91Ó°ÊÓ

Quadratic functions are a fundamental part of algebra and play a crucial role in calculus as well. A quadratic function is a polynomial function of degree two, generally represented as \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). In this particular exercise, the function is given as \( f(x) = 3x^2 \). This is a simple quadratic function with \( a = 3 \) and \( b = c = 0 \).
The graph of a quadratic function is a parabola. Since \( a \) is positive in our function, it opens upwards. Quadratic functions are symmetric around their vertex, which is the peak or the lowest point of the graph, depending on the sign of \( a \).
This concept is important because it helps us understand the behavior of the function and allows us to calculate things like the average rate of change, as we will be doing in the exercise.

In the study of calculus, a secant line is a line that intersects a curve at two or more points. The average rate of change of a function over an interval corresponds to the slope of the secant line between two points on that function.
When dealing with a specific continuous function like our quadratic function \( f(x) = 3x^2 \), understanding how to find this line is pivotal. In this exercise, you are comparing the function values at \( x = 2 \) and \( x = 2 + h \). The secant line provides a linear approximation of the curve over this interval.
The formula for the slope of the secant line, which is the average rate of change, is found using the expression \( \frac{f(x_2) - f(x_1)}{x_2 - x_1} \). This gives a straight line that effectively connects the two points on the parabola.

The difference quotient is an expression that provides a way to calculate the average rate of change of a function over a specified interval. This quotient has a format that resembles the formula for the slope of a line, indicating the change in function values relative to the change in input values.
For the function \( f(x) = 3x^2 \), and considering the interval from \( x = 2 \) to \( x = 2 + h \), the difference quotient is expressed as\[\frac{f(2 + h) - f(2)}{h}. \]
In the solution provided, we replaced \( f(2) \) and \( f(2 + h) \) with their calculated values, leading to the expression \( \frac{(12 + 12h + 3h^2) - 12}{h} \). By simplifying, we obtained \( 12 + 3h \), which reflects the average rate of change. As \( h \) approaches 0, the difference quotient becomes the derivative, which further implies the instantaneous rate of change.