91Ó°ÊÓ

Quadratic functions are a fundamental concept in mathematics, commonly represented by the general form \( f(x) = ax^2 + bx + c \). Here, \(a\), \(b\), and \(c\) are constants, and \(aeq0\). This ensures the graph of the function is a parabola - a U-shaped curve.

• The leading coefficient \(a\) affects the width and direction of the parabola. Positive values of \(a\) make it open upwards, and negative values make it open downwards.
• The constant term \(c\) is the y-intercept of the function, indicating where the parabola crosses the y-axis.

The specific quadratic function in this exercise is \(f(x) = x^2 + 1\), which signifies a parabola that opens upwards and is shifted one unit up from the origin along the y-axis without any horizontal shift. Understanding the shape and position of quadratic functions is crucial for various real-world applications, such as optimizing areas and growth patterns.

Function evaluation involves substituting a specific value for the variable \(x\) in the function's equation and simplifying to find the result. This process helps in understanding the behavior of the function at specific points.In the given problem, we evaluated the function \(f(x) = x^2 + 1\) at \(x = 3\). By substituting \(3\) for \(x\), we calculated \(f(3) = 3^2 + 1 = 10\). Subsequently, we used the outcome \(f(3) = 10\) as the new input for evaluating \(f(x)\) again to find \(f(f(3))\), resulting in \(f(10) = 101\).This process illustrates how to use output from one evaluation as input for another, a common practice in function compositions. Evaluating functions is essential in fields such as physics and engineering, where predictions are made based on variable conditions.

Algebraic expressions consist of constants, variables, and operations like addition, subtraction, multiplication, and division. Understanding how to manipulate these expressions is a key algebra skill.In this exercise, the expressions derived from function evaluation include substitution and simplification. For example, evaluating \(f(3)\) required replacing \(x\) with \(3\) to get \(3^2 + 1\). Simplifying \(3^2\) led us to calculate \(9 + 1 = 10\).Similarly, \(f(10) = 10^2 + 1\) involved calculating \(10^2\) to get \(100\), then adding \(1\) to get \(101\). These steps highlight how algebraic manipulation is applied in function evaluation.Being proficient in handling algebraic expressions is crucial for solving more complex mathematical problems, as they form the backbone of algebra allowing us to express relationships and make computations.