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Quadratic functions are often introduced in **standard form**. This form is expressed as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.

• \( a \) is the coefficient of \( x^2 \), which determines the parabola's direction and width.
• \( b \) is the coefficient of \( x \), contributing to the line's slope.
• \( c \) is the constant term, defining the parabola's y-intercept.

For the given exercise, the function \( k(x) = 3x^2 - 6x - 9 \) is already in standard form, with \( a = 3 \), \( b = -6 \), and \( c = -9 \).
The advantage of the standard form is that it gives immediate information about the parabola's orientation. Since \( a = 3 \) is positive, the parabola opens upwards. Knowing how to identify and utilize the standard form is crucial for understanding more complex transformations of quadratic functions.

The **vertex form** of a quadratic equation provides a way to easily determine the vertex of the parabola, which is its peak or trough point. The vertex form is written as \( a(x-h)^2 + k \), where:

• \((h, k)\) is the vertex of the parabola.
• \( a \) indicates the parabola's opening direction and width, similar to the standard form.

To find the vertex from a quadratic function, it is often necessary to convert the function from standard form to vertex form.
The completed example shows this with the final vertex form being \( 3(x-1)^2 - 12 \). From this, it is easy to see that the vertex is at \((1, -12)\).
Using the vertex form can be quite useful when graphing, as knowing the vertex allows for an easy starting point on the graph. It tells at a glance where the minimum or maximum of the function occurs.

**Completing the Square** is a method used to convert a quadratic function from standard form to vertex form. This method involves manipulating the equation into a perfect square trinomial.Here's how it works:

• First, factor out the coefficient of \( x^2 \) from the quadratic and linear terms.
• Next, find half of the linear coefficient (\( b \)), square it, and add/subtract this square inside the parentheses.
• The expression inside the parentheses now becomes a perfect square trinomial. Simplify this expression.

In the given problem, starting with \( 3x^2 - 6x \), factoring out \( 3 \) gives \( 3(x^2 - 2x) \).
Taking half of \(-2\), squaring it to \(1\), and adding/subtracting it results in \( 3((x-1)^2 - 1) \). This finally simplifies to vertex form \( 3(x-1)^2 - 12 \).
Completing the square is essential for accurately graphing and understanding the properties of quadratic functions. It allows us to convert the function into a form that directly reveals the vertex, making analyzing parabolas much simpler.