Finding The Minimum Of A Quadratic Function At (4, -3)

Quadratic functions are super interesting, aren't they? They draw these beautiful U-shaped curves called parabolas. These parabolas can either open upwards, giving us a lowest point (a minimum), or open downwards, giving us a highest point (a maximum). Today, we're diving deep into which graph of which function has a minimum located at (4, -3). It's like a fun detective mission where we use a bit of algebra to uncover the secret! Understanding these functions isn't just for math class; it's incredibly useful in real-world scenarios, from physics to economics, helping us figure out optimal values, whether it's the lowest cost or the highest profit. So, let's roll up our sleeves and explore how we can pinpoint that exact minimum point for a given quadratic equation. We'll look at several examples and break down the process step by step, making sure you grasp the core concepts behind identifying a parabola's vertex. Get ready to transform those abstract formulas into clear, actionable insights!

Understanding Quadratic Functions and Their Vertices

To begin our quest to find the minimum of a quadratic function at (4, -3), we first need to understand what quadratic functions are all about. A quadratic function is any function that can be written in the standard form: f(x) = ax² + bx + c, where 'a', 'b', and 'c' are constants, and 'a' cannot be zero. The 'a' coefficient is super important because it tells us two crucial things about our parabola. First, if 'a' is positive (a > 0), the parabola opens upwards, like a smile, and it will have a minimum point. This minimum point is the lowest point on the entire graph. Second, if 'a' is negative (a < 0), the parabola opens downwards, like a frown, and it will have a maximum point, which is the highest point. Since our goal is to find a minimum located at (4, -3), we can immediately eliminate any function where 'a' is negative. This narrows down our search considerably, making our detective work much easier and more efficient!

The special point where the parabola changes direction – either from decreasing to increasing (for a minimum) or increasing to decreasing (for a maximum) – is called the vertex. The vertex is precisely what we're looking for, specifically a minimum vertex. There's a fantastic formula that helps us find the x-coordinate of this vertex: x = -b / (2a). Once we have the x-coordinate, finding the y-coordinate is simply a matter of plugging that x-value back into the original quadratic function: y = f(x). So, if we want to determine which graph of which function has a minimum located at (4, -3), our strategy will be to check each function. We'll first ensure its 'a' value is positive, then calculate its vertex using the formula, and finally compare it to our target point (4, -3). This systematic approach ensures we don't miss any steps and arrive at the correct answer confidently. This process is fundamental in algebra and pre-calculus, providing the foundation for understanding optimization problems across various scientific and engineering disciplines. Knowing how to quickly identify and calculate the vertex can save you a lot of time and effort when dealing with quadratic equations, making you a quadratic function pro in no time!

Analyzing the Given Functions

Now, let's put our knowledge to the test and analyze each of the given quadratic functions to find the minimum of a quadratic function at (4, -3). Remember our first filter: for a function to have a minimum, its 'a' coefficient must be positive. If 'a' is negative, we know it will have a maximum, so we can disregard it for our current goal. Let's go through each option one by one, carefully examining its potential.

First up, we have f(x) = -3x² + 4x - 11. Here, the 'a' value is -3. Since -3 is a negative number, this parabola opens downwards, meaning it has a maximum point, not a minimum. So, this function is immediately out of the running. No need to calculate its vertex for a minimum! This quick check saves us valuable time and effort, allowing us to focus on the functions that actually meet our criteria.

Next, consider f(x) = -3x² + 16x - 35. Again, the 'a' value is -3, which is negative. Just like the previous function, this parabola will open downwards and have a maximum point. Therefore, it cannot have a minimum at (4, -3) or anywhere else. Another function swiftly eliminated from our list of candidates! It's satisfying how just looking at one number can tell us so much about the overall shape and behavior of the graph. This initial check is a powerful tool in your mathematical toolkit.

Moving on to f(x) = -4x² + 4x + 5. The 'a' value here is -4. Once more, this is a negative number, indicating that this parabola also opens downwards and will possess a maximum point. So, this function, too, is not the one we're looking for. It's interesting how many of these options are designed to have maximums, perhaps to test our understanding of this fundamental rule! This reinforces the importance of always checking the 'a' coefficient first, as it's the gatekeeper to whether a minimum or maximum even exists.

Finally, we arrive at f(x) = 2x² + 11x + 35. Aha! Here, the 'a' value is 2, which is a positive number. This is great news because it means this parabola opens upwards and does have a minimum point. This function is our only potential candidate, and now we need to perform the detailed calculations to see if its vertex truly matches (4, -3). This is where the real work begins, but we've already done a fantastic job narrowing down our choices. The positive 'a' value gives us the green light to proceed with the vertex formula, confident that we are searching for a minimum as requested. Without this initial positive 'a' value, any subsequent calculations for a minimum would be pointless, demonstrating the efficiency of our systematic approach.

Step-by-Step Breakdown: The Quest for (4, -3)

Now that we've narrowed our search to f(x) = 2x² + 11x + 35, it's time to dive into the detailed calculations to confirm if its minimum is indeed located at (4, -3). This is the crucial stage where we apply the vertex formula we discussed earlier. Remember, for a quadratic function in the form f(x) = ax² + bx + c, the x-coordinate of the vertex is given by the formula x = -b / (2a). Once we find this x-value, we'll plug it back into the original function to get the corresponding y-coordinate, thus revealing the vertex point. Let's break it down step by step with our potential candidate.

For the function f(x) = 2x² + 11x + 35, we can identify our coefficients: 'a' = 2, 'b' = 11, and 'c' = 35. Since 'a' is positive (a = 2), we've already established that this parabola opens upwards and has a minimum. Now, let's find the x-coordinate of this minimum point using our formula: x = -b / (2a). Plugging in our values, we get: x = - (11) / (2 * 2). This simplifies to x = -11 / 4. So, the x-coordinate of the vertex for this function is -11/4, or -2.75. Immediately, we can see a problem. We were looking for a minimum at an x-coordinate of 4, but this function's vertex has an x-coordinate of -11/4. This means that f(x) = 2x² + 11x + 35 does not have a minimum located at (4, -3). It's close, in the sense that it has a minimum, but not at the specific point we're targeting. This illustrates the importance of precision in mathematics; even if the 'a' value is correct, the vertex coordinates must match exactly.

It appears that none of the provided functions actually have a minimum at (4, -3) based on our analysis. This is a vital outcome of our investigation. While f(x) = 2x² + 11x + 35 was the only one that met the minimum criteria (a > 0), its calculated vertex was at (-11/4, f(-11/4)), not (4, -3). Let's quickly calculate the y-coordinate for completeness, even though we know the x-coordinate is already wrong. If x = -11/4, then f(-11/4) = 2(-11/4)² + 11(-11/4) + 35 = 2(121/16) - 121/4 + 35 = 121/8 - 242/8 + 280/8 = (121 - 242 + 280)/8 = 159/8. So, the vertex is at (-11/4, 159/8), which is clearly not (4, -3). This thorough examination confirms that none of the given options are the answer to which graph of which function has a minimum located at (4, -3). This highlights that sometimes, the answer is that none of the given options fit the criteria, which is a perfectly valid and important conclusion. Our systematic approach allowed us to confidently arrive at this conclusion, providing clear evidence for each step of the process. It's a great example of how mathematical reasoning helps us eliminate possibilities and pinpoint precise answers, even if that answer is