Alright, guys, let's dive into transforming the quadratic equation y = 3x^2 + 30x + 82 into its vertex form. Vertex form is super useful because it instantly tells us the vertex (the highest or lowest point) of the parabola. It also makes graphing a whole lot easier. So, buckle up, and let's get started!

Understanding Vertex Form

Before we jump into the conversion, let's quickly recap what vertex form actually looks like. The vertex form of a quadratic equation is given by:

y = a(x - h)^2 + k

Where:

• (h, k) is the vertex of the parabola.
• a is the same leading coefficient as in the standard form (which tells us if the parabola opens upwards or downwards and how wide or narrow it is).

Our goal is to manipulate the given equation, y = 3x^2 + 30x + 82, to fit this form. This involves a process called completing the square. Don't worry; it's not as scary as it sounds!

Step-by-Step Conversion

Step 1: Factor out the Leading Coefficient

The first thing we need to do is factor out the leading coefficient (the number in front of the x^2 term) from the x^2 and x terms. In our equation, the leading coefficient is 3. So, we factor it out like this:

y = 3(x^2 + 10x) + 82

Notice that we only factored the 3 from the terms with 'x' in them. The constant term, 82, stays outside the parentheses for now. This is a crucial step, so make sure you've got it right.

Step 2: Completing the Square

Now comes the fun part – completing the square! Inside the parentheses, we have x^2 + 10x. To complete the square, we need to add and subtract a value that turns this expression into a perfect square trinomial. A perfect square trinomial is something that can be factored into the form (x + something)^2 or (x - something)^2.

To find this value, we take half of the coefficient of the x term (which is 10), square it, and that's what we add and subtract. Half of 10 is 5, and 5 squared is 25. So, we add and subtract 25 inside the parentheses:

y = 3(x^2 + 10x + 25 - 25) + 82

Why do we add and subtract the same number? Because we're essentially adding zero, which doesn't change the value of the equation. We're just changing its form. Remember, balance is key!

Step 3: Rewrite as a Perfect Square

Now, we can rewrite the first three terms inside the parentheses as a perfect square. The expression x^2 + 10x + 25 is the same as (x + 5)^2. So, our equation becomes:

y = 3((x + 5)^2 - 25) + 82

See how we've created a squared term? That's the essence of completing the square. We're getting closer to vertex form!

Step 4: Distribute and Simplify

Next, we need to distribute the 3 back into the parentheses. Remember to distribute it to both the (x + 5)^2 term and the -25:

y = 3(x + 5)^2 - 3(25) + 82

y = 3(x + 5)^2 - 75 + 82

Now, combine the constant terms:

y = 3(x + 5)^2 + 7

Step 5: Identify the Vertex

Tada! We've successfully converted the equation into vertex form: y = 3(x + 5)^2 + 7. Now, we can easily identify the vertex. Comparing this to the standard vertex form y = a(x - h)^2 + k, we can see that:

• h = -5 (note the minus sign in the formula!)
• k = 7

So, the vertex of the parabola is (-5, 7). Isn't that neat?

Quick Recap and Key Points

Let's quickly go over the steps we took:

1. Factor out the leading coefficient: y = 3(x^2 + 10x) + 82
2. Complete the square: y = 3(x^2 + 10x + 25 - 25) + 82
3. Rewrite as a perfect square: y = 3((x + 5)^2 - 25) + 82
4. Distribute and simplify: y = 3(x + 5)^2 + 7
5. Identify the vertex: Vertex is (-5, 7)

Key points to remember:

• Always factor out the leading coefficient correctly.
• Remember to add and subtract the same value when completing the square.
• Pay attention to the signs when identifying the vertex coordinates.

Why is Vertex Form Useful?

You might be wondering, why bother converting to vertex form in the first place? Well, here are a few reasons:

• Finding the Vertex: As we've seen, vertex form makes it incredibly easy to find the vertex of the parabola. This is the maximum or minimum point of the quadratic function, which is useful in many applications.
• Graphing: Vertex form simplifies graphing. You know the vertex, and you know whether the parabola opens upwards (if 'a' is positive) or downwards (if 'a' is negative). This gives you a great starting point for sketching the graph.
• Understanding Transformations: Vertex form shows how the basic parabola y = x^2 has been transformed. The 'h' value represents a horizontal shift, the 'k' value represents a vertical shift, and the 'a' value represents a vertical stretch or compression.

Common Mistakes to Avoid

• Forgetting to factor: Make sure you factor the leading coefficient only from the x^2 and x terms, not the constant term.
• Incorrectly completing the square: Double-check that you're taking half of the coefficient of the x term and squaring it.
• Distributing incorrectly: When distributing the leading coefficient back in, make sure you multiply it by both terms inside the parentheses.
• Sign errors: Pay close attention to the signs when identifying the vertex coordinates. Remember that the x-coordinate of the vertex is the opposite of the value inside the parentheses.

Real-World Applications

Quadratic equations and parabolas pop up in all sorts of real-world situations. Here are a few examples:

• Projectile Motion: The path of a projectile (like a ball thrown in the air) can be modeled by a parabola. The vertex represents the highest point the projectile reaches.
• Optimization Problems: Many optimization problems involve finding the maximum or minimum value of a quadratic function. For example, you might want to find the dimensions of a rectangular garden that maximize its area, given a certain amount of fencing.
• Engineering: Parabolas are used in the design of bridges, antennas, and other structures.

Practice Problems

To really master converting to vertex form, it's important to practice. Here are a couple of problems for you to try:

1. Convert y = 2x^2 - 8x + 5 to vertex form.
2. Convert y = -x^2 + 4x + 1 to vertex form.

Solution to Problem 1:

1. Factor out the leading coefficient: y = 2(x^2 - 4x) + 5
2. Complete the square: y = 2(x^2 - 4x + 4 - 4) + 5
3. Rewrite as a perfect square: y = 2((x - 2)^2 - 4) + 5
4. Distribute and simplify: y = 2(x - 2)^2 - 8 + 5
5. Vertex form: y = 2(x - 2)^2 - 3
6. Vertex: (2, -3)

Solution to Problem 2:

1. Factor out the leading coefficient: y = -(x^2 - 4x) + 1
2. Complete the square: y = -(x^2 - 4x + 4 - 4) + 1
3. Rewrite as a perfect square: y = -((x - 2)^2 - 4) + 1
4. Distribute and simplify: y = -(x - 2)^2 + 4 + 1
5. Vertex form: y = -(x - 2)^2 + 5
6. Vertex: (2, 5)

Conclusion

So, there you have it! Converting a quadratic equation to vertex form might seem a bit tricky at first, but with a little practice, you'll become a pro in no time. Remember the steps, watch out for those common mistakes, and you'll be finding vertices and graphing parabolas like a boss! Keep practicing, and good luck!