Problem 33 - 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]. $$q(x)=\frac{1}{2} x$$

Step-by-Step Solution

Answer

The area of the region is 9 square units.

Step 1: Identify the Function

The given function to graph is \( q(x) = \frac{1}{2}x \). This is a linear function with a slope of \( \frac{1}{2} \) and a y-intercept of 0.

Step 2: Determine Key Points for the Graph

To graph \( q(x) = \frac{1}{2}x \), select key points within the interval \([0, 7]\). When \( x = 0 \), \( q(x) = 0 \). When \( x = 7 \), \( q(x) = \frac{1}{2} \times 7 = 3.5 \). Plot these points on the coordinate plane.

Step 3: Draw the Graph

Using the key points \((0, 0)\) and \((7, 3.5)\), draw a straight line to represent the function within the interval \([0, 7]\). The graph should be a straight line starting at the origin and rising with a slope of \( \frac{1}{2} \).

Step 4: Define the Interval for Area Calculation

We need to calculate the area under the graph of \( q(x) \) and above the \( x \)-axis within the interval \([0, 6]\). This area forms a right triangle on the graph.

Step 5: Identify the Triangle's Dimensions

The triangle's base extends from \( x = 0 \) to \( x = 6 \). The height at \( x=6 \) is calculated as \( q(6) = \frac{1}{2} \times 6 = 3 \). So, the triangle has a base of 6 units and a height of 3 units.

Step 6: Calculate the Area Using Geometry

The area of a right triangle is given by \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). Plug in the calculated values: \( \text{Area} = \frac{1}{2} \times 6 \times 3 = 9 \).

Key Concepts

Area Under a Curve

When dealing with functions, especially in calculus and precalculus, a fundamental concept is finding the area under a curve. This area represents the space between the curve on a graph and the x-axis over a specified interval. It can have practical applications, such as calculating distance traveled or total quantity accrued over time. To find the area under the curve for linear functions over a given interval, you often form geometric shapes, like triangles or rectangles, and use basic geometry to calculate the area. In our case, the function is a straight line, which simplifies the process. For the function \( q(x) = \frac{1}{2}x \) over the interval \([0, 6]\), you form a right triangle. The base of this triangle extends from \( x = 0 \) to \( x = 6 \), and the height is determined by the function value at \( x = 6 \), which is 3 units. The area is calculated using the formula for the area of a triangle: \( \frac{1}{2} \times \text{base} \times \text{height} \), providing an area of 9 square units. Understanding this process is a vital step in grasping how calculus uses integration to generalize this concept.

Linear Equation

A linear equation is a type of equation that represents a straight line when graphed on a coordinate plane. It has the general form \( y = mx + c \), where \( m \) is the slope, and \( c \) is the y-intercept. The slope describes how steep the line is, while the y-intercept shows where the line crosses the y-axis.In the exercise, the linear equation given is \( q(x) = \frac{1}{2}x \). This particular equation has:

• A slope \( m = \frac{1}{2} \), indicating the line rises one unit vertically for every two units it moves horizontally.
• A y-intercept of 0, meaning the line passes through the origin (point \((0, 0)\)).

When graphing such equations, it helps to determine key points — values like \( x = 0 \) and \( x = 7 \) — that help draw a precise graph. Once these points are plotted, you can use them to sketch the line accurately. Understanding linear equations is crucial as they commonly appear in various math fields and real-world applications.

Geometry in Precalculus

Geometry plays a significant role in precalculus, especially when dealing with graphing and analyzing functions. It helps bridge the gap to calculus concepts by offering a tangible way to comprehend abstract mathematical ideas. In the task of graphing linear functions, geometry provides tools and principles for visualizing and calculating areas, as seen with the calculation of the area under the curve. The use of shapes like rectangles and triangles makes it easier to apply basic mathematical principles to solve more complex problems involving functions. Understanding the geometric relationships between elements such as slopes, intercepts, and areas helps in forming a deeper understanding of how functions behave. This knowledge aids in making the transition to calculus smoother, as calculus often requires one to think in terms of areas, slopes of tangents, and the behavior of curves.