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Chapter 2: Problem 54

Graph each equation in a rectangular coordinate system. \(x-0\)

Short Answer

Expert verified
The graph of the equation \(x = 0\) is a vertical line that coincides with the y-axis.

Step by step solution

01

Understand the equation

The equation given is \(x = 0\). This equation is in the form of \(x = k\) where \(k\) is a constant. The graph of any equation in this form is a vertical line.
02

Find the interception points with axes

For the equation \(x = 0\), the line intersects the x-axis at 0 and all other y-axis values. Thus, this line coincides with the y-axis.
03

Graph the line

Plot the line \(x = 0\) on the rectangular coordinate system. The graph is a vertical line that coincides with the y-axis.

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Key Concepts

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

Rectangular Coordinate System
The rectangular coordinate system, also known as the Cartesian coordinate system, is a two-dimensional plane that allows us to graph equations using coordinates. This system consists of two axes:
  • The horizontal axis, known as the x-axis.
  • The vertical axis, known as the y-axis.
These two axes intersect at the origin, which is the point (0, 0). Each point on the plane is represented by a pair of numbers \(x, y\), where 'x' indicates the horizontal position and 'y' indicates the vertical position.

This system is essential for graphing linear equations, as it gives us a visual way to see the relationships between different variables.
Vertical Line
A vertical line in the rectangular coordinate system is a line that goes straight up and down. It does not tilt or lean. This kind of line has a special characteristic: all the points on the line have the same x-coordinate.

For instance, the equation \(x = 0\) describes a vertical line. This means no matter what y-value you choose, the x-value will always be 0. Consequently, for every point on this line, x remains constant while y can be any real number.

Vertical lines are unique because they cannot be described by the usual y = mx + b form, as their slope is undefined.
Interception Points
Interception points are where a line crosses the axes on the coordinate plane. In the equation \(x = 0\), the line intersects the x-axis at the origin (0,0) because that is the point where x is zero.

Since this is a vertical line, it does not cross the y-axis at any single point but rather runs directly along the y-axis itself. The points of intersection on the coordinate axes help in visualizing and graphing the line more accurately.
  • On the x-axis: Intersection at (0,0).
  • On the y-axis: Line coincides along the y-axis.
Equation of the Form x = k
Equations of the form \(x = k\) describe vertical lines where 'k' is a constant. Here, x has a fixed value, meaning for any point on this line, x always equals k, regardless of y.

For example, when given the equation \(x = 0\), this tells us that the line is vertical and passes through the x-coordinate of 0.

This form simplifies graphing since you only need to know the constant value of x to draw the line. It's an easy way to plot a straight vertical line on the coordinate plane.

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Most popular questions from this chapter

Use a graphing utility to graph each function. Use a \([-5,5,1]\) by \([-5,5,1]\) viewing rectangle. Then find the intervals on which the function is increasing, decreasing, or constant. $$g(x)=x^{\frac{2}{3}}$$

Begin by graphing the cube root function, \(f(x)-\sqrt[3]{x} .\) Then use transformations of this graph to graph the given function. $$ g(x)-\sqrt[3]{x}-2 $$

Begin by graphing the standard cubic function, \(f(x)-x^{3} .\) Then use transformations of this graph to graph the given function. $$ g(x)-(x-2)^{3} $$

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The graph of \((x-4)+(y+6)=25\) is a circle with radius 5 centered at \((4,-6)\)

Begin by graphing the absolute value function, \(f(x)-|x| .\) Then use transformations of this graph to graph the given function. $$ h(x)-2|x+4| $$
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