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 delves into the vertex form of a quadratic equation, a crucial concept in algebra. We'll explore its definition, how to convert between different forms, applications, and provide ample examples to solidify your understanding. This guide is designed to help you ace Unit 3, Homework 5, and beyond!

Understanding the Vertex Form

The vertex form of a quadratic equation provides a concise and insightful representation of a parabola. It highlights key features like the vertex (the parabola's turning point) and the axis of symmetry. The general form is:

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 parabola.
• (h, k): Represents the coordinates of the vertex. The x-coordinate of the vertex is 'h', and the y-coordinate is 'k'.
• x and y: Represent the variables of the equation.

Converting to Vertex Form: Completing the Square

The most common method for converting a standard quadratic equation (ax² + bx + c = 0) into vertex form is completing the square. Let's break down this process step-by-step:

1. Isolate the x terms: Group the terms containing 'x' together.

2. Factor out 'a': If 'a' is not equal to 1, factor it out from the x terms.

3. Complete the square: Take half of the coefficient of the x term (b/2a), square it ((b/2a)²), and add and subtract this value within the parentheses. This ensures we haven't changed the equation's value.

4. Factor the perfect square trinomial: The terms within the parentheses should now form a perfect square trinomial, which can be factored as (x + b/2a)².

5. Simplify and rewrite: Simplify the equation and rearrange it into the vertex form y = a(x - h)² + k.

Example:

Convert the quadratic equation y = 2x² + 8x + 5 into vertex form.

1. Isolate x terms: y = 2(x² + 4x) + 5

2. Factor out 'a': (Note: 'a' is already factored out in this example.)

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

4. Factor the perfect square trinomial: y = 2((x + 2)² - 4) + 5

5. Simplify and rewrite: y = 2(x + 2)² - 8 + 5 => y = 2(x + 2)² - 3

Therefore, the vertex of this parabola is (-2, -3).

Identifying Key Features from Vertex Form

Once the equation is in vertex form, extracting key features becomes straightforward:

• Vertex: The vertex is directly identified as (h, k).
• Axis of Symmetry: The axis of symmetry is a vertical line that passes through the vertex. Its equation is x = h.
• y-intercept: To find the y-intercept, substitute x = 0 into the equation and solve for y.
• x-intercepts (roots): To find the x-intercepts, set y = 0 and solve for x. This often involves using the square root property or the quadratic formula.
• Direction of opening: Determined by the value of 'a'. If a > 0, the parabola opens upwards; if a < 0, it opens downwards.

Applications of Vertex Form

The vertex form's concise nature makes it incredibly useful in various applications:

• Graphing parabolas: Knowing the vertex, axis of symmetry, and direction of opening allows for quick and accurate sketching of the parabola.
• Modeling real-world phenomena: Quadratic equations frequently model projectile motion, area optimization, and other scenarios. The vertex form helps determine the maximum or minimum value in these applications. For example, in projectile motion, the vertex represents the maximum height.
• Solving optimization problems: Many optimization problems, such as finding the maximum area given a fixed perimeter, involve quadratic equations. The vertex form simplifies the process of finding the optimal solution.
• Understanding transformations: The vertex form clearly shows how the parabola is transformed from its basic form, y = x². The values of 'a', 'h', and 'k' represent vertical stretches/compressions, horizontal shifts, and vertical shifts, respectively.

Advanced Techniques and Problem Solving

Let's delve into some more complex scenarios and problem-solving strategies:

1. Dealing with fractions and decimals: The completing-the-square process might involve fractions or decimals. Careful attention to arithmetic is essential to avoid errors.

2. Equations without a constant term: If the quadratic equation lacks a constant term (c = 0), completing the square becomes slightly simpler. You will still follow the same process but without the step of adding and subtracting a value to maintain the equation’s balance.

3. Finding the equation given the vertex and a point: If you're given the vertex (h, k) and another point (x, y) on the parabola, you can substitute these values into the vertex form to solve for 'a'. Then you can write the complete equation.

4. Word problems: Many real-world problems involve translating word descriptions into quadratic equations, then converting to vertex form to solve for optimal values.

Example of a word problem:

A farmer wants to fence a rectangular area using 100 meters of fencing. What dimensions will maximize the area of the enclosed region?

• Let x be the length and y be the width.
• Perimeter: 2x + 2y = 100 => y = 50 - x
• Area: A = xy = x(50 - x) = 50x - x²
• Converting to vertex form: A = - (x² - 50x) = -(x² - 50x + 625 - 625) = -(x - 25)² + 625

The vertex is (25, 625), indicating that the maximum area (625 square meters) occurs when x = 25 meters. Therefore, the dimensions that maximize the area are 25 meters by 25 meters (a square).

Practice Problems

To solidify your understanding, try these practice problems:

1. Convert y = x² - 6x + 11 to vertex form.
2. Find the vertex, axis of symmetry, and y-intercept of y = -3(x + 1)² + 4.
3. A ball is thrown upwards with an initial velocity of 40 m/s from a height of 2 meters. Its height (h) after t seconds is given by h = -5t² + 40t + 2. Convert this equation to vertex form and find the maximum height reached by the ball.
4. Find the equation of a parabola with vertex (2, -1) that passes through the point (4, 3).

Solutions to these problems will be provided in a follow-up guide. Remember to consult your textbook and notes for additional support.

Conclusion

Mastering the vertex form of a quadratic equation is essential for a strong understanding of quadratic functions. By understanding its definition, conversion techniques, and applications, you can confidently tackle a wide range of problems, from simple graphing to complex optimization scenarios. Practice regularly and don't hesitate to seek help when needed. Remember that consistent effort is key to success in mastering this crucial algebraic concept. Good luck with your Unit 3, Homework 5!