Problem 17 - Algebra A Combined Function

Question

Graph each hyperbola. $$ x^{2}-4 y^{2}=16 $$

Step-by-Step Solution

Answer

Graph the hyperbola with vertices at \((4, 0)\) and \((-4, 0)\), and asymptotes \(y = \pm \frac{1}{2}x\).

Step 1: Identify Standard Form

First, identify the standard form of a hyperbola. The standard form for a hyperbola centered at the origin is either: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\). In this case, we have the equation \(x^2 - 4y^2 = 16\). We need to rearrange this equation to match the standard form.

Step 2: Divide Equation by Constant

To convert the equation \(x^2 - 4y^2 = 16\) into standard form, divide the entire equation by 16 to get \(\frac{x^2}{16} - \frac{4y^2}{16} = 1\). This simplifies to \(\frac{x^2}{16} - \frac{y^2}{4} = 1\).

Step 3: Identify Parameters of the Hyperbola

From the equation \(\frac{x^2}{16} - \frac{y^2}{4} = 1\), identify \(a^2 = 16\) and \(b^2 = 4\). Thus, \(a = 4\) and \(b = 2\). The hyperbola is horizontal since the \(x^2\) term is positive.

Step 4: Determine Vertices and Asymptotes

The vertices of the hyperbola are determined by \(\pm a\), which gives us the points \((4, 0)\) and \((-4, 0)\). The equations of the asymptotes for a horizontal hyperbola are \(y = \pm \frac{b}{a}x = \pm \frac{2}{4}x\), simplifying to \(y = \pm \frac{1}{2}x\).

Step 5: Sketch the Hyperbola

To graph the hyperbola, plot the vertices \((4, 0)\) and \((-4, 0)\) on the coordinate plane. Draw the asymptotes \(y = \frac{1}{2}x\) and \(y = -\frac{1}{2}x\). Sketch the two branches of the hyperbola that open horizontally and approach these asymptotes, ensuring the center of the hyperbola is at the origin \((0, 0)\).

Key Concepts

Equation Rearrangement

Rearranging equations is a crucial step in solving problems involving conic sections like hyperbolas. The goal here is to convert the given hyperbola equation into its standard form. We start with the equation given in the problem: \( x^2 - 4y^2 = 16 \). The standard form of a hyperbola centered at the origin is either \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \). To achieve this, we need to have a 1 on the right side of the equation. Hence, we divide the entire equation by 16:

• \( \frac{x^2}{16} - \frac{4y^2}{16} = 1 \)

This simplifies to \( \frac{x^2}{16} - \frac{y^2}{4} = 1 \). Now we have the equation in standard form, ready to identify other essential parameters of the hyperbola.

Vertices of Hyperbola

Finding the vertices of a hyperbola is an integral step in understanding its structure. Once the equation is in standard form, \( \frac{x^2}{16} - \frac{y^2}{4} = 1 \), you can easily identify the values for \(a\) and \(b\). Here, we have:

• \( a^2 = 16 \), so \( a = 4 \)
• \( b^2 = 4 \), so \( b = 2 \)

Since the \(x^2\) term is positive, this hyperbola opens horizontally. Therefore, the vertices are located along the \(x\)-axis at \( \pm a \).Thus, the vertices are at the points \( (4, 0) \) and \( (-4, 0) \). These points are crucial not just for the shape of the hyperbola, but also for sketching its behavior on a graph.

Asymptotes

Asymptotes play an important role in defining the boundaries within which a hyperbola resides. They are straight lines that the hyperbola will approach but never meet.For a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the equations of the asymptotes are given by:

• \( y = \pm \frac{b}{a}x \)

Substituting the values for \(a\) and \(b\): \( b = 2 \) and \( a = 4 \), leads us to:

• \( y = \pm \frac{2}{4}x \)

This simplifies to:

• \( y = \pm \frac{1}{2}x \)

These lines give the directions in which the branches of the hyperbola open, acting as boundaries for the curve as it extends to infinity. By drawing these lines first, we can better visualize how the hyperbola fits on the graph.

Graphing Hyperbolas

Graphing hyperbolas involves plotting not only the vertices and asymptotes but also understanding the overall behavior of the hyperbolic branches. To begin, place the vertices \( (4, 0) \) and \( (-4, 0) \) on the coordinate plane. Then, draw the asymptotes using the lines \( y = \frac{1}{2}x \) and \( y = -\frac{1}{2}x \). These lines serve as the guiding edges for each branch of the hyperbola.The hyperbola consists of two separate curves, or branches, which move outward from the vertices. Because this is a horizontal hyperbola, the curves will open left and right from the origin, approaching the asymptotes as they extend.Remember to ensure that the center of the hyperbola is correctly placed at the origin \( (0, 0) \), with the shape symmetrically aligned based on its orientation. Following these steps will help anyone graph hyperbolas more accurately and understand their geometric properties.