/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Graph the linear function \(f\) ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 60

Graph the linear function \(f\) on a domain of \([-10,10]\) for the function whose slope is \(\frac{1}{8}\) and \(y\) -intercept is \(\frac{31}{16} .\) Label the points for the input values of \(-10\) and \(10 .\)

Short Answer

Expert verified
The function is graphed with points at \((-10, \frac{11}{16})\) and \((10, \frac{51}{16})\).

Step by step solution

01

Identify the Linear Function Equation

The generic form of a linear function is \( y = mx + b \), where \( m \) is the slope and \( b \) is the \( y \)-intercept. Given \( m = \frac{1}{8} \) and \( b = \frac{31}{16} \), the equation of the function becomes \( y = \frac{1}{8}x + \frac{31}{16} \).
02

Calculate the Output for Input \( x = -10 \)

Substitute \( x = -10 \) into the equation \( y = \frac{1}{8}x + \frac{31}{16} \). Compute \( y = \frac{1}{8}(-10) + \frac{31}{16} = -\frac{10}{8} + \frac{31}{16} = -\frac{5}{4} + \frac{31}{16} = \frac{-20}{16} + \frac{31}{16} = \frac{11}{16} \). The point is \((-10, \frac{11}{16})\).
03

Calculate the Output for Input \( x = 10 \)

Substitute \( x = 10 \) into the equation \( y = \frac{1}{8}x + \frac{31}{16} \). Compute \( y = \frac{1}{8}(10) + \frac{31}{16} = \frac{10}{8} + \frac{31}{16} = \frac{5}{4} + \frac{31}{16} = \frac{20}{16} + \frac{31}{16} = \frac{51}{16} \). The point is \((10, \frac{51}{16})\).
04

Plot the Points and Graph the Line

On a graph, mark the points \((-10, \frac{11}{16})\) and \((10, \frac{51}{16})\). Draw a straight line through these points, extending it to cover the domain \([-10, 10]\). Make sure the slope of the line visually matches \(\frac{1}{8}\) and that it crosses the \( y\)-axis at \(\frac{31}{16}\).
05

Label the Points and Verify the Graph

Label the points \((-10, \frac{11}{16})\) and \((10, \frac{51}{16})\) on the graph. Verify that the line passes through these points and aligns with the calculated slope and \(y\)-intercept. This ensures the accuracy of the graph.

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Key Concepts

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

linear equations
Linear equations are the backbone of algebra, describing a straight line on a graph. These equations are typically written in the form \( y = mx + b \), where \( m \) represents the slope, and \( b \) is the y-intercept. Each variable and number in this equation holds a specific meaning:
  • \( y \) and \( x \) are the dependent and independent variables, respectively.
  • The slope \( m \) indicates how steep the line is.
  • The y-intercept \( b \) shows where the line crosses the y-axis.
Linear equations are useful because they model relationships where one quantity changes at a constant rate with another. For example, if you have a linear function describing time versus distance on a trip, the \( m \) could be speed, and \( b \) could be the starting distance, like a head start.
slope-intercept form
The slope-intercept form of a linear equation makes it easy to graph. It鈥檚 given as \( y = mx + b \). This form is popular because it straightforwardly shows both the slope and the y-intercept, simplifying the graphing process. Here鈥檚 a breakdown of the components:
  • Slope \( m \): It measures the rise over the run, or how much \( y \) increases as \( x \) increases. For instance, if \( m = \frac{1}{8} \), every 8 units you move to the right on the x-axis, you move up by 1 unit on the y-axis.
  • Y-intercept \( b \): This is the point where the line touches the y-axis. If \( b = \frac{31}{16} \), the line starts above the origin at this value on the y-axis.
By placing a line on a graph using these two values, you can picture its behavior long before sketching. Understanding this helps you see how variables relate to each other.
plotting points
Plotting points on a graph is a fundamental step in visualizing linear equations. Each point has an \( x \)-coordinate and a \( y \)-coordinate, expressed as \( (x, y) \). Here鈥檚 how you plot a point:
  • Find the \( x \)-coordinate on the horizontal x-axis.
  • From there, move vertically to the \( y \)-coordinate, and mark the point.
For example, with the linear function \( y = \frac{1}{8}x + \frac{31}{16} \), you calculate points such as \((-10, \frac{11}{16})\) and \((10, \frac{51}{16})\). Each point confirms that when you substitute \( x \) into the equation, you get a precise \( y \) value. After acquiring these points, connecting them with a line will represent the entire function graphically. It鈥檚 like finding breadcrumbs that lead you, step-by-step, to the complete path described by the function.
domain and range
Understanding the domain and range of a function is crucial to graphing correctly. The domain is the set of all possible \( x \)-values that you can input into a function, while the range is all the possible \( y \)-values it can output. For linear functions, the domain is straightforward unless otherwise restricted.
  • Domain: In this problem, the domain is \([-10, 10]\), meaning the function only includes \( x \)-values from -10 to 10.
  • Range: Once you substitute these \( x \)-values into the linear equation, the resulting \( y \)-values determine the range. In our case, the range is from \( \frac{11}{16} \) to \( \frac{51}{16} \) based on our calculated points.
Graphing over a specific domain ensures that you only focus on the region that matters for your problem, highlighting how the function behaves in a given interval. This precision helps in avoiding information overload and keeping the focus on relevant details.

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Most popular questions from this chapter

Does the following table represent a linear function? If so, find the linear equation that models the data. $$\begin{array}{|c|c|c|c|c|}\hline x & {-4} & {0} & {2} & {10} \\ \hline g(x) & {18} & {-2} & {-12} & {-52} \\ \hline\end{array}$$

For the following exercises, consider the data in Table 2.18, which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year. $$\begin{array}{|c|c|c|c|c|c|}\hline \text { Year } & {2000} & {2002} & {2005} & {2007} & {2010} \\ \hline \text { Percent Graduates } & {6.5} & {7.0} & {7.4} & {8.2} & {9.0} \\ \hline\end{array}$$ Based on the set of data given in Table 2.20, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. $$\begin{array}{|c|c|c|c|c|c|}\hline x & {10} & {12} & {15} & {18} & {20} \\\ \hline y & {36} & {34} & {30} & {28} & {22} \\ \hline\end{array}$$

For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population and the year over the ten-year span, (population, year) for specific recorded years: $$(2500,2000),(2650,2001),(3000,2003),(3500,2006),(4200,2010)$$ Use linear regression to determine a function \(y\), where the year depends on the population. Round to three decimal places of accuracy.

A clothing business finds there is a linear relationship between the number of shirts, \(n,\) it can sell and the price, \(p,\) it can charge per shirt. In particular, historical data shows that \(1,000\) shirts can be sold at a price of \(\$ 30\) , while \(3,000\) s shirts can be sold at a price of \(\$ 22 .\) Find a linear equation in the form \(p(n)=m n+b\) that gives the price \(p\) they can charge for \(n\) shirts.

When temperature is 0 degrees Celsius, the Fahrenheit temperature is \(32 .\) When the Celsius temperature is \(100,\) the corresponding Fahrenheit temperature is \(212 .\) Express the Fanrenheit temperature as a linear function of \(C,\) the Celsius temperature, \(F(C) .\) a. Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius. b. Find and interpret \(F(28)\) . c. Find and interpret \(F(-40)\) .
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