Problem 18 - A Graphical Approach to College Algebra

Question

For each plane curve, use a graphing calculator to generate the curve over the interval for the parameter \(t\), in the window specified. Then, find a rectangular equation for the curve. \(x=2 t-1, y=\frac{1}{t},\) for \(t\) in \((-\infty, 0) \cup(0, \infty)\) window: \([-6,6]\) by \([-4,4]\)

Step-by-Step Solution

Answer

The rectangular equation is \(y = \frac{2}{x+1}\) with \(x \neq -1\).

Step 1: Understand the Parametric Equations

We have two parametric equations: \(x = 2t - 1\) and \(y = \frac{1}{t}\). These equations define a plane curve based on the parameter \(t\). The interval given for \(t\) is \(( -\infty, 0 ) \cup ( 0, \infty )\); this means \(t\) can take any value except 0. The window bounds for the graph are \([-6,6]\) for \(x\) and \([-4,4]\) for \(y\).

Step 2: Eliminate the Parameter to Find a Rectangular Equation

To eliminate the parameter \(t\), express \(t\) in terms of \(x\) from one of the parametric equations. From \(x = 2t - 1\), solve for \(t\):\[t = \frac{x + 1}{2}.\]Substitute \(t\) in the equation for \(y\), which gives:\[y = \frac{1}{\left(\frac{x + 1}{2}\right)} = \frac{2}{x + 1}.\]This equation \(y = \frac{2}{x + 1}\) is the rectangular form of the curve.

Step 3: Consider the Domain and Range

The original restriction for \(t\) was \((-\infty, 0) \cup (0, \infty)\). Since \(t\) cannot be zero, observe the rectangular form \(y = \frac{2}{x + 1}\). The denominator \(x + 1\) cannot be zero, so \(x eq -1\). The domain of the rectangular equation is therefore \(x \in (-\infty, -1) \cup (-1, \infty)\).

Step 4: Graph the Rectangular Equation

Using a graphing calculator, input the rectangular equation \(y = \frac{2}{x + 1}\) and adjust the viewing window to \([-6, 6]\) by \([-4, 4]\). Remember that there will be a vertical asymptote at \(x = -1\) because the function is undefined there.

Key Concepts

Rectangular equation

Parametric equations can often be converted to rectangular equations, which are more familiar to many students. A rectangular equation expresses a relationship between two variables, usually x and y, without involving a third parameter.

In the given exercise, the parametric equations are:

• \(x = 2t - 1\)
• \(y = \frac{1}{t}\)

To find the rectangular equation, we need to eliminate the parameter \(t\).
First, solve for \(t\) in terms of \(x\):

• \(t = \frac{x + 1}{2}\)

Then, substitute \(t\) in the \(y\) equation:

• \(y = \frac{1}{\left(\frac{x + 1}{2}\right)} = \frac{2}{x + 1}\)

This transformation results in the rectangular equation \(y = \frac{2}{x + 1}\). Understanding this kind of conversion is key in tackling parametric equations, as it allows the use of traditional graphing techniques.

Graphing calculator

A graphing calculator is an essential tool for visualizing equations, including both parametric and rectangular forms. It can handle complex calculations and produce high-quality graphs, making it easier to understand the relationships represented by the equations.

For this task, the graphing calculator helps us visualize the curve represented by the rectangular equation \(y = \frac{2}{x+1}\). When inputting this equation, set the viewing window to the specified bounds:

• \(x\) from \([-6, 6]\)
• \(y\) from \([-4, 4]\)

These settings ensure you see the relevant portions of the graph and understand its behavior across the given domain.

The graphing calculator will show a hyperbolic curve with a vertical asymptote at \(x = -1\), highlighting points where the function is undefined. Visually interpreting these graphs helps solidify your understanding of the equation's characteristics.

Domain and range

Understanding the domain and range of an equation is crucial, as it tells you where the equation is valid or meaningful. For both parametric and rectangular equations, constraints on the variables can significantly affect the solutions.

Initially, the parameter \(t\) was restricted to \((-\infty, 0) \cup (0, \infty)\), meaning \(t\) could never be zero. After conversion, the rectangular equation becomes \(y = \frac{2}{x + 1}\).

The domain of this rectangular equation is affected by the restriction that the denominator cannot be zero, which results in:

• \(x eq -1\)

Thus, the domain for \(x\) becomes \((-\infty, -1) \cup (-1, \infty)\). The function has a vertical asymptote at \(x = -1\), where it is undefined.

The range of the equation is all real numbers except zero – \((-\infty, 0) \cup (0, \infty)\) – because \(y = \frac{2}{x + 1}\) can take any value except zero, which corresponds to no real value of \(x\) yielding \(y = 0\). Understanding these limitations helps in sketching accurate graphs and solving equations effectively.