Problem 8
Question
Use the \(x\) - and \(y\)-intercepts to graph each linear equation. \(3 x=2 y+6\)
Step-by-Step Solution
Verified Answer
The \(y\)-intercept is \(-3\) and the \(x\)-intercept is \(2\).
1Step 1: Find the \(y\)-intercept
To find the \(y\)-intercept, set \(x=0\) in the equation. So, start with the original equation \(3 x=2 y+6\), and substitute \(0\) for \(x\). Simplify to find the value of \(y\) which is \(y=-3\). Thus, the \(y\)-intercept is \(-3\). This is the point where the line crosses the \(y\)-axis.
2Step 2: Find the \(x\)-intercept
To find the \(x\)-intercept, set \(y=0\) in the equation. Start with the original equation \(3 x=2 y+6\), substitute \(0\) for \(y\) and simplify to find the value of \(x\) which is \(x=2\). Thus, the \(x\)-intercept is \(2\). This is the point where the line crosses the \(x\)-axis.
3Step 3: Plot the intercepts and the line
Plot the \(y\)-intercept at \((0, -3)\) on the graph and the \(x\)-intercept at \((2, 0)\). Now, draw a straight line that passes through these two points. This line is the graph of the equation \(3 x=2 y+6\).
Key Concepts
Understanding X-InterceptsUnderstanding Y-InterceptsSolving Linear Equations for Graphing
Understanding X-Intercepts
When we talk about the x-intercept, we are focusing on the point where a line crosses the x-axis in a graph. At this particular point, the value of y is always zero.
This means to find the x-intercept of a linear equation, you need to set y to zero in the equation and solve for x.
For the equation given in the exercise, which is \(3x = 2y + 6\), setting y to zero simplifies it to \(3x = 6\).
By solving this equation, you'll find that the x-intercept is at \(x = 2\).
This means to find the x-intercept of a linear equation, you need to set y to zero in the equation and solve for x.
For the equation given in the exercise, which is \(3x = 2y + 6\), setting y to zero simplifies it to \(3x = 6\).
By solving this equation, you'll find that the x-intercept is at \(x = 2\).
- Remember, x-intercept always happens at \((x, 0)\) on the graph.
- It represents the value of x when y is zero.
Understanding Y-Intercepts
The y-intercept is a key concept just like the x-intercept, but it represents where the line crosses the y-axis. At this special point, the value of x is always zero.
So, to find the y-intercept of a linear equation, you need to set x to zero in the equation and solve for y.
In the case of the equation \(3x = 2y + 6\) from the exercise, setting x to zero changes the equation to \(-2y = 6\). This simplifies the problem, showing that \(y = -3\).
Thus, the y-intercept for this equation is at \((0, -3)\).
So, to find the y-intercept of a linear equation, you need to set x to zero in the equation and solve for y.
In the case of the equation \(3x = 2y + 6\) from the exercise, setting x to zero changes the equation to \(-2y = 6\). This simplifies the problem, showing that \(y = -3\).
Thus, the y-intercept for this equation is at \((0, -3)\).
- Y-intercept will always appear at \((0, y)\) on the graph.
- It shows the value of y when x is zero.
Solving Linear Equations for Graphing
To graph linear equations effectively, it's very useful to start with finding both the x-intercept and y-intercept. They provide two specific points through which you can draw the line of the equation.
By doing this, you turn an abstract equation, like \(3x = 2y + 6\), into something visual and comprehensible.
Having a firm grasp on intercepts and their use in graphing equips you with a solid foundation for working with more complex functions later on.
By doing this, you turn an abstract equation, like \(3x = 2y + 6\), into something visual and comprehensible.
- First, you will determine the y-intercept by setting x to zero and solve for y.
- Second, you will find the x-intercept by setting y to zero and solve for x.
- With these two intercepts plotted on the graph, you can connect them with a straight path, revealing the line of the equation.
Having a firm grasp on intercepts and their use in graphing equips you with a solid foundation for working with more complex functions later on.
Other exercises in this chapter
Problem 8
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