91Ó°ÊÓ

Understanding the vertex of a quadratic function is crucial for analyzing its graph and behavior. For a quadratic function in standard form, \( f(x) = ax^2 + bx + c \), the vertex is the high or low turning point on its parabolic graph.
The vertex can be found using the formula: \( x = -\frac{b}{2a} \). This formula helps locate the x-coordinate of the vertex. Substituting this value back into the equation gives you the y-coordinate. Thus, the vertex is \( \left( -\frac{b}{2a}, f\left( -\frac{b}{2a} \right) \right) \).
In our specific function, \( f(x) = (1 + b^2)x^2 + 2bx + 1 \), with \( a = 1 + b^2 \) and \( b = 2b \), applying the vertex formula gives the vertex at \( x = -\frac{b}{1 + b^2} \).
This gives insight into where the function's minimum or maximum value will occur. With quadratic functions, this point usually represents the minimum value when \( a > 0 \) and a maximum when \( a < 0 \).

The minimum value of a quadratic function occurs at its vertex when the parabola opens upwards, which is determined by the leading coefficient \( a \) being positive. In the case of our function \( f(x) = (1 + b^2)x^2 + 2bx + 1 \), \( a = 1 + b^2 \), this condition is met since the expression is always positive.
To find the minimum value of \( f(x) \), we evaluate the function at the vertex found earlier. Substituting \( x = -\frac{b}{1 + b^2} \) into \( f(x) \) gives us:

• First, \( (1 + b^2)\left( -\frac{b}{1 + b^2} \right)^2 = \frac{b^2}{1+b^2} \)
• Second, the linear term simplifies: \( 2b\left( -\frac{b}{1 + b^2} \right) = -\frac{2b^2}{1 + b^2} \)
• Add the constant term 1.

The simplified expression results in the minimum value: \( m(b) = \frac{1}{1 + b^2} \).
This value depends solely on \( b \) and showcases how the quadratic behaves at its lowest point depending on the value of \( b \).

Completing the square is a technique used to rewrite a quadratic expression, making it easier to identify its properties, such as the vertex. The process involves rearranging the terms of the quadratic to fit the form \( (x - d)^2 + e \).
For the quadratic \( f(x) = (1 + b^2)x^2 + 2bx + 1 \), we want to express it in vertex form: \( (1 + b^2)(x - \text{something})^2 + \text{something else} \).
Let's go through this process:

• Start with \( (1 + b^2)x^2 + 2bx \). Divide everything by \( a = 1+b^2 \) to help complete the square.
• Rewrite \( x^2 + \frac{2b}{1+b^2}x \) by adding and subtracting \( \left(\frac{b}{1+b^2}\right)^2 \) inside the expression.
• Now, the expression becomes \( \left(x + \frac{b}{1+b^2}\right)^2 \). Subtract what was added and adjust the constant term.

This rewrites the quadratic in a form where the vertex can be identified at \( \left(-\frac{b}{1+b^2}, \frac{1}{1+b^2}\right) \).
The process of completing the square clarifies the vertex, helping students understand the geometry of quadratics better.