Problem 32 - Precalculus: Functions and Graphs

Question

Exer. 31-34: Examine the expression for the given set of data points of the form \((x, y) .\) Find the constant of variation and a formula that describes how \(y\) varles with respect to \(x .\) $$\begin{aligned} &x y ;\\{(0.2,-26.5),(0.4,-13.25),(0.8,-6.625)\\\ &(1.6,-3.3125),(3.2,-1.65625)\\} \end{aligned}$$

Step-by-Step Solution

Answer

The constant of variation is \(-5.3\) and the formula is \(y = \frac{-5.3}{x}\).

Step 1: Observe Pattern

Begin by observing that the relationship appears to be a type of variation. When each value of \(x\) is doubled, the value of \(y\) is halved. This suggests an inverse variation where \(y\) varies inversely with \(x\).

Step 2: Identify the Variation

In an inverse variation, the product \(xy = k\) is constant for all data points. Check if \(xy\) is constant for each pair.

Step 3: Calculate Constant of Variation

Compute the product \(xy\) for each pair:- For \((0.2, -26.5)\), \(xy = 0.2(-26.5) = -5.3\)- For \((0.4, -13.25)\), \(xy = 0.4(-13.25) = -5.3\)- For \((0.8, -6.625)\), \(xy = 0.8(-6.625) = -5.3\)- For \((1.6, -3.3125)\), \(xy = 1.6(-3.3125) = -5.3\)- For \((3.2, -1.65625)\), \(xy = 3.2(-1.65625) = -5.3\)Since the product is constant, \(k = -5.3\).

Step 4: Write the Equation

Based on the inverse variation where the product \(xy = k\) is constant, we write the equation: \(xy = -5.3\). So, the variation equation is \(y = \frac{-5.3}{x}\).

Step 5: Verify the Equation

To ensure accuracy, substitute a pair (for example, \(x = 0.4\)) back into the equation: \(y = \frac{-5.3}{0.4}\) simplifies to \(-13.25\), which matches the data set, confirming the equation \(y = \frac{-5.3}{x}\).

Key Concepts

Inverse Variation

Inverse variation is a fascinating mathematical concept. It occurs when two variables are related in such a way that their product is a constant. In simpler terms, as one variable increases, the other decreases, and vice versa. This relationship can be expressed as \(xy = k\), where \(k\) is a constant.

What does this mean practically? Imagine that if you double one quantity, the other halves. It's like they are doing a balancing act to keep their product steady.

• For our data set, we noticed that when values of \(x\) doubled, \(y\) values halved.
• This suggested an inverse relationship in the arrangement of data.

Identifying inverse variation helps us understand and predict how one variable changes with the other. It is crucial to check that the product \(xy\) remains constant across different data points to conclude an inverse variation.

Variation Equations

Variation equations are mathematical expressions that describe how two variables are related to each other. They are of several types, but we focus here on inverse variation equations, where \(y\) varies inversely with \(x\).

In an inverse variation equation, the relationship is described as \(y = \frac{k}{x}\), where \(k\) is the constant of variation and \(x\) must not be zero, as division by zero is not possible in mathematics.

For the given exercise, we computed that \(k = -5.3\). This was determined because for all data points, their products \(xy\) equaled \(-5.3\). We wrote the variation equation as \(y = \frac{-5.3}{x}\).

• This formula tells us how to find \(y\) for any given \(x\) value within the context of our data.
• We can use this equation to predict other pairs (\(x, y\)) that would satisfy this inverse variation relationship.

Data Points Analysis

Analyzing data points is essential to uncover the relationship between variables and find constants like \(k\). By examining how pairs of data fit into an inverse variation model, we can draw conclusions about the variables' behavior.

Let's look at this step-by-step:

• First, we observed that doubling \(x\) resulted in halving \(y\). This initial observation hinted at an inverse variation.
• Next, by multiplying each \(x\) and \(y\) together, we discovered that their products were consistently equal to \(-5.3\).

This consistent product confirmed that we indeed had inverse variation, with \(k\) being \(-5.3\).

By plugging in one of the data points, like \((0.4, -13.25)\), into our variation equation \(y = \frac{-5.3}{x}\), we verified that everything matched, adding further credibility to the equation. This type of analysis helps ensure that the model is not only theoretically sound, but accurate according to the data at hand.