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Chapter 0: Problem 18

Find the slope and \(y\)-intercept (if possible) of the equation of the line. Sketch the line. \(y+4=0\)

Short Answer

Expert verified
The slope of the line \(y + 4 = 0\) is 0 and the y-intercept is -4. The graph of this line is a horizontal line crossing the y-axis at point (0, -4).

Step by step solution

01

Identifying the Slope

First, identify the slope of this equation. Since there is no variable \(x\) in the equation, the slope which is represented by the coefficient of \(x\) is zero. Therefore, the slope of the line can be written as \(m = 0\).
02

Identifying the y-intercept

Next, identify the y-intercept. The y-intercept is the value of \(y\) when \(x = 0\). In this equation \(y + 4 = 0\), solving for \(y\), we get \(y = -4\). Therefore, the line crosses the y-axis at \(y = -4\). Thus, the y-intercept can be written as (0, -4).
03

Sketching the Line

Finally, sketch the line using the identified slope and y-intercept. As the slope is zero, it indicates a horizontal line. The horizontal line crosses the y-axis at -4. Plot (0, -4) on the y-axis and draw a straight horizontal line through this point to sketch the line.

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Key Concepts

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

Understanding Slope
The concept of slope is crucial when analyzing linear equations. The slope of a line describes its steepness, which can be determined from the equation of the line. In general, slope is often represented by the letter \(m\) and calculated as the ratio of the change in the \(y\) values to the change in the \(x\) values (rise over run). More simply, this is expressed as:
  • Slope \(m = \frac{\Delta y}{\Delta x}\)
If \(m = 0\), the line is horizontal, meaning it has no inclination whatsoever. This results in all the \(y\) coordinates being the same regardless of the \(x\) value. It moves purely horizontally across the \(x\)-axis.
A slope of zero is a defining feature of horizontal lines, which we'll talk more about later.
Grasping the y-intercept
The \(y\)-intercept is the point where a line crosses the \(y\)-axis. In the equation of a line given in the form \(y = mx + b\), the \(y\)-intercept is \(b\). It represents the value of \(y\) when \(x\) equals zero.
A line can only have one \(y\)-intercept. This point is crucial for graphing because it serves as the starting point to sketch the line. In the equation \(y+4=0\), solving it gives us \(y = -4\), which means the line crosses the \(y\)-axis at \((0, -4)\).
  • The \(y\)-intercept is an essential point for drawing the line on a graph as it grounds the line's position relative to the axes.
  • Even without a varying \(x\), a line still needs this anchor point.
Graphing Lines Made Easy
Graphing lines involves plotting points and connecting them with a straight path. When graphing, it's essential to know the slope and \(y\)-intercept of the line. For the equation \(y + 4 = 0\), where \(m = 0\) and the \(y\)-intercept is \((0, -4)\):
  • Start by plotting the \(y\)-intercept on the graph.
  • Since the slope is zero, the line will be horizontal, spanning across the plane parallel to the \(x\)-axis.
  • Draw a straight line through the point \((0, -4)\).
This approach gives a visual representation of the equation and helps in understanding the underlying relationship between \(x\) and \(y\) coordinates.
Exploring Horizontal Lines
Horizontal lines have unique characteristics that set them apart from other lines. They occur when the slope \(m\) is zero, which implies no vertical change as you move along the line. In essence, these lines maintain a constant \(y\) value.
  • Equation form: \(y = k\).
  • No matter how \(x\) changes, \(y\) remains unchanged.
For instance, in our example \(y+4=0\) simplifies to \(y=-4\). This indicates a horizontal line where every point on the line has \(y = -4\).
Horizontal lines are simple to plot. Just remember they are parallel to the \(x\)-axis and fixed at a particular \(y\) value.

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Most popular questions from this chapter

(a) use the position equation \(s=-16 t^{2}+v_{0} t+s_{0}\) to write a function that represents the situation, (b) use a graphing utility to graph the function, (c) find the average rate of change of the function from \(t_{1}\) to \(t_{2}\), (d) interpret your answer to part (c) in the context of the problem, (e) find the equation of the secant line through \(t_{1}\) and \(t_{2}\), and (f) graph the secant line in the same viewing window as your position function. An object is thrown upward from ground level at a velocity of 120 feet per second. $$ t_{1}=3, t_{2}=5 $$

In Exercises 39-54, (a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=\frac{6 x+4}{4 x+5} $$

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$ \begin{array}{cc} \text { Function } & x \text {-Values } \\ f(x)=-2 x+15 & x_{1}=0, x_{2}=3 \end{array} $$

In Exercises 39-54, (a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=3 x+1 $$

The total numbers \(f\) (in billions) of miles traveled by motor vehicles in the United States from 1995 through 2002 are shown in the table. The time (in years) is given by \(t\), with \(t=5\) corresponding to 1995 . (Source: U.S. Federal Highway Administration) $$ \begin{array}{|c|c|} \hline 0 \text { Year, } t & \text { Miles traveled, } f(t) \\ \hline 5 & 2423 \\ 6 & 2486 \\ 7 & 2562 \\ 8 & 2632 \\ 9 & 2691 \\ 10 & 2747 \\ 11 & 2797 \\ 12 & 2856 \\ \hline \end{array} $$ (a) Does \(f^{-1}\) exist? (b) If \(f^{-1}\) exists, what does it mean in the context of the problem? (c) If \(f^{-1}\) exists, find \(f^{-1}\) (2632). (d) If the table was extended to 2003 and if the total number of miles traveled by motor vehicles for that year was 2747 billion, would \(f^{-1}\) exist? Explain.
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