Inverse Variation: Find The Equation Through (4,5) And (10,2)

Hey guys! Today we're diving deep into the world of inverse variation, and trust me, it's going to be a blast. We're tackling a specific problem: figuring out which equation represents an inverse variation function that gracefully passes through the points $(4,5)$ and $(10,2)$. Stick around, because we're going to break this down step-by-step, making sure you totally get it. We'll explore what inverse variation means, how to use those given points to unlock the secret of the equation, and then we'll put our detective hats on to find the right answer among the options. So, grab your notebooks, maybe a snack, and let's get this mathematical party started! We're not just solving a problem; we're building your understanding from the ground up. Get ready to feel like a math whiz!

Understanding Inverse Variation: The Core Concept

Alright, let's get down to the nitty-gritty of inverse variation. What does it even mean for two variables, say x and y, to be in inverse variation? It's super straightforward once you wrap your head around it. When two quantities are in inverse variation, it means that as one quantity goes up, the other quantity goes down, and vice versa. Think about it like this: the more you study, the less time you have for video games (bummer, I know!). Or, the faster you drive, the less time it takes to get to your destination. This inverse relationship is mathematically represented by a fundamental equation. If y varies inversely with x, then their product is always a constant. We call this constant of variation, and we usually represent it with the letter k. So, the golden equation for inverse variation is:

y = k/x

Or, rearranging it, you'll often see it written as:

xy = k

This k is the key! It's the secret sauce that links x and y in an inverse variation. Finding the value of k is usually the first big step in solving any inverse variation problem. Once you know k, you can plug it back into the general equation ($y = k/x$) to get the specific equation that describes the relationship between x and y for that particular scenario. So, remember: inverse variation means the product of the two variables is a constant ($k$). This constant k is what we need to find using the information given in the problem. It's the unique fingerprint of any inverse variation relationship.

Using the Given Points to Find 'k'

Now, we're armed with the definition of inverse variation and its signature equation, $y = k/x$. The problem has handed us a massive clue: our inverse variation function passes through two specific points, $(4,5)$ and $(10,2)$. These aren't just random numbers, guys; they are pairs of x and y values that satisfy the equation. This means we can substitute these coordinate pairs into our inverse variation equation to solve for that all-important constant, k.

Let's take the first point, $(4,5)$. Here, $x=4$ and $y=5$. Plugging these values into $xy = k$, we get:

$$(4)(5) = k$$

$$20 = k$$

Boom! We found a value for k. But hold up, we have a second point, $(10,2)$. A true inverse variation function should satisfy the relationship for all points it passes through. So, let's check if this second point gives us the same k value. Here, $x=10$ and $y=2$. Plugging these into $xy = k$:

$$(10)(2) = k$$

$$20 = k$$

Fantastic! Both points give us the exact same value for k, which is 20. This confirms that these points are indeed part of an inverse variation relationship, and the constant of variation for this specific function is k = 20. If we had gotten different k values from the two points, it would mean that the function isn't a true inverse variation, or there was a mistake in our calculations. But here, everything aligns perfectly, giving us confidence in our findings. This consistency is what makes mathematical models reliable; they hold true across all applicable data points.

Constructing the Specific Inverse Variation Equation

We've done the heavy lifting, guys! We know that the general form of an inverse variation equation is $y = k/x$, and we've successfully determined that the constant of variation, k, for the function passing through $(4,5)$ and $(10,2)$ is 20. Now comes the satisfying part: plugging this value of k back into the general equation to get the specific equation that represents this particular inverse variation.

So, we take our general equation:

$$y = \frac{k}{x}$$

And we substitute $k=20$:

$$y = \frac{20}{x}$$

And there you have it! The equation that represents the inverse variation function passing through the points $(4,5)$ and $(10,2)$ is $y = \frac{20}{x}$. This equation perfectly encapsulates the relationship between x and y for any point on this specific curve. It tells us that as x increases, y decreases proportionally, maintaining a constant product of 20. It's like the unique DNA of this particular inverse relationship. We've transformed a general mathematical concept into a specific, applicable formula using just two data points.

Evaluating the Options: Finding the Correct Match

We've worked hard to derive the equation $y = \frac{20}{x}$. Now, it's time to compare our findings with the multiple-choice options provided in the question. This is where we see if our detective work paid off! Let's list the options and see which one matches our derived equation:

• A. $y=\frac{5}{4} x$ This equation represents a direct variation, where y is directly proportional to x. If we plug in $(4,5)$, we get $5 = \frac{5}{4}(4)$, which simplifies to $5=5$. This works for the first point. However, if we plug in $(10,2)$, we get $2 = \frac{5}{4}(10)$, which is $2 = \frac{50}{4}$ or $2 = 12.5$. This is false. So, option A is incorrect.

• B. $y=20 x$ This equation also represents a direct variation. Let's check our points. For $(4,5)$: $5 = 20(4)$, which is $5=80$. This is false. Option B is incorrect.

• C. $y=\frac{20}{x}$ This is the equation we derived! Let's double-check. For $(4,5)$: $5 = \frac{20}{4}$, which simplifies to $5=5$. True! For $(10,2)$: $2 = \frac{20}{10}$, which simplifies to $2=2$. True! This option perfectly matches our derived equation and satisfies both points. This is our winner!

• D. (Assuming there was a D option, but since it was not provided, we proceed with the options given.)

See? By systematically applying the definition of inverse variation and using the given points, we were able to pinpoint the correct equation. It's all about understanding the underlying principles and performing the calculations accurately. The process of elimination using the given points is a crucial step in confirming our answer and ensuring we haven't made any mistakes along the way. It's a satisfying confirmation when our calculated equation matches one of the choices perfectly.

Conclusion: Mastering Inverse Variation

So there you have it, math enthusiasts! We've successfully navigated the concept of inverse variation, learned how to identify its defining equation ($y = k/x$), and crucially, how to use given points to solve for the constant of variation, k. By applying these principles to the specific problem of finding the equation passing through $(4,5)$ and $(10,2)$, we not only found k to be 20 but also arrived at the correct equation: $y = \frac{20}{x}$. We then confidently identified this as option C among the choices.

Remember, the core idea behind inverse variation is that the product of the two variables ($x$ and $y$) remains constant ($k$). When you're given points, you use them to find that specific constant. It’s like finding the unique fingerprint of that particular relationship. This skill is super valuable, not just for passing tests, but for understanding how quantities can relate to each other in the real world – think about how pressure and volume of a gas are related, or how speed and travel time.

Keep practicing, guys! The more you work through these types of problems, the more intuitive inverse variation will become. Don't be afraid to jot down the formulas, check your work, and most importantly, understand why the math works the way it does. You've got this! Keep exploring the fascinating world of mathematics, one equation at a time. Stay curious and keep those math skills sharp!