Problem 58 - College Algebra with Corequisite Support

Question

For the following exercises, express the equation for the hyperbola as two functions, with \(y\) as a function of \(x\). Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. \(\frac{(x-2)^{2}}{16}-\frac{(y+3)^{2}}{25}=1\)

Step-by-Step Solution

Answer

The two functions are \( y_1 = -3 + \sqrt{\frac{25(x-2)^2}{16} - 25} \) and \( y_2 = -3 - \sqrt{\frac{25(x-2)^2}{16} - 25} \).

Step 1: Identify the Standard Form

The given equation \( \frac{(x-2)^2}{16} - \frac{(y+3)^2}{25} = 1 \) is in the standard form of a hyperbola \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). For this equation, \( h = 2 \), \( k = -3 \), \( a^2 = 16 \) (so \( a = 4 \)), and \( b^2 = 25 \) (so \( b = 5 \)).

Step 2: Rearrange to Solve for \( y \)

Starting from the equation \( \frac{(x-2)^2}{16} - \frac{(y+3)^2}{25} = 1 \), isolate \( \frac{(y+3)^2}{25} \) on one side: \( \frac{(y+3)^2}{25} = \frac{(x-2)^2}{16} - 1 \).

Step 3: Multiply through by 25

Clear the fraction by multiplying all terms by 25: \( (y+3)^2 = 25 \left( \frac{(x-2)^2}{16} - 1 \right) \).

Step 4: Simplify and Isolate the Square Root

This results in \( (y+3)^2 = \frac{25}{16}(x-2)^2 - 25 \). Take the square root of both sides: \( y + 3 = \pm \sqrt{\frac{25(x-2)^2}{16} - 25} \).

Step 5: Solve for \( y \)

Isolate \( y \) by subtracting 3 from both sides: \( y = -3 \pm \sqrt{\frac{25(x-2)^2}{16} - 25} \). Now you have two functions: \( y_1 = -3 + \sqrt{\frac{25(x-2)^2}{16} - 25} \) and \( y_2 = -3 - \sqrt{\frac{25(x-2)^2}{16} - 25} \).

Step 6: Use a Graphing Calculator

Enter both functions \( y_1(x) \) and \( y_2(x) \) into a graphing calculator. Adjust the viewing window to capture the key features of the hyperbola, such as vertices and asymptotes, to visualize how the functions represent the upper and lower portions of the hyperbola.

Key Concepts

Standard Form of Hyperbola

In order to work with hyperbolas, it's essential to understand their standard form. The given equation \( \frac{(x-2)^2}{16} - \frac{(y+3)^2}{25} = 1 \) is in a format that resembles \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). This form helps in identifying the properties and orientation of the hyperbola.

• The center of the hyperbola is at \((h, k)\), which is \((2, -3)\) in this case.
• The values \(a\) and \(b\) can be found using \(a^2 = 16\) giving \(a = 4\), and \(b^2 = 25\) giving \(b = 5\).
• The transverse axis (along the direction of opening) indicates whether the hyperbola opens horizontally or vertically.

The standard form provides a blueprint for converting the equation to its graphable form. Recognizing this form enables one to isolate \(y\) as a function of \(x\) and further analyze the hyperbola.

Graphing Calculator

A graphing calculator is incredibly useful for visualizing equations and understanding their real-world implications quickly. When working with a hyperbola, inputting equations like \( y_1 = -3 + \sqrt{\frac{25(x-2)^2}{16} - 25} \) and \( y_2 = -3 - \sqrt{\frac{25(x-2)^2}{16} - 25} \) into a graphing calculator allows students to see the hyperbola's shape.

• Ensure both functions are entered correctly, representing the upper and lower branches of the hyperbola.
• It's important to set an appropriate window that captures the essence of the hyperbola's structure, including the vertices and asymptotes, to get a full view of the graph.
• Experiment with the zoom and pan features to enhance understanding and gain a clearer view of key features.

Using a graphing calculator enhances learning by providing a tangible representation of algebraic equations.

Functions of x

To express the hyperbola in terms of \(y\), we need to find \(y\) as two separate functions of \(x\). Starting with the standard form \( \frac{(x-2)^2}{16} - \frac{(y+3)^2}{25} = 1 \), the steps lead us to two functions: \( y_1 = -3 + \sqrt{\frac{25(x-2)^2}{16} - 25} \) and \( y_2 = -3 - \sqrt{\frac{25(x-2)^2}{16} - 25} \).

• Both functions derive from isolating \(y\) to reveal how changes in \(x\) affect the \(y\) values and, subsequently, the graph of the hyperbola.
• The process includes rearranging terms, solving for one variable, in this case, \(y\), and taking the square root, which introduces the positive and negative components.
• This conversion into functions shows the symmetrical nature of the hyperbola above and below the center point \((2, -3)\).

Understanding the concept of functions of \(x\) is vital for graphing and performing further calculations with hyperbolas.

Vertices and Asymptotes

Vertices and asymptotes are key features of a hyperbola that define its shape and direction. Identifying these helps in sketching the hyperbola accurately.

• Vertices are easy to find once you have \(a\) from the equation. For this hyperbola, the vertices lie on the transverse axis at points \((2 \pm a, -3)\), hence at \((6, -3)\) and \((-2, -3)\).
• Asymptotes are straight lines the branches of the hyperbola approach but never meet. To express these, use the formula for horizontal hyperbolas: \( y = k \pm \frac{b}{a}(x - h) \).
• For the given equation, the asymptotes are \( y = -3 \pm \frac{5}{4}(x - 2) \). These guide how the hyperbola opens and widens as it extends away from the center.

Recognizing vertices and asymptotes is crucial for graphing hyperbolas and understanding their behavior in mathematical modeling.