Find The Toy's Height Function: A Geometric Sequence

Hey guys! Ever wondered how to describe the path of a bouncing toy mathematically? We're diving deep into the cool world of geometric sequences today, and we've got a super fun problem to break down. Imagine a toy that bounces, reaching specific heights at each peak. We're talking about its first peak hitting 64 inches, the second peak at 48 inches, and the third peak at 36 inches. Your mission, should you choose to accept it, is to figure out the explicit function that perfectly represents this geometric sequence of heights. This isn't just about numbers; it's about understanding patterns and how they can be expressed with elegant mathematical formulas. So, grab your notebooks, sharpen those pencils, and let's get this mathematical adventure started!

Unpacking the Bouncing Toy Problem: Identifying a Geometric Sequence

Alright, let's get down to business, folks. The first thing we need to do when faced with a problem like this is to carefully identify the type of sequence we're dealing with. We're given the heights of a bouncing toy at its successive peaks: 64 inches, 48 inches, and 36 inches. To determine if this is a geometric sequence, we need to check if there's a constant ratio between consecutive terms. Remember, in a geometric sequence, each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Let's do the math, shall we?

For our sequence of heights, let's denote the first term ($a_1$) as 64, the second term ($a_2$) as 48, and the third term ($a_3$) as 36.

• Ratio between the second and first term: $\frac{a_2}{a_1} = \frac{48}{64}$. We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 16. So, $\frac{48 \div 16}{64 \div 16} = \frac{3}{4}$.
• Ratio between the third and second term: $\frac{a_3}{a_2} = \frac{36}{48}$. Again, let's simplify this fraction. The greatest common divisor of 36 and 48 is 12. So, $\frac{36 \div 12}{48 \div 12} = \frac{3}{4}$.

See that? The ratio between consecutive terms is consistently $\frac{3}{4}$. This confirms that the heights of the bouncing toy form a geometric sequence with a common ratio ($r$) of $\frac{3}{4}$. This is a crucial step, guys, because understanding the nature of the sequence dictates the type of formula we'll use to find the explicit function. If the ratios weren't constant, we'd be looking at a different type of sequence altogether, and our approach would change dramatically. So, high five for recognizing the geometric pattern!

Deriving the Explicit Function for a Geometric Sequence

Now that we've established that our toy's heights follow a geometric sequence, let's talk about finding the explicit function. An explicit function, in the context of sequences, is a formula that allows you to find any term in the sequence directly, without having to calculate all the preceding terms. For a geometric sequence, the general form of the explicit function is given by: $a_n = a_1 \cdot r^{(n-1)}$, where:

• $a_n$ represents the nth term of the sequence.
• $a_1$ is the first term of the sequence.
• $r$ is the common ratio.
• $n$ is the term number (which usually corresponds to our input variable, like $x$ in $f(x)$).

In our bouncing toy scenario:

• The first term, $a_1$, is 64 inches.
• The common ratio, $r$, is $\frac{3}{4}$.

So, to create our explicit function, we just need to substitute these values into the general formula. Let's use $x$ to represent the term number (or the peak number), so $n=x$. Our function will be denoted as $f(x)$ instead of $a_n$. Thus, the explicit function representing the heights of the toy is:

$f(x) = 64 \cdot \left(\frac{3}{4}\right)^{(x-1)}$

This formula is super powerful, guys. It tells you the height of the toy at any peak just by plugging in the peak number. For instance, if you wanted to know the height at the 5th peak, you'd just calculate $f(5) = 64 \cdot \left(\frac{3}{4}\right)^{(5-1)} = 64 \cdot \left(\frac{3}{4}\right)^4$. We'll explore some examples of this later, but the core idea is that this formula encapsulates the entire bouncing pattern.

Evaluating the Proposed Function and Verifying Our Result

Okay, so we've derived our explicit function: $f(x) = 64 \cdot \left(\frac{3}{4}\right)^{(x-1)}$. Now, it's super important to verify that this function actually works with the given data. The problem stated the heights for the first three peaks. Let's plug in $x=1$, $x=2$, and $x=3$ into our function and see if we get the expected heights.

• For the first peak ($x=1$): $f(1) = 64 \cdot \left(\frac{3}{4}\right)^{(1-1)}$ $f(1) = 64 \cdot \left(\frac{3}{4}\right)^0$ Since any non-zero number raised to the power of 0 is 1, we have: $f(1) = 64 \cdot 1 = \mathbf{64}$ inches. This matches the given first peak height. Perfect!

• For the second peak ($x=2$): $f(2) = 64 \cdot \left(\frac{3}{4}\right)^{(2-1)}$ $f(2) = 64 \cdot \left(\frac{3}{4}\right)^1$ $f(2) = 64 \cdot \frac{3}{4}$ To calculate this, we can multiply 64 by 3 and then divide by 4, or divide 64 by 4 first, which is 16, and then multiply by 3. $f(2) = 16 \cdot 3 = \mathbf{48}$ inches. This matches the given second peak height. Awesome!

• For the third peak ($x=3$): $f(3) = 64 \cdot \left(\frac{3}{4}\right)^{(3-1)}$ $f(3) = 64 \cdot \left(\frac{3}{4}\right)^2$ $f(3) = 64 \cdot \left(\frac{3^2}{4^2}\right)$ $f(3) = 64 \cdot \left(\frac{9}{16}\right)$ Again, we can simplify before multiplying. Notice that 64 is a multiple of 16 ($64 = 4 \times 16$). $f(3) = \frac{64}{1} \cdot \frac{9}{16} = \frac{64}{16} \cdot 9 = 4 \cdot 9 = \mathbf{36}$ inches. This matches the given third peak height. Fantastic!

So, our derived function $f(x) = 64 \cdot \left(\frac{3}{4}\right)^{(x-1)}$ correctly predicts all the given heights. This confirms that our understanding of geometric sequences and the formula for an explicit function is spot on. It’s always good practice to check your work, especially when dealing with mathematical models, guys. It ensures accuracy and builds confidence in your problem-solving abilities!

Comparing with the Provided Option and Finalizing the Answer

Now, let's take a look at the option provided in the question: $f(x)=64\left(\frac{5}{6}\right)^{x-1}$. We need to determine if this is the correct explicit function for our toy's bouncing heights. We've already found our derived function to be $f(x) = 64 \cdot \left(\frac{3}{4}\right)^{(x-1)}$. Let's compare the two.

Both functions share the same first term, $a_1 = 64$, which is great. This means they both start at the correct initial height. However, the common ratios are different:

• Our derived function has a common ratio $r = \frac{3}{4}$.
• The provided option has a common ratio $r = \frac{5}{6}$.

Since the common ratios are different, these two functions will produce different sequences of heights after the first peak. Let's quickly check the second peak using the provided function $f(x)=64\left(\frac{5}{6}\right)^{x-1}$:

• For the second peak ($x=2$) using the provided function: $f(2) = 64 \cdot \left(\frac{5}{6}\right)^{(2-1)}$ $f(2) = 64 \cdot \left(\frac{5}{6}\right)^1$ $f(2) = 64 \cdot \frac{5}{6}$ $f(2) = \frac{320}{6}$ $f(2) = \frac{160}{3} \approx 53.33$ inches.

This value, approximately 53.33 inches, is not equal to the given second peak height of 48 inches. Therefore, the provided function $f(x)=64\left(\frac{5}{6}\right)^{x-1}$ is incorrect.

Our carefully derived function, $f(x) = 64 \cdot \left(\frac{3}{4}\right)^{(x-1)}$, accurately represents the geometric sequence of the toy's heights, as it produces the correct heights for the first, second, and third peaks. So, the explicit function that represents the geometric sequence of the heights of the toy is indeed $f(x) = 64 \cdot \left(\frac{3}{4}\right)^{(x-1)}$. It's awesome how a small difference in the common ratio can lead to such different outcomes! Keep practicing, and you'll master identifying and working with these sequences in no time, guys!

Practical Applications and Further Exploration

This problem, while seemingly simple, touches upon fundamental concepts in mathematics with real-world applications. Understanding geometric sequences is not just for textbook problems; it's a key to modeling various phenomena. Think about it: radioactive decay, compound interest, population growth (or decline), and even the spread of information online often follow geometric patterns. For instance, if a certain percentage of a population is vaccinated each year, the number of unvaccinated individuals might decrease geometrically. Or, consider how a popular social media post might initially reach a certain number of people, and then each subsequent share reaches a fraction of those it reached before – that's a geometric decay!

The explicit function we found, $f(x) = 64 \cdot \left(\frac{3}{4}\right)^{(x-1)}$, allows us to predict the height of the toy at any future peak without needing to calculate each one individually. This predictive power is incredibly useful. Imagine you want to know how high the toy will bounce on its 10th peak. You simply plug $x=10$ into the formula: $f(10) = 64 \cdot \left(\frac{3}{4}\right)^{(10-1)} = 64 \cdot \left(\frac{3}{4}\right)^9$. Calculating this would give you a precise prediction. This ability to model and predict based on initial conditions and a rate of change (the common ratio) is what makes mathematics such a powerful tool for understanding the world around us.

For further exploration, guys, consider these points: What happens to the height as $x$ becomes very large? In our case, since the common ratio $\frac{3}{4}$ is less than 1, the heights will get progressively smaller, approaching zero. This makes physical sense – a toy won't bounce forever! You could also explore scenarios where the common ratio is greater than 1 (leading to exponential growth) or where the initial term is different. Experimenting with different values for $a_1$ and $r$ will deepen your understanding of how these parameters influence the sequence. You might even find yourself creating your own bouncing toy scenarios and deriving the functions to model them. Keep exploring, keep questioning, and keep that mathematical curiosity alive!