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In Exercises 1 to \(8,\) draw the graph of each equation. Name any intencepts. $$x-3 y=4$$

Short Answer

Expert verified
The x-intercept is (4, 0) and the y-intercept is (0, -4/3).

Step by step solution

01

Write the Equation in Slope-Intercept Form

Start with the given equation: \(x - 3y = 4\).Rearrange it to the slope-intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.Subtract \(x\) from both sides: \(-3y = -x + 4\).Finally, divide the entire equation by \(-3\) to solve for \(y\):\[y = \frac{1}{3}x - \frac{4}{3}\].
02

Identify the Slope and Y-Intercept

From the slope-intercept form \(y = \frac{1}{3}x - \frac{4}{3}\), identify the slope and y-intercept:- Slope \(m = \frac{1}{3}\).- Y-intercept \(b = -\frac{4}{3}\), meaning at \((0, -\frac{4}{3})\).
03

Plot the Y-Intercept

Plot the point \((0, -\frac{4}{3})\) on the graph, which is the y-intercept.
04

Use the Slope to Find Another Point

Use the slope \(\frac{1}{3}\), which means a rise of 1 unit and a run of 3 units:From \((0, -\frac{4}{3})\), move up 1 unit to \(-\frac{1}{3}\) and right 3 units to \(3\), giving the point \((3, -\frac{1}{3})\).Plot this second point \((3, -\frac{1}{3})\) on the graph.
05

Draw the Line Through the Points

Draw a straight line through the points \((0, -\frac{4}{3})\) and \((3, -\frac{1}{3})\) to represent the equation on the graph.
06

Find the X-Intercept

To find the x-intercept, set \(y = 0\) in the original equation \(x - 3y = 4\), and solve for \(x\):\(x - 3(0) = 4\) gives \(x = 4\).Thus, the x-intercept is \((4, 0)\).Plot the x-intercept point on the graph and verify that the line passes through \((4, 0)\).
07

Name the Intercepts

List the intercepts found:- Y-intercept: \((0, -\frac{4}{3})\).- X-intercept: \((4, 0)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope-Intercept Form
The slope-intercept form is an essential way of expressing a linear equation. It allows you to quickly identify important features of a line. The standard form of the slope-intercept equation is \(y = mx + b\). Here, \(m\) represents the slope of the line, which indicates how steep the line is, and \(b\) is the y-intercept, the point where the line crosses the y-axis.

Converting equations to slope-intercept form is a common task. Start with any linear equation, such as \(x - 3y = 4\). Your goal is to solve this for \(y\). First, rearrange the equation: subtract \(x\) from both sides to get \(-3y = -x + 4\). Then, divide every term by \(-3\) to isolate \(y\), resulting in \(y = \frac{1}{3}x - \frac{4}{3}\).

This final equation visually prepares you for graphing by clearly showing the slope \(m = \frac{1}{3}\) and y-intercept \(b = -\frac{4}{3}\).
X-Intercept
Finding the x-intercept is an integral part of graphing linear equations. The x-intercept is where the line crosses the x-axis, meaning the y-value is zero at this point. To find it, you substitute \(y = 0\) into the equation and solve for \(x\).

For example, using the equation \(x - 3y = 4\), set \(y = 0\):
  • Substitute \(y = 0\) into the equation: \(x - 3(0) = 4\).
  • Solve for \(x\): \(x = 4\).

The x-intercept here is \((4, 0)\). Remember this point as it's critical for understanding the line's behavior. It is also essential for accurately drawing the graph, as it provides a point where the line "starts" on the right or left depending on its slope and direction.
Y-Intercept
The y-intercept is another crucial aspect of graphing linear equations. This is the point on the graph where the line crosses the y-axis. To locate the y-intercept from the slope-intercept form, simply look at the constant term \(b\) in the equation \(y = mx + b\). This \(b\) value is the y-coordinate of the intercept.

For the equation \(y = \frac{1}{3}x - \frac{4}{3}\), the y-intercept is \(-\frac{4}{3}\). This gives you the point \((0, -\frac{4}{3})\), which you can plot directly on the y-axis.
  • Plotting this point is your starting step in graphing.
  • The y-intercept allows you to "anchor" the line.

The y-intercept provides a foundation. From this point, you use the slope to find other points and complete the line.

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