the discriminant is negative.

• Quadratic Inequalities: Solving inequalities involving quadratic expressions.

By mastering the vertex form of a quadratic equation, you gain a powerful tool for analyzing and manipulating parabolas, unlocking a deeper understanding of this fundamental mathematical concept and its diverse applications. Remember to practice consistently and explore different problem types to solidify your comprehension. Good luck with Unit 3 Homework 5!