Homework 5 Vertex Form Of A Quadratic Equation

Homework 5: Mastering the Vertex Form of a Quadratic Equation

This comprehensive guide delves into the vertex form of a quadratic equation, equipping you with the knowledge and skills to confidently tackle any related problem. We'll explore its structure, applications, and various problem-solving techniques. By the end, you’ll not only understand the why behind the vertex form but also the how – mastering its manipulation to solve complex quadratic equations.

Understanding the Vertex Form: y = a(x - h)² + k

The vertex form of a quadratic equation is expressed 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 parabola.

• (h, k) represents the vertex of the parabola – the lowest (minimum) or highest (maximum) point on the curve. 'h' is the x-coordinate, and 'k' is the y-coordinate of the vertex.

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

This form is incredibly useful because it directly reveals the vertex, a crucial element in understanding the parabola's behavior and solving related problems. Unlike the standard form (y = ax² + bx + c), the vertex form provides immediate visual information.

Why is the Vertex Form Important?

The vertex form offers several advantages over the standard form:

• Easy Identification of the Vertex: The vertex (h, k) is explicitly stated, eliminating the need for calculations using the formula x = -b/2a (used in the standard form).

• Simple Graphing: Knowing the vertex and the direction (determined by 'a') allows for quick and accurate sketching of the parabola.

• Solving Real-World Problems: Many applications, such as projectile motion, optimization problems, and modeling parabolic curves in engineering and physics, benefit directly from the vertex form's clear depiction of the maximum or minimum point.

• Transformations: The vertex form clearly showcases transformations applied to the parent function y = x². The values of 'a', 'h', and 'k' directly indicate vertical stretches/compressions, horizontal shifts, and vertical shifts respectively.

Working with the Vertex Form: Examples and Problem-Solving

Let's dive into examples illustrating how to use the vertex form effectively.

Example 1: Finding the Vertex and Graphing

Given the equation y = 2(x - 3)² + 1, find the vertex and sketch the parabola.

Solution:

1. Identify 'a', 'h', and 'k': Here, a = 2, h = 3, and k = 1.

2. Determine the Vertex: The vertex is (h, k) = (3, 1).

3. Determine the Parabola's Direction: Since a = 2 (positive), the parabola opens upwards.

4. Sketch the Graph: Plot the vertex (3, 1). Since the parabola opens upwards, it will have a minimum point at the vertex. You can plot additional points by substituting x-values into the equation to find corresponding y-values.

Example 2: Finding the Equation from the Vertex and a Point

A parabola has a vertex at (-2, 4) and passes through the point (0, 0). Find its equation in vertex form.

Solution:

1. Substitute the Vertex: We know the vertex is (-2, 4), so h = -2 and k = 4. Our equation so far is y = a(x + 2)² + 4.

2. Use the Point to Find 'a': Substitute the coordinates of the point (0, 0) into the equation: 0 = a(0 + 2)² + 4. Solving for 'a', we get a = -1.

3. Write the Equation: The equation of the parabola is y = -(x + 2)² + 4.

Example 3: Converting from Standard Form to Vertex Form

Convert the quadratic equation y = x² - 4x + 7 into vertex form.

Solution:

This involves a process called completing the square.

1. Group the x terms: y = (x² - 4x) + 7

2. Complete the Square: Take half of the coefficient of x (-4), square it ((-2)² = 4), and add and subtract this value inside the parentheses: y = (x² - 4x + 4 - 4) + 7

3. Factor the Perfect Square Trinomial: y = (x - 2)² - 4 + 7

4. Simplify: y = (x - 2)² + 3

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

Example 4: Real-World Application: Projectile Motion

A ball is thrown upwards with an initial velocity of 40 m/s from a height of 2 meters. Its height (in meters) after t seconds is given by the equation h(t) = -5t² + 40t + 2. Find the maximum height reached by the ball and the time it takes to reach this height.

Solution:

1. Convert to Vertex Form (Completing the Square): While we could use the formula t = -b/2a to find the time of maximum height, converting to vertex form provides a clearer understanding. Completing the square for h(t) = -5t² + 40t + 2 gives h(t) = -5(t - 4)² + 82.

2. Identify the Vertex: The vertex is (4, 82).

3. Interpret the Vertex: The vertex represents the maximum height (82 meters) reached at time t = 4 seconds.

Advanced Techniques and Considerations

Transformations of the Parabola

Understanding how 'a', 'h', and 'k' affect the graph is crucial.

• 'a': Controls the vertical stretch or compression and the direction of opening. |a| > 1 makes the parabola narrower, 0 < |a| < 1 makes it wider. a < 0 reflects the parabola across the x-axis.

• 'h': Represents a horizontal shift. If h > 0, the parabola shifts to the right; if h < 0, it shifts to the left.

• 'k': Represents a vertical shift. If k > 0, the parabola shifts upwards; if k < 0, it shifts downwards.

Finding the x-intercepts (Roots)

The x-intercepts are where the parabola intersects the x-axis (y = 0). To find them, set y = 0 in the vertex form and solve for x. This often involves taking the square root and considering both positive and negative solutions.

Discriminant and Nature of Roots

The discriminant, b² - 4ac (from the standard form), helps determine the nature of the roots:

• b² - 4ac > 0: Two distinct real roots (two x-intercepts).
• b² - 4ac = 0: One real root (the vertex touches the x-axis).
• b² - 4ac < 0: No real roots (the parabola does not intersect the x-axis).

Homework Problems: Putting Your Knowledge to the Test

1. Find the vertex and sketch the graph of y = -3(x + 1)² - 2.

2. A parabola passes through the points (1, 2) and (3, 2) and has a vertex at (2, 0). Find its equation in vertex form.

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

4. A rocket is launched vertically upwards from the ground with an initial velocity of 60 m/s. Its height (in meters) after t seconds is given by h(t) = -5t² + 60t. Find the maximum height reached and the time it takes to reach this height.

5. Explain the effect of changing the value of 'a' in the vertex form y = a(x - h)² + k on the parabola's graph.

By diligently working through these examples and homework problems, you'll solidify your understanding of the vertex form and its applications. Remember, practice is key to mastering this crucial concept in quadratic equations. The more you work with the vertex form, the more intuitive its applications will become, empowering you to solve a wide array of problems efficiently and accurately. Good luck!