Problem 35 - Precalculus

Question

Graph each function over the interval [0,7]. Then use geometry to find the area of the region below the graph, and above the \(x\) -axis in the interval [0,6]. $$t(x)=\frac{1}{2} x+1$$

Step-by-Step Solution

Answer

The area is 15 square units.

Step 1: Understand the Function

The given function is a linear function \( t(x) = \frac{1}{2}x + 1 \). This means it is a straight line with a slope of \(\frac{1}{2}\) and a y-intercept at \(1\).

Step 2: Create a Table of Values

Calculate the values of \( t(x) \) at different points to plot the graph. For example, at \( x = 0 \), \( t(0) = 1 \); at \( x = 6 \), \( t(6) = \frac{1}{2}(6) + 1 = 4 \). These values will help in plotting the points and understanding the shape of the graph.

Step 3: Plot the Graph

On a graph, plot the points calculated previously, such as (0,1), (3,2.5), and (6,4). Then draw a straight line through these points as the function describes a linear relationship. Extend this line slightly beyond \( x = 6 \) to cover the interval \([0,7]\).

Step 4: Determine the Shape for Area Calculation

The area under the function \( t(x) \) from \( x = 0 \) to \( x = 6 \) forms a trapezoid. The vertices of this trapezoid on the graph are (0,1), (6,4), (6,0), and (0,0).

Step 5: Use Geometry to Find the Area of the Trapezoid

The trapezoid height is \( 6 \) units (the difference in the x-values from 0 to 6). The two bases are \( 1 \) unit and \( 4 \) units, which are the values of \( t(x) \) at \( x = 0 \) and \( x = 6 \), respectively. The area \( A \) of a trapezoid is given by \( A = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \). Substituting the values, we get: \[ A = \frac{1}{2} \times (1+4) \times 6 = 15 \] square units.

Key Concepts

Linear Functions

Linear functions are a fundamental concept in algebra. They form the simplest type of functions, represented by straight lines on a graph. The standard form of a linear function is \( f(x) = mx + c \), where \( m \) is the slope of the line and \( c \) is the y-intercept.

• Slope \( m \): This determines the steepness and direction of the line. A higher value means a steeper incline. If \( m \) is positive, the line slopes upwards; if negative, it slopes downwards.
• Y-intercept \( c \): This is where the line crosses the y-axis. It's the value of \( f(x) \) when \( x = 0 \).

Understanding these parts helps in graphing the function and identifying trends over an interval. In the exercise, the linear function given is \( t(x) = \frac{1}{2}x + 1 \). The slope \( \frac{1}{2} \) means for every 2 units increase in \( x \), \( t(x) \) increases by 1. The y-intercept, 1, indicates where the line starts on the y-axis. Linear functions are prevalent for modeling situations with constant rates of change and serve as a foundational block in mathematics.

Area Calculation

Area calculation involves determining the size of a surface enclosed within certain boundaries on a graph. It's a crucial part of geometry taught in various fields like mathematics, physics, and engineering. The area can be calculated for various shapes such as rectangles, triangles, and trapezoids using different formulas. The process includes:

• Identifying the Shape: Recognize the shape of the region whose area needs to be calculated.
• Applying the Formula: Use the correct geometric formula to calculate the area.

In graphing problems, the area might lie under a function between specified intervals on the \( x \)-axis. Understanding how to calculate areas helps solve real-world problems, like finding the space within boundaries or determining quantities of materials needed for construction.

Trapezoid Area Formula

A trapezoid is a four-sided shape with at least one pair of parallel sides. The area of a trapezoid is found using the formula:\[A = \frac{1}{2} \, \times \, (\text{Base}_1 \, + \, \text{Base}_2) \, \times \, \text{Height}\]Here's how to apply the formula step-by-step:

• Identify the Bases: The parallel sides are \( \text{Base}_1 \) and \( \text{Base}_2 \).
• Measure the Height: This is the perpendicular distance between the two bases.
• Substitute into the Formula: Plug in the lengths of the bases and the height into the area formula to get the result.

In the exercise, the linear function from \( x = 0 \) to \( x = 6 \) forms a trapezoid. The values at these points are the bases: \( \text{Base}_1 \) is 1 (the value at \( x = 0 \)) and \( \text{Base}_2 \) is 4 (the value at \( x = 6 \)). The height, the horizontal span, is 6 units. Substituting, we calculate an area of 15 square units. This method of finding the area is especially useful in calculus and physics for determining the region under curves or lines.