Find The Function With Minimum At (4,-3)

Hey guys! Today, we're diving deep into the wild world of quadratic functions and how to find their absolute lowest points, also known as the minimum. We're tackling a super common math problem that pops up a lot: figuring out which function's graph has its minimum vertex at a specific point. In this case, we're on the hunt for the function whose graph kisses the point (4,-3) on the coordinate plane. This isn't just about memorizing formulas; it's about understanding the underlying structure of parabolas and how the coefficients of a quadratic equation dictate their shape and position. So, grab your calculators, maybe a snack, and let's break down these options, shall we? We'll be looking at four different functions, and it's our job to meticulously analyze each one to see which one fits the bill. Get ready to flex those math muscles because we're going to unravel this puzzle piece by piece.

Understanding Quadratic Functions and Their Minimums

Alright, let's get down to business. What exactly are we dealing with when we talk about quadratic functions? Simply put, they're functions that have the highest power of x as 2. Their general form is often written as $f(x) = ax^2 + bx + c$, where 'a', 'b', and 'c' are just numbers. The magic of these functions lies in their graphs, which are always U-shaped curves called parabolas. Now, these parabolas can either open upwards or downwards. If the coefficient 'a' (the number in front of $x^2$) is positive, the parabola opens upwards, and guess what? It has a lowest point, a minimum, which is our target! If 'a' is negative, the parabola opens downwards, and it has a highest point, a maximum. Our problem specifically asks for a minimum, so we're immediately looking for functions where 'a' is positive. This is a crucial first filter, guys. It cuts our options in half right off the bat. The minimum (or maximum) point of a parabola is called its vertex. The coordinates of the vertex are incredibly important because they tell us the exact location of this extreme point. Our target vertex is (4, -3). This means the x-coordinate of the minimum is 4, and the y-coordinate (the function's value at that x) is -3. We need to find the function that satisfies both these conditions simultaneously. There are a couple of ways to find the vertex of a parabola. One common method is using the formula for the x-coordinate of the vertex: $x = -b / (2a)$. Once we have the x-coordinate, we can plug it back into the function to find the corresponding y-coordinate. Another approach, especially useful when dealing with multiple-choice questions like this, is to check if the given vertex coordinates satisfy the function's equation. We'll be using both these techniques to systematically eliminate the incorrect options and pinpoint the right one. Remember, understanding why a method works is just as important as knowing how to apply it. So, let's keep our minds engaged and our calculators handy as we embark on this analytical journey.

Analyzing Option A: f(x) = -rac{1}{2} x^2 + 4x - 11

First up on our investigative tour is option A: f(x) = -rac{1}{2} x^2 + 4x - 11. The very first thing we should notice here, guys, is the coefficient of the $x^2$ term. It's -rac{1}{2}. Since this value is negative, this parabola opens downwards. What does that mean for us? It means this function has a maximum, not a minimum. Our problem is specifically asking for a function with a minimum at (4,-3). Therefore, option A is immediately ruled out. It's like trying to fit a square peg into a round hole; it just doesn't work. Even if the vertex happened to be at (4,-3), it would be a maximum, not a minimum. It's super important to pay attention to these leading coefficients because they dictate the overall behavior of the parabola. For a minimum to exist, the parabola must open upwards, meaning the 'a' value needs to be positive. So, while we could technically calculate the vertex of this function to see where its maximum is, it's a wasted effort in this particular context because it can't possibly be the answer we're looking for. This initial check saves us valuable time and mental energy. Always look at the sign of the $x^2$ coefficient first when dealing with minimum or maximum problems. It’s your first and most powerful clue. This efficient screening process is key to tackling these kinds of problems without getting bogged down in calculations that are destined to lead nowhere.

Analyzing Option B: $f(x) = -2x^2 + 16x - 35$

Moving on to option B, we have $f(x) = -2x^2 + 16x - 35$. Just like with option A, let's cast our eyes upon the coefficient of the $x^2$ term. Here, it's -2. Again, this is a negative number. What does a negative leading coefficient tell us about the parabola? You guessed it – it opens downwards! This means the function has a maximum point, not a minimum. Since the question is specifically asking for a function whose graph has a minimum located at (4,-3), option B is also incorrect. It's a bit of a pattern emerging, isn't it? The question is designed to test your understanding of how the sign of the leading coefficient affects the parabola's orientation. A downward-opening parabola will never have a minimum value; it extends infinitely downwards. So, even if we were to calculate the vertex of this function, we'd find the location of its maximum, which is not what we need. This is a critical takeaway: always check the sign of the $x^2$ coefficient first. If it's negative, and the question asks for a minimum, you can confidently eliminate that option. This quick check is a huge time-saver and prevents you from getting lost in unnecessary calculations. It’s a fundamental concept in understanding quadratic functions, and recognizing it immediately simplifies the problem considerably. Don't let the other terms distract you; the sign of 'a' is your primary guide here.

Analyzing Option C: f(x) = rac{1}{2} x^2 - 4x + 5

Alright, let's examine option C: f(x) = rac{1}{2} x^2 - 4x + 5. Good news, guys! The coefficient of the $x^2$ term here is rac{1}{2}, which is positive. This means the parabola opens upwards, so this function does have a minimum. Now, we need to check if this minimum occurs at the point (4,-3). We can use the vertex formula to find the x-coordinate of the minimum: $x = -b / (2a)$. In this function, a = rac{1}{2} and $b = -4$. So, x = -(-4) / (2 * rac{1}{2}) = 4 / 1 = 4. This x-coordinate matches the x-coordinate of our target vertex, which is 4. Awesome! Now, let's find the y-coordinate by plugging this x-value back into the function: f(4) = rac{1}{2}(4)^2 - 4(4) + 5 = rac{1}{2}(16) - 16 + 5 = 8 - 16 + 5 = -8 + 5 = -3. Wow, the y-coordinate we calculated is -3, which also matches our target vertex (4, -3). So, option C has a minimum at (4,-3)! This looks like our winner, but let's be thorough and check option D just to be absolutely sure. It's always good practice to verify everything, especially in math, to build confidence in our answer. This process of using the vertex formula and then substituting the x-value back into the function is a robust way to find the exact vertex coordinates for any upward-opening parabola.

Analyzing Option D: $f(x) = 2x^2 - 16x + 35$

Finally, let's check out option D: $f(x) = 2x^2 - 16x + 35$. The coefficient of the $x^2$ term is 2, which is positive. This means the parabola opens upwards and has a minimum. Excellent, this is a candidate! Now, let's find the x-coordinate of its vertex using the formula $x = -b / (2a)$. Here, $a = 2$ and $b = -16$. So, $x = -(-16) / (2 * 2) = 16 / 4 = 4$. The x-coordinate of the minimum is 4, which matches our target vertex's x-coordinate. So far, so good! Now, let's find the y-coordinate by plugging $x = 4$ into the function: $f(4) = 2(4)^2 - 16(4) + 35 = 2(16) - 64 + 35 = 32 - 64 + 35 = -32 + 35 = 3$. Hmm, the y-coordinate we calculated is 3. Our target vertex has a y-coordinate of -3. Since $3 eq -3$, the minimum for option D is located at (4, 3), not (4, -3). Therefore, option D is incorrect. This highlights how precise we need to be. Both options C and D had the correct x-coordinate for the minimum, but only option C had the correct y-coordinate as well. This is why calculating the full vertex coordinates is essential, rather than just stopping after finding the x-value. It's the complete (x,y) pair that defines the vertex location.

Conclusion: The Winning Function!

So, after carefully analyzing all four options, guys, we've reached our conclusion! We systematically eliminated options A and B because their parabolas opened downwards, meaning they had maximums, not minimums. Then, we meticulously calculated the vertex for option C and found that its minimum was indeed located precisely at (4, -3). We also checked option D and found that while its minimum had the correct x-coordinate of 4, its y-coordinate was 3, not -3. Therefore, the only function whose graph has a minimum located at (4, -3) is Option C. It's a fantastic feeling when you work through a problem step-by-step and arrive at the correct answer with confidence. Remember these key takeaways: always check the sign of the leading coefficient ('a') first to determine if a minimum or maximum exists, use the vertex formula $x = -b / (2a)$ to find the x-coordinate of the vertex, and then substitute that x-value back into the function to find the corresponding y-coordinate. Keep practicing these skills, and you'll be a quadratic function whiz in no time! Happy graphing!