/*! 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 4 Graph each equation on your calc... [FREE SOLUTION] | 91Ó°ÊÓ

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Graph each equation on your calculator, and trace to find the approximate \(y\)-value for the given \(x\)-value. a. \(y=1.21-x\) when \(x=70.2\) b. \(y=6.02+44.3 x\) when \(x=96.7\) c. \(y=-0.06+0.313 x\) when \(x=0.64\) d. \(y=1183-2140 x\) when \(x=-111\)

Short Answer

Expert verified
a: \(y = -68.99\); b: \(y = 4286.41\); c: \(y = 0.14\); d: \(y = 238723\).

Step by step solution

01

Understanding the Problem

We need to find the approximate \(y\)-value for each equation provided when given specific \(x\)-values. The equations are linear and are in the form \(y = mx + b\). Our task is to use a graphing calculator to plot each equation and determine the \(y\)-value by tracing the calculator's graph to the requested \(x\)-value.
02

Graph Equation a

For equation a, \(y = 1.21 - x\), input the equation into the calculator. Enter the graphing mode and ensure the plot window setting allows you to see \(x = 70.2\). Use the trace function to move to \(x = 70.2\). The calculator will display the associated \(y\)-value.
03

Calculate Manually for Equation a

Instead of using a calculator, we can directly substitute \(x = 70.2\) into the equation \(y = 1.21 - x\). Thus, \(y = 1.21 - 70.2 = -68.99\).
04

Graph Equation b

For equation b, \(y = 6.02 + 44.3x\), enter the equation on the calculator. Ensure your window settings cover \(x = 96.7\) appropriately. Use the trace function to locate \(x = 96.7\) and find the associated \(y\)-value.
05

Calculate Manually for Equation b

Substitute \(x = 96.7\) into the equation \(y = 6.02 + 44.3x\). Thus, \(y = 6.02 + 44.3 \times 96.7 = 4286.41\).
06

Graph Equation c

For equation c, \(y = -0.06 + 0.313x\), input the equation on the calculator. Set the plotting window to cover \(x = 0.64\). Use the trace function to find the \(y\)-value when \(x = 0.64\).
07

Calculate Manually for Equation c

Substitute \(x = 0.64\) into the equation \(y = -0.06 + 0.313x\). Thus, \(y = -0.06 + 0.313 \times 0.64 = 0.14\).
08

Graph Equation d

For equation d, \(y = 1183 - 2140x\), input this into the graphing calculator. Ensure your plot window can display \(x = -111\). Use the trace function to find the \(y\)-value when \(x = -111\).
09

Calculate Manually for Equation d

Substitute \(x = -111\) into the equation \(y = 1183 - 2140x\). Thus, \(y = 1183 - 2140 \times (-111) = 238723\).

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Key Concepts

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

Using a Graphing Calculator
When it comes to graphing linear equations, graphing calculators are powerful tools that make the process straightforward and efficient. Whether you're plotting a single line or comparing different equations, a graphing calculator can help you visualize mathematical relationships quickly.

To begin, enter the equation in the standard linear form, often y = mx + b, into the calculator. Start by switching it to the graphing mode. Make sure the window settings are properly adjusted to show the desired range for the x-axis. This will help in tracing specific x-values with ease. Typically, you will use the 'Trace' function on the calculator to move along the graph. By inputting the x-value you're interested in, the calculator can give you the corresponding y-value in mere seconds.

Remember that having a well-adjusted window is crucial; you want to ensure that the x-values needed are not off-screen, which can lead to errors when tracing. This tool is incredibly useful for checking your work after doing manual calculations.
Linear Functions
Linear functions are one of the fundamental concepts in algebra and pre-calculus. Represented in the form y = mx + b, these equations describe a straight line. Understanding the components of this form is key to interpreting linear relationships.

  • x and y are the variables representing the input and output, respectively.
  • m is the slope, which tells us how steep the line is and the direction it moves. A positive slope indicates the line moves upwards, while a negative slope indicates a downward movement.
  • b is the y-intercept, showing where the line crosses the y-axis. It's the output when x equals zero.

The behavior of linear functions is predictable. They represent constant rates of change, making them easier to work with compared to other polynomial functions. Graphically, the line maintains an equal distance between all points on it, reinforcing the concept of a linear relationship. Moreover, linear functions provide the foundation for understanding more complex mathematical concepts.
Substitution Method
An important algebraic technique is the substitution method, which can be used both to solve systems of equations and to find specific values within a single equation. This method involves substituting a known value into an equation to determine an unknown, often leveraging the linear form.

Here's how to use substitution for linear functions:
  • First, identify the linear equation and the value of x for which you want to calculate y.
  • Substitute this x-value directly into the equation in place of the x-variable.
  • Solve the equation to find the exact y-value, giving you a specific point (x, y) on the line.

For example, if given the equation y = 3x + 2 and asked to find y when x = 4, replace x with 4, resulting in y = 3(4) + 2.

This equals 12 + 2, so y = 14. By using substitution, you obtain precise values without needing a graph. This method is especially useful for handling multiple equations simultaneously or verifying results from graphing calculators.

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Most popular questions from this chapter

Consider the expression $$ \frac{5.4+3.2(x-2.8)}{1.2}-2.3 $$ a. Use the order of operations to find the value of the expression if \(x=7.2\). b. Set the expression equal to \(3.8\). Solve for \(x\) by undoing the sequence of operations you listed in \(11 \mathrm{a}\).

APPLICATION Sonja bought a pair of \(210 \mathrm{~cm}\) cross-country skis. Will they fit in her ski bag, which is \(6 \frac{1}{2} \mathrm{ft}\) long? Why or why not?

Jo mows lawns after school. She finds that she can use the equation \(P=-300+15 \mathrm{~N}\) to calculate her profit. a. Give some possible real-world meanings for the numbers \(-300\) and 15 and the variable \(N\). b. Invent two questions related to this situation and then answer them. c. Solve the equation \(P=-300+15 N\) for the variable \(N\). d. What does the equation in \(7 \mathrm{c}\) tell you?

APPLICATION A long-distance telephone carrier charges \(\$ 1.38\) for international calls of 1 minute or less and \(\$ 0.36\) for each additional minute. a. Write a recursive routine using calculator lists to find the cost of a 7 -minute phone call. (A) b. Without graphing the sequence, give a verbal description of the graph showing the costs for calls that last whole numbers of minutes. Include in your description all the important values you need in order to draw the graph.

You can use the equation \(d=24-45 t\) to model the distance from a destination for someone driving down the highway, where distance \(d\) is measured in miles and time \(t\) is measured in hours. Graph the equation and use the trace function to find the approximate time for each distance given in \(2 \mathrm{a}\) and \(\mathrm{b}\). a. \(d=16 \mathrm{mi}\) (a) b. \(d=3 \mathrm{mi}\) c. What is the real-world meaning of 24 ? (a) d. What is the real-world meaning of 45 ? e. Solve the equation \(24-45 t=16\).

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