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Give the equation and graph for a line with y-intercept and slope given in Exercises. $$y \text { -intercept }=-2.5 ; \text { slope }=5$$

Short Answer

Expert verified
Answer: The equation of the line is y = 5x - 2.5.

Step by step solution

01

Understand the problem

We need to find the equation of a line given its slope and y-intercept. The slope (\(m\)) is given as \(5\), and the y-intercept (\(b\)) is given as \(-2.5\).
02

Use the slope-intercept form

The slope-intercept form of a linear equation is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. We will plug the given values for slope and y-intercept into this equation.
03

Substitute the given values

We are substituting the slope, \(m = 5\), and the y-intercept, \(b = -2.5\), into the equation \(y = mx + b\). This gives us the equation: $$y = 5x -2.5$$
04

Graph the line

To graph the line, we will: 1. Plot the y-intercept, \(-2.5\), on the y-axis. 2. Use the slope to find another point on the line. The slope is the ratio of the change in \(y\) to the change in \(x\) (i.e., rise over run). With a slope of \(5\), this means that for every 1 unit increase in \(x\), \(y\) increases by 5 units. So, if we start at our y-intercept point \((-2.5)\), we can go up 5 units and to the right 1 unit to obtain another point on the line: \((1, 2.5)\). 3. Draw a line through these two points, extending in both directions, to complete the graph. The graph of the line with y-intercept \(-2.5\) and slope \(5\) is a straight line passing through the points \((0, -2.5)\) and \((1, 2.5)\).

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Key Concepts

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

Slope-Intercept Form
The slope-intercept form is a quick method to write the equation of a straight line given the slope and y-intercept. The general format is given by the equation
\[ y = mx + b \]
where \( m \) represents the slope and \( b \) stands for the y-intercept. This form is especially useful because it readily provides both the gradient and the starting point of the line on the y-axis.
To illustrate, let's connect it with the problem at hand. Given the slope of \( 5 \) and a y-intercept of \( -2.5 \), one simply substitutes these values into the slope-intercept form to arrive at:
\[ y = 5x - 2.5 \]
This equation is practical because with any value for \( x \), one can instantly solve for \( y \), making it straightforward to graph or to calculate points along the line.
Y-Intercept
Understanding the y-intercept is crucial for graphing and interpreting linear equations. The y-intercept is where the line crosses the y-axis, pinpointing the exact moment when \( x \) equals zero. This feature makes it an invaluable reference point in graphing because you can always begin at a known location on the graph.
With a y-intercept of \( -2.5 \), as in our example, this means that when \( x = 0 \), the value of \( y \) is \( -2.5 \). Therefore, the initial point you would plot on a graph is \( (0, -2.5) \). It signifies the starting height of the line when moving from left to right across the graph. The y-intercept also gives us a glimpse into the function's behavior; in this case, indicating that the line will pass below the origin since the y-intercept is negative.

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Most popular questions from this chapter

Use the data given in Exercises 6-7 (Exercises 17-18, Section 12.1). Construct the ANOVA table for a simple linear regression analysis, showing the sources, degrees of freedom, sums of squares, and mean sauares. $$\begin{aligned}&\text { Six points have these coordinates: }\\\&\begin{array}{l|llllll}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 9.7 & 6.5 & 6.4 & 4.1 & 2.1 & 1.0\end{array}\end{aligned}$$ a. Find the least-squares line for the data. b. Plot the six points and graph the line. Does the line appear to provide a good fit to the data points? c. Use the least-squares line to predict the value of \(y\) when \(x=3.5\) d. Fill in the missing entries in the MS Exce/ analysis of variance table.

Use the information given (reproduced below) to find a prediction interval for a particular value of \(y\) when \(x=x_{0} .\) Is the interval wider than the corresponding confidence interval from Exercises \(3-4 ?\) $$ \begin{array}{l} n=10, \mathrm{SSE}=24, \Sigma x_{i}=59, \Sigma x_{i}^{2}=397, \\ \hat{y}=.074+.46 x, x_{0}=5,90 \% \text { prediction interval } \end{array} $$

Give the y-intercept and slope for the line. $$y=2 x+3$$

Is there any relationship between these two variables? To find out, we randomly selected 12 people from a data set constructed by Allen Shoemaker (Journal of Statistics Education) and recorded their body temperature and heart rate. \({ }^{19}\) $$\begin{array}{lllllll}\hline \text { Person } & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline \text { Temperature } & 96.3 & 97.4 & 98.9 & 99.0 & 99.0 & 96.8 \\ \text { (degrees) } & & & & & & \\\\\text { Heart Rate } & 70 & 68 & 80 & 75 & 79 & 75 \\\\\text { (beats per } & & & & & & \\\\\text { minute) } & & & & & & \\\ \hline\end{array}$$ $$\begin{array}{lllllll}\hline \text { Person } & 7 & 8 & 9 & 10 & 11 & 12 \\\\\hline \text { Temperature } & 98.4 & 98.4 & 98.8 & 98.8 & 99.2 & 99.3 \\\\\multicolumn{2}{l} {\text { (degrees) }} & & & & & & \\\\\text { Heart Rate } & 74 & 84 & 73 & 84 & 66 & 68 \\\\\text { (beats per } & & & & & & \\\\\text { minute) } & & & & & & & \\\& & & & \\\\\hline\end{array}$$ a. Find the correlation coefficient \(r\), relating body temperature to heart rate. b. Is there sufficient evidence to indicate that there is a correlation between these two variables? Test at the 5\% level of significance.

Use the information given to find a confidence interval for the average value of \(y\) when \(x=x_{0}\). $$ \begin{array}{l} n=10, \mathrm{SSE}=24, \Sigma x_{i}=59, \Sigma x_{i}^{2}=397, \\ \hat{y}=.074+.46 x, x_{0}=5,90 \% \text { confidence level } \end{array} $$

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