Problem 57 - 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. \(\frac{y^{2}}{9}-\frac{x^{2}}{1}=1\)

Step-by-Step Solution

Answer

The two functions are \( y = 3\sqrt{x^2 + 1} \) and \( y = -3\sqrt{x^2 + 1} \).

Step 1: Identify the Standard Form of the Hyperbola

The given equation of the hyperbola is \( \frac{y^{2}}{9} - \frac{x^{2}}{1} = 1 \). This is in the standard form for a vertical hyperbola: \( \frac{y^{2}}{a^2} - \frac{x^{2}}{b^2} = 1 \) where \( a^2 = 9 \) and \( b^2 = 1 \).

Step 2: Solve for y in terms of x

From the equation \( \frac{y^{2}}{9} - \frac{x^{2}}{1} = 1 \), solve for \( y^2 \):\[ y^2 = 9(x^2 + 1) \]Next, take the square root of both sides to solve for \( y \):\[ y = \pm \sqrt{9(x^2 + 1)} \] Simplify the expression:\[ y = \pm 3\sqrt{x^2 + 1} \]

Step 3: Write the Equations for y as Two Functions of x

The hyperbola can be expressed as two separate functions: 1. \( y = 3\sqrt{x^2 + 1} \)2. \( y = -3\sqrt{x^2 + 1} \)

Step 4: Graph the Functions

Using a graphing calculator, graph the two functions:- \( y = 3\sqrt{x^2 + 1} \) represents the upper branch of the hyperbola.- \( y = -3\sqrt{x^2 + 1} \) represents the lower branch of the hyperbola.These functions should be sketched on the same set of axes to complete the graph of the hyperbola.

Key Concepts

vertical hyperbola

A vertical hyperbola is a type of hyperbola where the variable 'y' is emphasized, and its transverse axis is parallel to the y-axis. The standard form of a vertical hyperbola is:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]In this form, the coefficients of 'y' determine the height or 'opening' of the hyperbola, while the coefficients of 'x' affect the position along the x-axis. Here, 'a' is the distance from the center to the vertex along the y-axis, and 'b' is the distance along the x-axis. This ensures that the hyperbola opens vertically upwards and downwards.In our equation \( \frac{y^2}{9} - \frac{x^2}{1} = 1 \), we see that this is indeed a vertical hyperbola. We identified 'a' by setting \(a^2 = 9\), so \(a = 3\), and 'b' by setting \(b^2 = 1\), so \(b = 1\). This configuration leads to two branches of the hyperbola, opening both upwards and downwards along the y-axis.

graphing calculator

A graphing calculator is a helpful tool for visualizing mathematical functions, especially those involving more complex shapes like hyperbolas. By inputting the expressions derived for 'y', namely:

• \( y = 3\sqrt{x^2 + 1} \)
• \( y = -3\sqrt{x^2 + 1} \)

you can graph both branches of the hyperbola simultaneously on a graphing calculator.When using a graphing calculator:- Enter each function separately into the calculator to see them plotted.- Make sure your calculator is set to an appropriate scale to view both branches clearly.- Depending on your device, it might be necessary to adjust window settings for better visualization, capturing a full view of the hyperbola along both axes.Graphing these functions helps in understanding the separation between each branch and their symmetry regarding the center of the hyperbola.

functions of x

Functions of 'x' are equations where 'y' is expressed purely in terms of 'x'. In the case of our hyperbola, we derived these functions to show the relation between 'x' and 'y' for both branches of the hyperbola:

• \( y = 3\sqrt{x^2 + 1} \)
• \( y = -3\sqrt{x^2 + 1} \)

Here, 'x' is the independent variable that can take any value, while 'y' depends on the value chosen for 'x'.For these specific forms:- The use of \( \pm \) in \( y = \pm 3\sqrt{x^2 + 1} \) indicates two functions: one for the upper branch (positive) and one for the lower branch (negative).- The expression \( \sqrt{x^2 + 1} \) adds a non-linear aspect to the functions, creating the characteristic curves of the hyperbola.Understanding these expressions helps in analyzing how changes in 'x' impact 'y', enhancing comprehension on how hyperbolas behave.