Problem 3
Question
Graph each inequality. $$y>1$$
Step-by-Step Solution
Verified Answer
Graph a dashed horizontal line at \( y = 1 \) and shade above.
1Step 1: Understand the Inequality
The inequality given is \( y > 1 \). This means we are considering all the points in the coordinate plane where the \( y \)-value is greater than 1.
2Step 2: Graph the Boundary Line
Start by graphing the line \( y = 1 \), which serves as the boundary for the inequality. This line is a horizontal line passing through all points with a \( y \)-coordinate of 1.
3Step 3: Determine the Type of Line
Since the inequality is \( y > 1 \) (and not \( y \geq 1 \)), the boundary line should be dashed. This indicates that points on the line \( y = 1 \) are not solutions to the inequality.
4Step 4: Shade the Solution Area
Shade the area above the dashed horizontal line. This region represents all points where the \( y \)-coordinate is greater than 1, satisfying the inequality \( y > 1 \).
Key Concepts
Coordinate PlaneBoundary LineShading Solution AreaHorizontal Line
Coordinate Plane
The coordinate plane, also known as the Cartesian plane, is a two-dimensional surface on which we can plot points, lines, and graphs. It consists of two axes: the horizontal axis (x-axis) and the vertical axis (y-axis). These axes intersect at the origin, which is the point (0, 0).
The coordinate plane is divided into four quadrants:
The coordinate plane is divided into four quadrants:
- Quadrant I: Both x and y are positive.
- Quadrant II: x is negative and y is positive.
- Quadrant III: Both x and y are negative.
- Quadrant IV: x is positive and y is negative.
Boundary Line
The boundary line is a crucial part of graphing inequalities on the coordinate plane. It represents the set of points that satisfy the equation derived from the inequality. In the exercise, we derived the equation \(y = 1\) from the inequality \(y > 1\).
This line acts as a dividing path between the solution area and the non-solution area. It defines the edge of the region we are interested in. Depending on the inequality, the boundary line can be:
This line acts as a dividing path between the solution area and the non-solution area. It defines the edge of the region we are interested in. Depending on the inequality, the boundary line can be:
- Solid, if the inequality is \(\geq\) or \(\leq\), which means the points on the line are included in the solution set.
- Dashed, like in our exercise with \(y > 1\), indicating that points on this line are not part of the solution.
Shading Solution Area
Shading the solution area is the final step in graphing an inequality, such as \(y > 1\). The shading represents all points on the coordinate plane where the inequality holds true, showing the region that satisfies it.
For our exercise:
For our exercise:
- The boundary line is dashed, indicating it's not part of the solution.
- We shade the region above the line \(y = 1\), reflecting all points where the \(y\)-value is greater than 1.
Horizontal Line
Horizontal lines are unique because they run parallel to the x-axis. In graphing, they have the equation form \(y = c\), where \(c\) is a constant. In our inequality graph \(y > 1\), the horizontal line is \(y = 1\).
Important characteristics of horizontal lines include:
Important characteristics of horizontal lines include:
- All points along the line have the same y-value.
- They are perpendicular to vertical lines, which have the form \(x = c\).
Other exercises in this chapter
Problem 2
Find the dimension of each matrix. Identify any square, column, or rove matrices. Do not use a calculator. $$\left[\begin{array}{rrr}2 & -8 & 6 \\ 1 & 0 & -5 \\
View solution Problem 2
Verify that the given ordered triple is a solution of the system. Do not use a calculator. $$\begin{aligned} \left(\frac{1}{2},-\frac{3}{4}, \frac{1}{6}\right)
View solution Problem 3
Find each determinant. Do not use a calculator. $$\operatorname{det}\left[\begin{array}{rr}-1 & -2 \\\5 & 3\end{array}\right]$$
View solution Problem 3
Determine whether A and B are imerses by calculating AB and BA. Do not use a calculator. $$A=\left[\begin{array}{rr} -1 & 2 \\ 3 & -5 \end{array}\right] ; B=\le
View solution