91Ó°ÊÓ

A quadratic expression is a polynomial of degree 2, which can be written in the general form as \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). In the exercise above, the expression \(4x^2 - 4x + 109\) is a quadratic expression.
Quadratic expressions can represent parabolas when graphed: the curve with a U shape, which could open upwards or downwards.

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), the parabola opens downwards.

The goal in analyzing quadratic expressions is often to find their minimum or maximum values when they represent a function. To do this, methods like completing the square and finding the vertex of the parabola are used.

Derivatives in calculus allow us to determine the rate at which a function changes. For a function \(f(x)\), the derivative is usually denoted as \(f'(x)\) and it provides valuable information about the slope of the curve at any given point \(x\).
To solve for the minimum or maximum points of a function, we set its derivative equal to zero. These points are called critical points. In the example with function \(H(x) = 4x^4 - 4x^2 + 109\), the derivative is found to be \(H'(x) = 16x^3 - 8x\).
This derivative is used to find where the slope of \(H(x)\) is zero, revealing potential spots for minimum or maximum values.

Critical points are specific values of \(x\) where the derivative of a function equals zero or is undefined. These points signal potential locations for the minimum or maximum values of the function.
Once identified, each critical point must be analyzed to determine its role within the behavior of the function.

• If the derivative changes from positive to negative at the point, it is a maximum.
• If it changes from negative to positive, it represents a minimum.

In the exercise, this was applied to the function \(H(x)\), at points where \(H'(x) = 0\), resulting in solutions \(x = 0, \pm \frac{1}{2}\). These are then further analyzed to identify which provides a minimum value.

Completing the square is a technique used to simplify a quadratic expression and find its vertex form. This makes identifying the minimum or maximum values of a quadratic function straightforward. In our example, the expression \(4x^2 - 4x + 109\) is completed as follows: first, extract the coefficient of \(x^2\), leading to \(4(x^2 - x) + 109\). Then, to complete the square for \(x^2 - x\), add and subtract \((\frac{1}{2})^2 = \frac{1}{4}\), resulting in:

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

Substituting back leads to \(4(x - \frac{1}{2})^2 + 108\), indicating that the minimum occurs at the point where the square term is zero, specifically \(x = \frac{1}{2}\). This technique simplifies finding minimum values in quadratic functions or function components, which was crucial in analyzing \(F(x)\) and \(G(x)\).