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Completing the square is a mathematical technique used to simplify quadratic expressions and equations. It involves making a quadratic expression into a perfect square trinomial, which is a necessary step to identify the type of conic section an equation represents.
To complete the square for an expression like \(ax^2 + bx\), begin by factoring out the coefficient of \(x^2\) if it's not 1. Then, take the coefficient of \(x\), halve it, and square the result. This squared value is added and subtracted within the expression to form a perfect square trinomial. For example, if you have \(x^2 - 6x\):

• Halve \(-6\) to get \(-3\), and then square it to get \(9\).
• Add \(9\) and simultaneously subtract \(9\) to maintain equation balance: \(x^2 - 6x + 9 - 9\).
• Re-organize it to form \((x - 3)^2 - 9\).

This method transforms the expression into a form that is easier to work with when identifying conic sections in an equation.

A hyperbola is one of the four types of conic sections, which also include circles, ellipses, and parabolas. It is a curve formed by the intersection of a double cone and a plane where the angle of the plane is less steep than that of the cone. The standard equation for a hyperbola is:
\[ -\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
Here, \((h, k)\) represents the center of the hyperbola. A hyperbola has two branches that mirror each other about a central axis. Unlike ellipses, hyperbolas open outward either horizontally or vertically. In this context, the hyperbola's symmetry and shape are determined by the terms \(a^2\) and \(b^2\).

• If a hyperbola opens horizontally, the equation form is \(-\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).
• If it opens vertically, the equation becomes \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\).
• In our given problem, it opens horizontally since the \(x\)-term is negative.

The graph of a hyperbola consists of two separate curves that move farther apart as they extend.

In the realm of conic sections, vertices and foci are critical to understanding the shape and orientation of these curves. Vertices are points where each branch of the conic has its sharpest turn. For hyperbolas, they are located on the axis that connects the two branches.

• For the hyperbola formed by the equation \( -\frac{(x-3)^2}{9} + \frac{y^2}{16} = 1 \), the vertices are found at \((3 \pm 3, 0)\) which results in points \((0, 0)\) and \((6, 0)\).
• The formula used here is \(\text{Vertices} = (h \pm a, k)\).
• The foci of a hyperbola are points that lie along the axis with greater distance from the center than the vertices, found using \(c^2 = a^2 + b^2\), giving \(c = 5\) in this scenario, leading to foci at \((3 \pm 5, 0)\), or \((-2, 0)\) and \((8, 0)\).
• The foci are essential in defining the hyperbola's shape since each point on the curve produces constant difference of distances to them.

Asymptotes can also be calculated for hyperbolas, which guide their opening direction.

The standard form of a conic section equation allows you to easily identify the type of conic section represented. Conics can be expressed in several standard forms, and identifying the correct one helps in graphing and solving related problems.
Hyperbolas, for example, have a standard form of
\[ -\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
This standard equation indicates that we are dealing with a hyperbola.

• For ellipses, the equation would be \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), resembling a hyperbola but instead involves addition.
• Inthe case of parabolas, the general form becomes \((x-h)^2 = 4p(y-k)\) or \((y-k)^2 = 4p(x-h)\), depending on orientation.

Transforming and arranging equations into their standard form is crucial for understanding each conic section's characteristics, as it allows one to determine key geometric properties like vertices, foci, and orientation at a glance.