Problem 19 - A Graphical Approach to Precalculus with Limits

Question

Work each problem related to linear functions. (a) Evaluate \(f(-2)\) and \(f(4)\) (b) Graph \(f\). How can the graph of \(f\) be used to determine the zero of \(f ?\) (c) Find the zero of \(f\) $$f(x)=\frac{1}{15} x+\frac{1}{3}$$

Step-by-Step Solution

Answer

(a) \( f(-2) = \frac{1}{5} \), \( f(4) = \frac{3}{5} \); (c) Zero is \( x = -5 \).

Step 1: Evaluate f(-2)

To find \( f(-2) \), substitute \( x = -2 \) into the function: \[f(-2) = \frac{1}{15}(-2) + \frac{1}{3} \]Calculate:\[f(-2) = -\frac{2}{15} + \frac{1}{3} \]Convert \( \frac{1}{3} \) to a fraction with denominator 15:\[f(-2) = -\frac{2}{15} + \frac{5}{15} = \frac{3}{15} = \frac{1}{5} \]

Step 2: Evaluate f(4)

To find \( f(4) \), substitute \( x = 4 \) into the function:\[f(4) = \frac{1}{15}(4) + \frac{1}{3} \]Calculate:\[f(4) = \frac{4}{15} + \frac{1}{3} \]Convert \( \frac{1}{3} \) to a fraction with denominator 15:\[f(4) = \frac{4}{15} + \frac{5}{15} = \frac{9}{15} = \frac{3}{5} \]

Step 3: Graph the function f(x)

The function \( f(x) = \frac{1}{15}x + \frac{1}{3} \) is a straight line with slope \( \frac{1}{15} \) and y-intercept \( \frac{1}{3} \). The graph will pass through the point (0, \frac{1}{3}) and rise \( \frac{1}{15} \) for every unit it moves to the right. Plot these points and draw a straight line through them.

Step 4: Use the graph to find the zero of f

To find the zero of the function using its graph, look for the x-coordinate where the line crosses the x-axis. This is where \( f(x) = 0 \).

Step 5: Calculate the zero analytically

Set the function equal to zero and solve for \( x \):\[0 = \frac{1}{15}x + \frac{1}{3} \]Subtract \( \frac{1}{3} \) from both sides:\[-\frac{1}{3} = \frac{1}{15}x \]Multiply both sides by 15:\[-5 = x \]

Key Concepts

Evaluating Functions

Evaluating functions is an essential skill when working with linear functions. This process involves plugging in a specific value for the variable \( x \) in the function and simplifying the expression to find \( f(x) \). Consider the function given: \( f(x) = \frac{1}{15} x + \frac{1}{3} \). To evaluate \( f(-2) \), substitute \( x = -2 \) into the function:

• First, multiply: \( \frac{1}{15} \times (-2) = -\frac{2}{15} \).
• Next, add \( \frac{1}{3} \) to \( -\frac{2}{15} \). To do this, convert \( \frac{1}{3} \) to \( \frac{5}{15} \):
• The result is \( f(-2) = -\frac{2}{15} + \frac{5}{15} = \frac{3}{15} = \frac{1}{5} \).

Similarly, for \( f(4) \), replace \( x \) with 4:

• Calculate \( \frac{1}{15} \times 4 = \frac{4}{15} \).
• Again, add \( \frac{1}{3} \) (or \( \frac{5}{15} \)) to \( \frac{4}{15} \):
• The final value is \( f(4) = \frac{9}{15} = \frac{3}{5} \).

The key to evaluating functions is ensuring that any fractions are manipulated to have a common denominator for easy addition or subtraction.

Graphing Functions

Graphing linear functions provides a visual representation of the equation and helps in understanding its behavior. The standard form of a linear function is \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. For the function \( f(x) = \frac{1}{15}x + \frac{1}{3} \), the slope \( m = \frac{1}{15} \) indicates a gentle rise, and the y-intercept \( b = \frac{1}{3} \) shows where the line crosses the y-axis.

To graph this function, follow these steps:

• Start at the y-intercept (0, \( \frac{1}{3} \)). This is your first point.
• Use the slope \( \frac{1}{15} \). This means for every 1 unit you move right on the x-axis, go up \( \frac{1}{15} \) units.
• Plot several points using this slope and draw a line through all the points. This line continues infinitely in both directions.

Graphing a function allows you to visually assess characteristics such as increasing or decreasing trends and the intercepts on each axis. It also provides clues about where the function may equal zero.

Finding Zeros of Functions

Finding the zeros of a function involves determining where the function crosses the x-axis, or where \( f(x) = 0 \). This is a fundamental concept in understanding the behavior of linear functions. For the function \( f(x) = \frac{1}{15}x + \frac{1}{3} \), you can find the zero either graphically or analytically.

Graphically, you can find the zero by looking at the graph and identifying the point where the line crosses the x-axis. This point is the x-intercept, where \( f(x) \) equals zero.

Analytically, find the zero by setting the function equal to zero and solving for \( x \):

• Start with \( 0 = \frac{1}{15}x + \frac{1}{3} \).
• Subtract \( \frac{1}{3} \) from both sides to isolate the \( x \) term: \( -\frac{1}{3} = \frac{1}{15}x \).
• Multiply both sides by 15 to clear the fraction: \( -5 = x \).

This means the zero of the function is \( x = -5 \). Knowing how to find zeros is crucial as it helps in predicting where functions intersect the x-axis, which is often an important aspect in solving real-world problems.