Jessica's Quadratic Equation Solution Analysis

Introduction: The Quadratic Quest

Hey guys! Today, we're diving into a quadratic equation problem that Jessica tackled. She's on a mission to create an equation that perfectly represents a function. This function has a specific point it passes through, $(8, -11)$, and a vertex sitting pretty at $(6, -3)$. Jessica's got a plan, and she's already started working through it, but let's put on our detective hats and see if we can follow her logic, spot any potential hiccups, and maybe even learn a thing or two about quadratic equations along the way. You know, quadratic functions are super important in math because they describe parabolas, those U-shaped curves you often see. They pop up everywhere, from the path of a ball thrown in the air to the design of suspension bridges. So, understanding how to work with them is a key skill in your mathematical toolkit. Jessica's challenge is a classic example of how we use information about a parabola – a specific point and its vertex – to figure out the exact equation that defines it. It's like being given a few clues and then piecing them together to reveal the whole picture. We'll be using the vertex form of a quadratic equation, which is a particularly handy way to write these equations because it directly incorporates the coordinates of the vertex. This form makes it easier to see how the parabola is shifted and stretched compared to the basic $y = x^2$ parabola. So, let's jump into Jessica's work, step by step, and see how she's using the vertex form to solve this problem. We'll check her calculations, think about the reasoning behind each step, and make sure everything makes sense. By the end of this exploration, we'll not only understand Jessica's approach but also have a deeper appreciation for the power and elegance of quadratic equations. So, grab your pencils, your thinking caps, and let's get started!

Jessica's Steps: Decoding the Equation

Okay, let's break down Jessica's work step-by-step. The goal here is to find the quadratic equation that fits the given conditions: passing through the point $(8, -11)$ and having a vertex at $(6, -3)$. Jessica's using the vertex form of a quadratic equation, which is a fantastic strategy. This form looks like this: $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola, and 'a' determines how stretched or compressed the parabola is and whether it opens upwards or downwards.

Step 1: $-11 = a(8 - 6)^2 - 3$

In this initial step, Jessica cleverly plugs in the given information into the vertex form. She substitutes the $x$ and $y$ coordinates of the point $(8, -11)$ for $x$ and $y$ in the equation, and she also plugs in the vertex coordinates, $h = 6$ and $k = -3$. This is a brilliant move because it leaves us with only one unknown, which is 'a'. The value of 'a' will tell us how the parabola is stretched or compressed vertically. Essentially, Jessica is saying, "Okay, if this parabola passes through $(8, -11)$ and has its vertex at $(6, -3)$, then this equation must be true." It's like setting up a puzzle where we know some of the pieces and need to find the missing one.

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

Here, Jessica simplifies the expression inside the parentheses. She calculates $(8 - 6)$ to get $2$. This is a straightforward arithmetic operation, but it's crucial for making the equation easier to work with. It's like taking a messy problem and cleaning it up a bit so we can see the next step more clearly. By squaring the 2 in the next step, we'll be one step closer to isolating 'a' and finding its value.

Step 3: $-11 = 4a - 3$

Jessica continues simplifying by squaring the $2$, resulting in $4$, and then multiplying it by $a$, giving us $4a$. This step is all about tidying up the equation and getting closer to isolating 'a'. It's like peeling away the layers of an onion – each simplification brings us closer to the core of the problem, which is finding the value of 'a'. Now we have a much simpler equation that we can manipulate to solve for our unknown.

Step 4: $-8 = 4a$

In this step, Jessica adds $3$ to both sides of the equation to isolate the term with 'a'. This is a fundamental algebraic manipulation – whatever we do to one side of the equation, we must do to the other to maintain the balance. Adding $3$ to both sides cancels out the $-3$ on the right side, leaving us with just $4a$. This step is like separating the variable we want to find from the other constants in the equation.

Step 5: $a = -2$

Finally, Jessica divides both sides of the equation by $4$ to solve for 'a'. This is the final step in isolating 'a' and finding its value. Dividing both sides by $4$ cancels out the multiplication, leaving us with 'a' by itself. So, Jessica has determined that $a = -2$. This value is super important because it tells us that the parabola opens downwards (since 'a' is negative) and is stretched vertically by a factor of 2. But before we celebrate, let's double-check everything to make sure this value of 'a' really works and gives us the correct quadratic equation. We'll want to plug it back into the original vertex form and see if it all lines up!

Analysis and Potential Pitfalls: Is Jessica on the Right Track?

Now that we've walked through Jessica's steps, it's time to put on our critical thinking caps and analyze her work. Did she make any mistakes? Does her solution make sense? Let's dive in! Jessica's approach of using the vertex form of a quadratic equation, $y = a(x - h)^2 + k$, is spot-on. It's the most efficient way to tackle this type of problem because it directly incorporates the vertex coordinates. Plugging in the vertex $(6, -3)$ gives us $y = a(x - 6)^2 - 3$. So far, so good! Next, she substitutes the point $(8, -11)$ into the equation. This is the crucial step where we use the fact that the parabola must pass through this point. The equation becomes $-11 = a(8 - 6)^2 - 3$. Again, this looks perfect. The substitution is done correctly, and we're setting up the equation to solve for 'a'. Now, let's trace the algebraic manipulations. Simplifying inside the parentheses, $(8 - 6)$ becomes $2$, so we have $-11 = a(2)^2 - 3$. Squaring the $2$ gives us $4$, leading to $-11 = 4a - 3$. Adding $3$ to both sides, we get $-8 = 4a$. And finally, dividing both sides by $4$, we arrive at $a = -2$. At first glance, everything seems to check out. The arithmetic is correct, and the algebraic steps are logical. But here's where we need to pause and ask ourselves: does this value of 'a' really make sense in the context of the problem? A negative value for 'a' means the parabola opens downwards, which is certainly possible. But let's dig a little deeper. Remember, the vertex is the highest or lowest point on the parabola. Since the vertex is at $(6, -3)$ and the point $(8, -11)$ lies below the vertex, a downward-opening parabola is consistent with the given information. However, we should always double-check our work, especially when dealing with signs. A small error in arithmetic can lead to a completely wrong answer. So, let's plug $a = -2$ back into the vertex form equation and see if it truly satisfies the given conditions. This is a crucial step in verifying our solution and ensuring that we haven't made any sneaky mistakes along the way. We're not just looking for the answer; we're looking for the correct answer!

Verification and Solution: Does It All Add Up?

Alright, let's put Jessica's solution to the ultimate test: verification! We need to make sure that the value of $a = -2$ she found actually works with the given information. This is like the detective double-checking their clues to make sure they fit the crime scene. So, we'll take $a = -2$ and plug it back into the vertex form of the equation, along with the vertex $(6, -3)$. This gives us: $y = -2(x - 6)^2 - 3$. Now, the big question: does this equation pass through the point $(8, -11)$? To find out, we'll substitute $x = 8$ into the equation and see if we get $y = -11$. Let's do it! Substituting $x = 8$, we get: $y = -2(8 - 6)^2 - 3$. Now we simplify: $y = -2(2)^2 - 3$ $y = -2(4) - 3$ $y = -8 - 3$ $y = -11$ Bingo! It works! When we plug in $x = 8$, we do get $y = -11$. This confirms that our equation, $y = -2(x - 6)^2 - 3$, accurately represents a parabola that passes through the point $(8, -11)$ and has a vertex at $(6, -3)$. So, Jessica's work is indeed correct! She successfully navigated the steps, avoided any algebraic pitfalls, and arrived at the right answer. This is a fantastic example of how to solve a quadratic equation problem using the vertex form. We started with the vertex form equation, plugged in the known values, solved for the unknown coefficient 'a', and then verified our solution. This methodical approach is key to success in math. But let's not stop here. Now that we've verified the solution in vertex form, it might be interesting to see what this equation looks like in standard form, which is $y = ax^2 + bx + c$. This will give us a different perspective on the same parabola and help us connect the different forms of quadratic equations. Plus, it's just good practice!

To convert from vertex form to standard form, we need to expand and simplify the equation. Let's take our vertex form equation, $y = -2(x - 6)^2 - 3$, and work our magic. First, we expand the squared term: $(x - 6)^2 = (x - 6)(x - 6) = x^2 - 12x + 36$. Now, we substitute this back into the equation: $y = -2(x^2 - 12x + 36) - 3$. Next, we distribute the $-2$: $y = -2x^2 + 24x - 72 - 3$. Finally, we combine the constant terms: $y = -2x^2 + 24x - 75$. So, the standard form of the quadratic equation is $y = -2x^2 + 24x - 75$. Notice that the coefficient of the $x^2$ term is $-2$, which is the same as the 'a' value we found earlier. This is a good check that our conversion is correct. We've now expressed the same parabola in two different forms: vertex form and standard form. Each form has its own advantages. Vertex form makes it easy to identify the vertex, while standard form is useful for other purposes, such as finding the y-intercept or using the quadratic formula. Understanding how to convert between these forms is a valuable skill in algebra.

Conclusion: Mastering the Quadratic Equation

So, guys, we've journeyed through Jessica's quadratic equation problem, and what a ride it's been! We've seen how she skillfully used the vertex form of a quadratic equation to find the specific equation that represents a parabola passing through a given point and having a given vertex. We meticulously analyzed each step of her work, making sure her logic was sound and her calculations were accurate. And most importantly, we didn't just stop at finding the solution; we verified it! We plugged the value of 'a' back into the equation and confirmed that it indeed satisfied the given conditions. This verification step is crucial in mathematics – it's the ultimate way to ensure that your answer is correct. But we didn't stop there either! We went the extra mile and converted the equation from vertex form to standard form. This allowed us to see the same parabola from a different perspective and to appreciate the connections between the different forms of quadratic equations. We saw how the 'a' value remains consistent across both forms, providing a nice check on our work. Through this exploration, we've not only solved a specific problem but also reinforced our understanding of quadratic equations in general. We've seen how powerful the vertex form can be, how to manipulate equations algebraically, and how important it is to verify our solutions. These are all valuable skills that will serve you well in your mathematical journey. Remember, quadratic equations are more than just abstract formulas; they represent real-world phenomena, from the trajectory of a ball to the shape of a satellite dish. By mastering these equations, you're unlocking a powerful tool for understanding the world around you. So, keep practicing, keep exploring, and keep challenging yourselves with new problems. And who knows, maybe one day you'll be the one solving quadratic equation mysteries like a mathematical detective! Keep up the awesome work, and I'll catch you in the next mathematical adventure! Remember, math is not just about getting the right answer; it's about the journey of discovery and the thrill of solving a puzzle. So, embrace the challenge, enjoy the process, and never stop learning!