Problem 45 - College Algebra

Question

For the following exercises, use a calculator to graph the equation implied by the given variation. \(y\) varies inversely as the square of \(x\) and when \(x=1, y=4\).

Step-by-Step Solution

Answer

Graph \(y = \frac{4}{x^2}\) using the calculator.

Step 1: Understand the Relationship

Since the problem states that \(y\) varies inversely as the square of \(x\), we interpret this as \(y = \frac{k}{x^2}\), where \(k\) is a constant of variation.

Step 2: Find the Constant of Variation

We are given that when \(x = 1\), \(y = 4\). Substitute these values into the equation: \(4 = \frac{k}{1^2}\) or simply \(k = 4\). Thus, the equation becomes \(y = \frac{4}{x^2}\).

Step 3: Graph the Equation

Using a calculator, input the equation \(y = \frac{4}{x^2}\). Set up a range of values for \(x\) around the given point (e.g., from \(-5\) to \(5\)), and observe how \(y\) behaves as \(x\) varies. The graph should show a hyperbola with the curve approaching the x-axis but never touching it.

Key Concepts

Graphing Equations

Graphing an equation involves plotting its points on a coordinate grid to visualize its behavior. When we graph the equation \(y = \frac{4}{x^2}\), we're observing how changes in \(x\) impact \(y\).
In this instance, given the equation is a form of inverse variation, you'll see that the graph is a curve called a hyperbola. To get started, you select certain values of \(x\), both positive and negative, and substitute them into the equation to solve for \(y\). For example:

• If \(x = 1\), then \(y = \frac{4}{1^2} = 4\).
• If \(x = 2\), then \(y = \frac{4}{2^2} = 1\).
• If \(x = 3\), then \(y = \frac{4}{3^2} \approx 0.44\).
• If \(x = -1\), the same calculations apply: \(y = \frac{4}{(-1)^2} = 4\), since squaring a negative number results in a positive.

Finally, plot these points on the grid, and draw the hyperbola which descends steeply to the x-axis but doesn't touch it.

Constant of Variation

The constant of variation, denoted by \(k\), defines the nature and identity of an inverse variation equation. For our equation \(y = \frac{k}{x^2}\), knowing \(k\) is crucial because it specifies how intense the effect of inverse variation is.
Calculating \(k\):- Given that when \(x = 1\), \(y = 4\), you solve for \(k\) as follows:
\[ 4 = \frac{k}{1^2} \Rightarrow k = 4 \]
Therefore, \(k = 4\) is the constant of variation here.Now, with \(k\) identified, the equation describes precisely how "strongly" \(y\) varies inversely with \(x^2\). Each time \(x\) increases or decreases, \(y\) changes accordingly while respecting this constant relationship.

Hyperbola

A hyperbola is a type of curve that appears on the graph of an inverse variation equation like \(y = \frac{4}{x^2}\). It has distinct features:

• Asymptotes: These are lines that the graph approaches but never touches. For this hyperbola, the x-axis (y = 0) serves as a horizontal asymptote.
• Symmetry: The curve is symmetrical concerning the y-axis. This means that plugging \(x\) or \(-x\) into the equation gives the same \(y\) value.

When graphing, you'll notice that as \(x\) moves far from 0, \(y\) becomes very small, indicating that the hyperbola "stretches" out, getting closer to the x-axis. Conversely, as \(x\) nears zero, \(y\) drastically increases, showing the steep rise characteristic of a hyperbola.
Understanding these features allows us to predict and comprehend the behavior of equations involving inverse variation.