Problem 18 - Algebra A Combined Function

Question

Graph each hyperbola. $$ 4 x^{2}-y^{2}=36 $$

Step-by-Step Solution

Answer

The hyperbola opens horizontally with vertices at (±3,0) and asymptotes at \(y = ±2x\).

Step 1: Rearrange the equation

Start by moving all terms to one side of the equation: \[ 4x^2 - y^2 = 36 \]. To make it suitable for graphing, rearrange to the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\): \[ 4x^2 - y^2 = 36 \] becomes \(\frac{x^2}{9} - \frac{y^2}{36} = 1\) by dividing every term by 36.

Step 2: Identify the components

From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), identify \(a^2 = 9\) and \(b^2 = 36\), thus \(a = 3\) and \(b = 6\). The center of the hyperbola is at the origin (0,0), the transverse axis is horizontal, and the vertices are at (±3,0).

Step 3: Calculate the asymptotes

The equations of the asymptotes for the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) with a horizontal transverse axis are \(y = \pm \frac{b}{a}x\). Substitute \(a = 3\) and \(b = 6\) to get the asymptotes: \[y = \pm 2x\].

Step 4: Sketch the graph

Plot the vertices at (3,0) and (-3,0), and sketch the asymptotes \(y = 2x\) and \(y = -2x\). The hyperbola will open horizontally along the \(x\)-axis (because \(x^2\) is positive). Draw the hyperbola approaching the asymptotes but never touching them.

Key Concepts

Standard form of a hyperbola

To effectively graph a hyperbola, it's crucial to understand its standard form. Typically, a hyperbola's standard form has one of two expressions:

• For a horizontal transverse axis: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
• For a vertical transverse axis: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)

The standard form is key because it shows the relationship between the hyperbola's variables. In our exercise, the original equation \(4x^2 - y^2 = 36\) was converted by dividing each term by 36, resulting in \(\frac{x^2}{9} - \frac{y^2}{36} = 1\). This equation is now in standard form, indicating a horizontal hyperbola. Comparing it to its general form, we recognized that \(a^2 = 9\) and \(b^2 = 36\), helping us identify essential graph components.

Hyperbola components

Hyperbolas have key components that define their shape and position. These include:

• Center: The point at the very middle of the hyperbola. For the equation \(\frac{x^2}{9} - \frac{y^2}{36} = 1\), the center is at the origin (0,0).
• Transverse Axis: This is the line through the center along which the hyperbola opens. Since \(x^2\) comes first, the transverse axis is horizontal.
• Vertices: These are the two points where the hyperbola intersects the transverse axis. For the current equation, the vertices are at \((\pm 3, 0)\), directly derived from \(a = 3\).
• Foci: Located beyond the vertices, the foci are further out and help define the shape. They can be determined using \(c = \sqrt{a^2 + b^2}\).

Understanding these components not only helps in graphing but also clarifies how the hyperbola is structured.

Equation of asymptotes

Asymptotes are diagonal lines that the branches of the hyperbola approach but never touch. They guide the shape of a hyperbola, creating a framework for where the branches will lie as they extend indefinitely. For horizontal hyperbolas, like \(\frac{x^2}{9} - \frac{y^2}{36} = 1\), the equations of the asymptotes are:\[y = \pm \frac{b}{a}x\]By substituting the values of \(a = 3\) and \(b = 6\), we get:

• \(y = \pm 2x\)

These lines are essential when sketching because they represent the ultimate boundary of the hyperbola's arms. The arms of the hyperbola will curve infinitely closer to these lines without ever intersecting them.

Sketching hyperbolas

Sketching a hyperbola involves marking out its key points and guidelines on a coordinate plane. Begin by plotting the center at the origin. Next, place the vertices at \( (3, 0) \) and \( (-3, 0) \), which are determined using \(a\). These vertices indicate where the arms of the hyperbola will start.
Draw the asymptote lines \(y = 2x\) and \(y = -2x\). These lines stretch diagonally across the quadrants and act as invisible boundaries for the hyperbola's arms.
Now, sketch the hyperbola's branches so they approach these asymptotes, extending left and right. They curve outward from the vertices along the horizontal axis and follow the asymptotes closely, forming the distinctive hyperbolic shape. Remember, the branches never touch or cross the asymptotes.