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A hyperbola is one of the four main types of conic sections, which are the curves formed by intersecting a plane with a cone. Unlike a circle or ellipse, a hyperbola consists of two separate branches. These branches resemble curved lines that mirror each other across center.
A hyperbola with a horizontal transverse axis is described by the standard form equation: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] Here,

• \(h\) and \(k\) are the coordinates of the center,
• \(a\) and \(b\) are the distances from the center to the vertices and co-vertices, respectively.

The equation provided, \(\frac{(x-2)^2}{16} - \frac{y^2}{9} = 1\), suggests a hyperbola with its center shifted 2 units to the right from the origin, to the point \((2, 0)\). In this setup, the transverse axis lies horizontally along the x-axis.

Finding the foci and vertices of a hyperbola is crucial for understanding its shape and behavior. The vertices of the hyperbola are the points where the hyperbola crosses the transverse axis. For the equation \(\frac{(x-2)^2}{16} - \frac{y^2}{9} = 1\),

• \(a^2 = 16\), which means \(a = 4\).

Thus, the vertices from the center \((2, 0)\) are at points \((6, 0)\) and \((-2, 0)\).
The foci are located further from the center than the vertices, providing another crucial indicator of the hyperbola's shape. To find the distance \(c\) to the foci from the center, use the relation: \[ c^2 = a^2 + b^2 \]Given \(b^2 = 9\) as well, we have \(c^2 = 25\), leading to \(c = 5\). Therefore, the foci are located at the coordinates \((7, 0)\) and \((-3, 0)\).
These markers help in graphing the hyperbola and understanding its dimensions.

Asymptotes of a hyperbola are lines that the branches of the hyperbola approach but never touch. They provide a "guideline" for drawing the hyperbola accurately. These lines extend straight through the center and depict the general direction the hyperbola will open.
The formula for asymptotes of a horizontal hyperbola is: \[y = k \pm \frac{b}{a}(x-h)\]For our specific hyperbola \(\frac{(x-2)^2}{16} - \frac{y^2}{9} = 1\), we plug in \(h = 2\), \(k = 0\), \(a = 4\), and \(b = 3\) into this equation:

• The first asymptote: \(y = \frac{3}{4}(x - 2)\)
• The second asymptote: \(y = -\frac{3}{4}(x - 2)\)

These lines cross at the center and extend in opposite diagonal directions. When sketching the hyperbola, the branches should become increasingly close to these asymptotes but never intersect them. The steepness and direction of these asymptotes give insight into how 'wide' or 'narrow' the hyperbola will appear on a graph.