91Ó°ÊÓ

Function evaluation is the process of determining the output of a function given a specific input value. This is often one of the first things you learn when dealing with functions. Let's break it down.
When you have a function like \(f(x) = \frac{x + 5}{15}\), and you need to evaluate it at \(x = -2\), you substitute \(-2\) for \(x\) in the equation. So, \(f(-2) = \frac{-2 + 5}{15}\), simplifying this gives \(\frac{3}{15} = \frac{1}{5}\). Hence, \(f(-2) = \frac{1}{5}\).
Similarly, for \(x = 4\), substitute \(4\) for \(x\): \(f(4) = \frac{4 + 5}{15} = \frac{9}{15} = \frac{3}{5}\). So, \(f(4) = \frac{3}{5}\). This tells us that each x-value input you test gives you a corresponding y-value output, which is essentially the function evaluation. Remember this process as it is a fundamental skill in working with functions.

Graphing a function provides a visual way to understand the behavior of functions. When you graph \(f(x) = \frac{x+5}{15}\), you observe the line defined by this function's equation.
For linear functions like these, it helps to identify the slope and the y-intercept. In slope-intercept form, \(y = mx + b\), the slope, \(m\), is the rate of change of the function, here \(\frac{1}{15}\), and \(b\) is the y-intercept. For this function, the y-intercept is \(\frac{1}{3}\), which is the point where the line crosses the y-axis. You plot this at \((0, \frac{1}{3})\).
To plot a second point using the slope, you move upward 1 unit and to the right 15 units from the y-intercept. Draw a straight line through these two points to complete the graph. This visual representation can help you quickly determine important characteristics of the function.

Finding the zeros of a function is a crucial concept. The zeros, also called roots or x-intercepts, are the x-values that make the function equal to zero, \(f(x) = 0\). For linear functions, this happens where the graph crosses the x-axis.
To find the zero of \(f(x) = \frac{x+5}{15}\), set the equation to zero and solve for x. So, \(\frac{x+5}{15} = 0\). By multiplying both sides by 15, you get \(x + 5 = 0\). Subtract 5 to isolate \(x\), giving you \(x = -5\).
This zero means that if you were to plug \(-5\) into the function, the output would be zero. Understanding zeros helps in analyzing and solving equations, as they represent solutions where a given function's value satisfies the equation exactly.