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Chapter 2: Problem 44

Exer. \(41-44:\) Use the slope-intercept form to find the slope and \(y\) -intercept of the given line, and sketch its graph. $$x-5 y=-15$$

Short Answer

Expert verified
The slope is \( \frac{1}{5} \) and the y-intercept is \( 3 \).

Step by step solution

01

Understand the Slope-Intercept Form

The slope-intercept form of a line is given by the equation \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
02

Solve for y to Convert to Slope-Intercept Form

Given the line equation \( x - 5y = -15 \), we need to solve for \( y \) to convert it to \( y = mx + b \) form. Subtract \( x \) from both sides:\[ -5y = -x - 15 \] Now divide each term by \(-5\):\[ y = \frac{1}{5}x + 3 \]
03

Identify the Slope and Y-Intercept

From the equation \( y = \frac{1}{5}x + 3 \), we can identify the slope \( m \) as \( \frac{1}{5} \) and the y-intercept \( b \) as \( 3 \).
04

Sketch the Graph

To sketch the graph, start by plotting the y-intercept on the y-axis at point \( (0, 3) \). Next, use the slope \( \frac{1}{5} \) to determine the rise over run. From \( (0, 3) \), move up 1 unit and right 5 units to plot another point at \( (5, 4) \). Finally, draw a line through the points \( (0, 3) \) and \( (5, 4) \) to complete the graph.

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Key Concepts

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

Understanding Slope
The slope of a line is a crucial aspect of linear equations. It is represented by the symbol \( m \) in the slope-intercept form \( y = mx + b \). The slope denotes how steep a line is, essentially measuring the rate at which \( y \) changes as \( x \) changes. In simple terms, it tells us how much \( y \) will increase or decrease when \( x \) increases by one unit. This is often referred to as "rise over run."
\( m = \text{rise} / \text{run} \)
When we have a positive slope, the line rises as we move from left to right. Conversely, a negative slope means the line falls. In this exercise, the slope \( \frac{1}{5} \) tells us that for every 5 units we move to the right, the line rises by 1 unit. This gradual ascent reflects the line's gentle slope.
Identifying the Y-Intercept
The y-intercept is where the line crosses the y-axis. It's denoted by \( b \) in the slope-intercept form \( y = mx + b \). The y-intercept provides valuable information as it marks the start of our graph on the y-axis, especially when \( x = 0 \).
In the equation \( y = \frac{1}{5}x + 3 \), the y-intercept is \( 3 \). This means that when \( x = 0 \), \( y \) will be \( 3 \). In practical terms, it's the point where you start plotting the line. It's always helpful to note this point first before utilizing the slope to draw more accurate graphs.
Graphing Lines with the Slope-Intercept Form
Graphing lines can be simplified using the slope-intercept form. Once the equation is converted into \( y = mx + b \) format, sketching becomes straightforward. You start by plotting the y-intercept on the y-axis.
For the equation \( y = \frac{1}{5}x + 3 \), the first point is \( (0, 3) \). After marking this point, use the slope to find another. With a slope of \( \frac{1}{5} \), move up 1 unit and 5 units to the right from the intercept. Repeat this to draw a second point at \( (5, 4) \).
Finally, draw a line through these points extending in both directions. This visually represents the equation, making it easier to comprehend relationships and intersections with other lines.
Basics of Linear Equations
Linear equations represent straight lines and take the general form \( ax + by = c \). Converting them to slope-intercept form \( y = mx + b \) provides a clearer understanding of their characteristics. Each linear equation consists of two components:
  • The slope \( m \) determines the angle's steepness.
  • The y-intercept \( b \) marks the starting point on the y-axis.
Through linear equations like \( x - 5y = -15 \), solving for \( y \) provides a comprehensive view. By rearranging to get \( y = \frac{1}{5}x + 3 \), one easily discerns both the slope and intercept.
Linear equations are foundational in mathematics, applicable in various scenarios like predicting trends, determining rates, and calculating distances. Mastering them offers a significant advantage in mathematics and related fields.

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Most popular questions from this chapter

Parabolic path Under ideal conditions an object thrown from level ground will follow a parabolic path of the form \(f(x)=a x^{2}+b x,\) where \(a\) and \(b\) are constants and \(x\) represents the horizontal distance traveled by the object. (a) Determine \(a\) and \(b\) so that the object reaches a maximum height of 100 feet and travels a horizontal distance of 150 feet before striking the ground. (b) Graph \(f(x)=a x^{2}+b x\) in the viewing rectangle \([0,180,50]\) by \([0,120,50]\) (c) Graph \(y=k a x^{2}+b x,\) where \(k=\frac{1}{4}, \frac{1}{2}, 1,2,4,\) in the same viewing rectangle of \([0,600,50]\) by \([0,400,50]\) How does the constant \(k\) affect the path of the object?

Sketch the graph of the equation. $$y=\left|4-x^{2}\right|$$

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(a) Use the quadratic formula to find the zeros of \(f .\) (b) Find the maximum or minimum value of \(f(x)\) (c) Sketch the graph of \(f\) $$f(x)=-2 x^{2}+16 x-26$$

Precipitation in Seattle The average monthly precipitation (in inches) for Seattle is listed in the following table. (Note: April average is not given.) (a) Plot the average monthly precipitation. (b) Model the data with a quadratic function of the form \(f(x)=a(x-h)^{2}+k .\) Graph \(f\) and the data on the same coordinate axes. (c) Use \(f\) to predict the average rainfall in April. Compare your prediction with the actual value of 2.55 inches. $$\begin{array}{|l|c|} \hline \text { Month } & \text { Precipitation } \\ \hline \text { Jan. } & 5.79 \\ \hline \text { Feb. } & 4.02 \\ \hline \text { Mar. } & 3.71 \\ \hline \text { April } & \\ \hline \text { May } & 1.70 \\ \hline \text { June } & 1.46 \\ \hline \text { July } & 0.77 \\ \hline \text { Aug. } & 1.10 \\ \hline \text { Sept. } & 1.72 \\ \hline \text { Oct. } & 3.50 \\ \hline \text { Nov. } & 5.97 \\ \hline \text { Dec. } & 5.81 \\ \hline \end{array}$$
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