Problem 37
Question
For each of the following exercises, construct a table and graph the equation by plotting at least three points. $$ 2 y=x+3 $$
Step-by-Step Solution
Verified Answer
Create a table with points \((-2, \frac{1}{2})\), \((0, \frac{3}{2})\), and \((2, \frac{5}{2})\), then plot and connect these to graph \( y = \frac{1}{2}x + \frac{3}{2} \).
1Step 1: Rearranging the Equation
The given equation is \( 2y = x + 3 \). To plot this equation, first, we should rearrange it to the form \( y = mx + b \), where \( m \) represents the slope and \( b \) the y-intercept. Divide the entire equation by 2: \( y = \frac{1}{2}x + \frac{3}{2} \).
2Step 2: Choosing Values for x
Next, choose values for \( x \) to substitute into the equation and find corresponding \( y \) values. Let's use \( x = -2, 0, \) and \( 2 \) as our values.
3Step 3: Calculating Corresponding y Values
For each chosen \( x \), calculate the corresponding \( y \):- If \( x = -2 \), \( y = \frac{1}{2}(-2) + \frac{3}{2} = -1 + \frac{3}{2} = \frac{1}{2} \).- If \( x = 0 \), \( y = \frac{1}{2}(0) + \frac{3}{2} = \frac{3}{2} \).- If \( x = 2 \), \( y = \frac{1}{2}(2) + \frac{3}{2} = 1 + \frac{3}{2} = \frac{5}{2} \).
4Step 4: Constructing the Table
Using the values calculated in Step 3, construct a table:\[\begin{array}{|c|c|}\hlinex & y \\hline-2 & \frac{1}{2} \0 & \frac{3}{2} \2 & \frac{5}{2} \\hline\end{array}\]
5Step 5: Graphing the Equation
Using the table, plot the points \((-2, \frac{1}{2})\), \((0, \frac{3}{2})\), and \((2, \frac{5}{2})\) on a coordinate plane. Draw a straight line through these points to represent the equation \( y = \frac{1}{2}x + \frac{3}{2} \). The graph should be a straight line with a slope of \( \frac{1}{2} \) and a y-intercept at \( \frac{3}{2} \).
Key Concepts
Rearranging EquationsPlotting GraphsCoordinate Plane GraphingSlope and Intercept
Rearranging Equations
To fully understand and solve linear equations, one essential skill is rearranging them. Often, equations appear in forms that are not immediately useful for graphing. A common goal is to convert them to the slope-intercept form, \( y = mx + b \). This form is particularly useful because it makes identifying the slope and y-intercept straightforward.
In our example, we start with the equation \( 2y = x + 3 \). Our aim is to isolate \( y \) on one side. Dividing both sides by 2 helps to achieve this, resulting in \( y = \frac{1}{2}x + \frac{3}{2} \). Now, the equation is in a clear format where:
In our example, we start with the equation \( 2y = x + 3 \). Our aim is to isolate \( y \) on one side. Dividing both sides by 2 helps to achieve this, resulting in \( y = \frac{1}{2}x + \frac{3}{2} \). Now, the equation is in a clear format where:
- \( m \) (the slope) is \( \frac{1}{2} \)
- \( b \) (the y-intercept) is \( \frac{3}{2} \)
Plotting Graphs
Plotting graphs is a way to visually represent mathematical relationships. Once an equation is in the slope-intercept form, plotting becomes simplified.
The first step is to calculate numbers that fit into the equation. These are your input-output pairs, or \( (x, y) \) coordinates. For our equation \( y = \frac{1}{2}x + \frac{3}{2} \), choosing manageable numbers for \( x \) helps calculate easily solved values for \( y \). In our exercise, we chose \( x = -2, 0, \text{ and } 2 \), resulting in \( y \) values of \( \frac{1}{2}, \frac{3}{2}, \text{ and } \frac{5}{2} \) respectively.
The first step is to calculate numbers that fit into the equation. These are your input-output pairs, or \( (x, y) \) coordinates. For our equation \( y = \frac{1}{2}x + \frac{3}{2} \), choosing manageable numbers for \( x \) helps calculate easily solved values for \( y \). In our exercise, we chose \( x = -2, 0, \text{ and } 2 \), resulting in \( y \) values of \( \frac{1}{2}, \frac{3}{2}, \text{ and } \frac{5}{2} \) respectively.
- For \( x = -2 \), \( y = \frac{1}{2} \)
- For \( x = 0 \), \( y = \frac{3}{2} \)
- For \( x = 2 \), \( y = \frac{5}{2} \)
Coordinate Plane Graphing
The coordinate plane is a two-dimensional surface on which each point is uniquely specified by a pair of numerical coordinates.
To graph a linear equation, we plot points derived from our calculations onto a grid. The horizontal axis (x-axis) and the vertical axis (y-axis) intersect each other perpendicularly, forming four quadrants. Each point plotted must align with the x and y values found from converting the equation.
Take the points we calculated earlier: \((-2, \frac{1}{2}), (0, \frac{3}{2})\), and \((2, \frac{5}{2})\). By placing these points accurately on the graph and connecting them with a line, we visualize the equation on the plane. The line will slope upwards or downwards depending on whether the slope \( m \) is positive or negative.
To graph a linear equation, we plot points derived from our calculations onto a grid. The horizontal axis (x-axis) and the vertical axis (y-axis) intersect each other perpendicularly, forming four quadrants. Each point plotted must align with the x and y values found from converting the equation.
Take the points we calculated earlier: \((-2, \frac{1}{2}), (0, \frac{3}{2})\), and \((2, \frac{5}{2})\). By placing these points accurately on the graph and connecting them with a line, we visualize the equation on the plane. The line will slope upwards or downwards depending on whether the slope \( m \) is positive or negative.
Slope and Intercept
The slope and y-intercept are integral aspects of understanding a line's behavior on a graph surface. The slope, represented by \( m \) in the equation format \( y = mx + b \), indicates the steepness and direction of the line.
For instance, in \( y = \frac{1}{2}x + \frac{3}{2} \), the slope is \( \frac{1}{2} \). This positive slope signifies that for each unit increase along the x-axis, \( y \) increases by \( \frac{1}{2} \). Furthermore, a gradual slant implies a slow increase.
The y-intercept, represented by \( b \), is where the line crosses the y-axis. In our exercise, the y-intercept is \( \frac{3}{2} \). This is the point where \( x \) is zero and provides a starting point for drawing the line. Understanding these two components imparts insight into the fundamental characteristics and orientation of the graph.
For instance, in \( y = \frac{1}{2}x + \frac{3}{2} \), the slope is \( \frac{1}{2} \). This positive slope signifies that for each unit increase along the x-axis, \( y \) increases by \( \frac{1}{2} \). Furthermore, a gradual slant implies a slow increase.
The y-intercept, represented by \( b \), is where the line crosses the y-axis. In our exercise, the y-intercept is \( \frac{3}{2} \). This is the point where \( x \) is zero and provides a starting point for drawing the line. Understanding these two components imparts insight into the fundamental characteristics and orientation of the graph.
Other exercises in this chapter
Problem 37
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