Solve Quadratic Equations: Vertex Form Explained

Hey guys! Today, we're diving into a quadratic equation problem that Jessica tackled. She was asked to create a quadratic equation that represents a function passing through the point (8, -11) and having a vertex at (6, -3). Let's break down her work step-by-step and see how she nailed it (or where we can help her out!). This is a classic example of how understanding the vertex form of a quadratic equation can help us solve real-world problems, and we'll make sure you're confident in tackling similar challenges by the end of this article.

Understanding the Vertex Form

Before we jump into Jessica's calculations, let's quickly recap the vertex form of a quadratic equation. This form is super useful when you know the vertex of the parabola (the U-shaped curve that quadratic equations create). The vertex form looks like this:

$$f(x) = a(x - h)^2 + k$$

Where:

• f(x) represents the y-value for a given x-value.
• a determines the stretch or compression (and direction) of the parabola.
• (h, k) are the coordinates of the vertex. This is the turning point of the parabola – its minimum or maximum point.

Knowing this form is like having a secret weapon for solving quadratic problems! It allows us to directly plug in the vertex coordinates and then use another point on the parabola to find the crucial 'a' value. This 'a' value tells us how stretched or compressed the parabola is and whether it opens upwards (if 'a' is positive) or downwards (if 'a' is negative). So, remember this formula – it's your friend!

The power of the vertex form lies in its ability to directly incorporate the vertex coordinates (h, k). This is a massive advantage because the vertex is a key feature of a parabola. It's the point where the parabola changes direction, and it gives us a central reference point for understanding the graph's shape and position. When you're given the vertex, using this form drastically simplifies the process of finding the quadratic equation. You're essentially plugging in known values and solving for the unknown 'a', which determines the parabola's stretch and direction. Think of it like fitting a puzzle piece – the vertex gives you a crucial piece that locks into place, making the rest of the puzzle easier to solve. Without understanding the vertex form, you might find yourself wrestling with more complex methods, especially when dealing with word problems or real-world scenarios where the vertex is a naturally given piece of information.

Let's consider why this form is so much more efficient when we know the vertex. Imagine trying to find the equation of a parabola using the standard form, which is $f(x) = ax^2 + bx + c$. If you only have the vertex and one other point, you'd have a system of three equations with three unknowns (a, b, and c). Solving that can be a bit of a headache! But with the vertex form, you immediately know two pieces of the puzzle (h and k), and you only need to solve for 'a'. It transforms a potentially complex algebraic problem into a straightforward one. This efficiency is particularly valuable in applications where time is of the essence, such as in engineering or physics problems involving parabolic trajectories. So, mastering the vertex form not only helps you solve quadratic equations but also equips you with a powerful tool for tackling real-world challenges. Keep this in mind as we dissect Jessica's work – you'll see how elegantly this form simplifies the entire process!

Jessica's Step 1: Plugging in the Values

Okay, let's look at Jessica's first step:

$$-11 = a(8 - 6)^2 - 3$$

What's she doing here? She's using the vertex form we just talked about! She correctly identified that the vertex (6, -3) gives us the 'h' and 'k' values (h = 6, k = -3). She also knows that the point (8, -11) lies on the parabola. This means when x = 8, f(x) (or y) = -11. So, she's plugging these values into the vertex form equation:

$$f(x) = a(x - h)^2 + k$$

Replacing f(x) with -11, x with 8, h with 6, and k with -3, we get exactly Jessica's equation. This is a fantastic start! She's setting up the equation to solve for 'a', which will tell us how much the parabola is stretched or compressed.

This initial step is crucial because it lays the foundation for the entire solution. Jessica's action of directly substituting the given values into the vertex form demonstrates a strong understanding of how the equation relates to the graphical representation of the parabola. By correctly identifying and plugging in the vertex coordinates (h, k) and the additional point (8, -11), she's essentially creating a specific instance of the vertex form equation that corresponds to the parabola we're trying to define. This is a powerful technique because it transforms the problem from a general equation-finding task into a concrete algebraic one where we only need to solve for a single unknown, which is the 'a' value. This 'a' value, as we discussed earlier, is the key to understanding the parabola's shape – its direction (opening upwards or downwards) and its vertical stretch or compression.

Think of it like fitting puzzle pieces together. The vertex form is the puzzle frame, and the vertex coordinates and the additional point are specific pieces that fit into that frame. By plugging them in correctly, Jessica is essentially building the structure within which the solution will emerge. A mistake at this stage could throw off the entire calculation, so it's important to appreciate the precision and conceptual understanding that Jessica demonstrates in this first step. She's not just blindly plugging in numbers; she's strategically using the information provided to create an equation that reflects the specific properties of the parabola we're interested in. This solid foundation makes the subsequent steps much smoother and more manageable, highlighting the importance of a strong start in any mathematical problem-solving endeavor.

Jessica's Step 2: Simplifying the Equation

Let's move on to Jessica's second step:

$$-11 = a(2)^2 - 3$$

Here, Jessica is simplifying the equation she set up in the first step. Inside the parentheses, she's correctly calculated (8 - 6) = 2. This is a straightforward arithmetic operation, but it's essential to perform it accurately to avoid errors down the line. Squaring the 2 will be the next step in isolating the 'a' variable. This step demonstrates Jessica's attention to detail and her understanding of the order of operations. She's carefully working towards isolating the 'a' variable, which is our ultimate goal.

This simplification step might seem small, but it's a crucial bridge between the initial setup and the final solution. By correctly simplifying the expression inside the parentheses, Jessica is reducing the complexity of the equation and making it easier to manipulate. This is a common strategy in algebra – to break down complex expressions into simpler ones before attempting to solve for the unknown variable. In this case, simplifying (8 - 6) to 2 not only makes the equation look less intimidating but also sets the stage for the next operation, which is squaring the 2. This step is a perfect example of how mathematical problem-solving often involves a series of small, incremental steps, each building upon the previous one.

Each small step contributes to the overall clarity and accuracy of the solution. Imagine trying to solve a complex puzzle; you wouldn't try to force pieces together randomly. Instead, you'd start by grouping similar pieces and assembling smaller sections first. Similarly, in algebra, simplifying expressions is like assembling those smaller sections of the puzzle. It makes the overall task more manageable and reduces the chance of making mistakes. Jessica's meticulous attention to detail in this step shows that she understands this principle. She's not rushing to the answer; she's carefully simplifying the equation to ensure that each subsequent step is built on a solid foundation. This methodical approach is a hallmark of strong mathematical problem-solving skills, and it's a valuable lesson for anyone learning algebra. By breaking down the problem into smaller, manageable steps, Jessica is paving the way for a successful solution.

Jessica's Step 3: Isolating 'a'

Now, let's examine Jessica's third step:

$$-11 = 4a - 3$$

Here, Jessica has squared the 2, resulting in 4, and multiplied it by 'a' to get 4a. This is another correct application of the order of operations. She's continuing to isolate the 'a' variable. The next step would involve adding 3 to both sides of the equation to further isolate the term with 'a'. It's clear she's on the right track and is methodically working towards solving for the unknown.

This step is a critical one because it represents the direct application of algebraic principles to isolate the variable we're trying to find. By squaring the 2 and multiplying it by 'a', Jessica is effectively unwinding the operations that are applied to 'a' in the equation. This process of unwinding is fundamental to solving any algebraic equation. Think of it like peeling back the layers of an onion – each operation is a layer, and Jessica is carefully removing them one by one to reveal the core, which is the value of 'a'. The multiplication of 4 and 'a' is a key relationship that Jessica is addressing here. It's important to recognize that '4a' means 4 times 'a', and the next step will involve undoing that multiplication to finally get 'a' by itself.

This step also highlights the importance of maintaining balance in an equation. Remember, an equation is like a balanced scale – whatever you do to one side, you must do to the other to keep it balanced. Jessica understands this implicitly as she prepares to add 3 to both sides in the subsequent step. This principle of maintaining balance is a cornerstone of algebraic manipulation, and Jessica's work demonstrates a solid grasp of this concept. The clarity and precision in this step are commendable, as they demonstrate a deep understanding of the underlying algebraic principles. By carefully executing this step, Jessica is setting herself up for a smooth and accurate solution to the problem. It's a testament to her problem-solving skills and her ability to apply mathematical concepts in a logical and methodical way. So far, she's doing great!

Continuing the Solution

Okay, so Jessica has reached the equation:

$$-11 = 4a - 3$$

Let's help her finish this up! To isolate 'a', we need to get rid of the -3 on the right side. How do we do that? We add 3 to both sides of the equation:

$$-11 + 3 = 4a - 3 + 3$$

This simplifies to:

$$-8 = 4a$$

Now, 'a' is being multiplied by 4. To isolate 'a', we need to do the opposite operation: divide both sides by 4:

$$-8 / 4 = 4a / 4$$

This gives us:

$$-2 = a$$

So, we've found that a = -2! Now we know the stretch/compression and direction of the parabola.

With the 'a' value in hand, we're now in the home stretch of solving this problem. Finding 'a' is like unlocking the final piece of the puzzle. Remember, 'a' not only tells us how much the parabola is stretched or compressed vertically, but it also tells us whether the parabola opens upwards (if 'a' is positive) or downwards (if 'a' is negative). In this case, since a = -2, we know that the parabola opens downwards, indicating that it has a maximum point rather than a minimum point. This is valuable information that helps us visualize the graph of the quadratic equation and understand its behavior.

Now that we have the 'a' value, we can confidently write the complete quadratic equation in vertex form. This is where all our previous work comes together. We started with the vertex form equation, plugged in the vertex coordinates (h, k), used the additional point to solve for 'a', and now we can substitute 'a' back into the vertex form equation. This process of systematically working through the equation, one step at a time, highlights the power of algebraic manipulation. Each step has a clear purpose, and by carefully executing each step, we arrive at the final solution.

Thinking about the big picture, we've not just found the equation; we've also gained a deeper understanding of how quadratic equations work. We've seen how the vertex form simplifies the process of finding the equation when the vertex is known, and we've seen how the 'a' value influences the shape and direction of the parabola. This kind of conceptual understanding is crucial for success in mathematics. It's not just about memorizing formulas; it's about understanding the underlying principles and how they apply to different problems. So, let's celebrate this milestone and move on to the final step of writing out the complete equation!

The Final Equation

Now that we know a = -2 and the vertex is (6, -3), we can plug these values back into the vertex form:

$$f(x) = a(x - h)^2 + k$$

$$f(x) = -2(x - 6)^2 - 3$$

This is the quadratic equation that represents the function passing through the point (8, -11) and having a vertex at (6, -3)! Awesome!

This final step is where all the pieces come together to form the complete picture. We've taken the 'a' value we calculated, the vertex coordinates we were given, and plugged them back into the vertex form equation. This final substitution is like putting the last piece in a jigsaw puzzle – it completes the image and confirms that our solution is correct. The equation $f(x) = -2(x - 6)^2 - 3$ is not just a jumble of symbols; it's a concise mathematical representation of a specific parabola. It tells us the vertex is at (6, -3), the parabola opens downwards, and it's vertically stretched by a factor of 2. All of this information is packed into this one elegant equation.

Think about the journey we've taken to arrive at this solution. We started with a problem statement, dissected the given information, understood the importance of the vertex form, systematically solved for 'a', and finally constructed the complete equation. This process demonstrates the power of a structured approach to problem-solving in mathematics. Each step built upon the previous one, and by carefully executing each step, we reached a clear and accurate solution. This is a valuable skill that can be applied not just to quadratic equations but to a wide range of mathematical problems.

Moreover, this final equation allows us to predict the behavior of the parabola. We can plug in any x-value and find the corresponding y-value. We can graph the parabola and visualize its shape. We can analyze its properties, such as its axis of symmetry and its maximum value. The equation is a powerful tool that gives us a complete understanding of the quadratic function. So, let's appreciate the elegance and power of this final equation, and recognize the journey we've taken to arrive at this point. It's a testament to our problem-solving skills and our understanding of quadratic functions.

Key Takeaways

• The vertex form of a quadratic equation is your friend when you know the vertex.
• Plug in known values and solve for the unknowns.
• Pay attention to the order of operations.
• Double-check your work to avoid errors.

By following these steps and understanding the underlying concepts, you can confidently tackle similar quadratic equation problems! Keep practicing, and you'll become a quadratic equation master in no time. You got this!

Repair Input Keyword

Could you please clarify the steps Jessica took to write the quadratic equation, especially how she used the given point and vertex? Is her method correct for finding the equation of the quadratic function?

Title

Quadratic Equation Solver: Vertex Form Example