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Chapter 3: Problem 4

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. \( y = 4297.23 \); c. \( y = 0.14032 \); d. \( y = 238723 \).

Step by step solution

01

Graph the Equation for a

Input the equation \( y = 1.21 - x \) into your graphing calculator. Ensure that you have the appropriate window settings to view the line on your screen.
02

Trace to Find y for x=70.2 in a

Use the trace function on your calculator to find the approximate value of \( y \) when \( x = 70.2 \). The calculator will help you find this by showing the intersection point on the graph.
03

Calculate y Directly for a

Instead of only relying on the trace function, you can directly compute by substituting \( x = 70.2 \) into the equation: \( y = 1.21 - 70.2 \).
04

Solve for y in a

Calculate the value: \( y = 1.21 - 70.2 = -68.99 \).
05

Graph the Equation for b

Input the equation \( y = 6.02 + 44.3x \) into your graphing calculator and ensure a wide enough window to clearly see the graph.
06

Trace to Find y for x=96.7 in b

Using the trace feature, find \( y \) when \( x = 96.7 \). Track the value on the calculator to get an accurate reading.
07

Calculate y Directly for b

Substitute \( x = 96.7 \) in the equation: \( y = 6.02 + 44.3 \times 96.7 \).
08

Solve for y in b

Calculate the value: \( y = 6.02 + 4291.21 = 4297.23 \).
09

Graph the Equation for c

Enter \( y = -0.06 + 0.313x \) into your calculator and make sure the window settings allow you to view the line.
10

Trace to Find y for x=0.64 in c

Use the trace function to find the approximate y-value for \( x = 0.64 \). Locate this value on the graph using the calculator.
11

Calculate y Directly for c

Substitute \( x = 0.64 \) into the equation: \( y = -0.06 + 0.313 \times 0.64 \).
12

Solve for y in c

Compute the expression: \( y = -0.06 + 0.20032 = 0.14032 \).
13

Graph the Equation for d

Input the equation \( y = 1183 - 2140x \) into your calculator and adjust the viewing window to visualize the line.
14

Trace to Find y for x=-111 in d

With the trace function, find \( y \) for \( x = -111 \). Follow the graph on the calculator screen for an accurate value.
15

Calculate y Directly for d

Substitute \( x = -111 \) into the equation: \( y = 1183 - 2140 \times (-111) \).
16

Solve for y in d

Compute the value: \( y = 1183 + 237540 = 238723 \).

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Key Concepts

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

Calculator Use
Using a graphing calculator to plot equations can simplify complex calculations and allow you to visualize results, making it easier to understand mathematical concepts. Calculators help:
  • Quickly plot curves and lines based on input equations.
  • Provide immediate results on whether your graph settings (like the window size) correctly display all relevant details of your graph.
  • Experiment with changes dynamically to understand how different values or expressions affect the graph.
To begin using your calculator for graphing, make sure you input the given equation correctly. Adjust the window settings to ensure all important parts of the graph are visible. This practice makes verification and learning through visualization far more efficient.
Trace Function
The trace function on a graphing calculator is a valuable tool for examining specific points on a graphed equation. It allows you to see coordinates at various points on the curve, helping you to find approximate values easily. The trace function helps:
  • Locate particular y-values for specified x-values directly on the graph.
  • Understand intersection points or other relevant points on complex graphs by following the plotted line.
  • See how values change along the curve, which enhances comprehension of the graph's behavior.
To use the trace function, first graph your equation properly. Move the cursor along the graph line corresponding to your desired x-value to read off the approximate y-value directly from the display. This hands-on tool aids in learning about graph shapes and their implications in applied settings.
Substitution Method
The substitution method is a straightforward algebraic approach to finding unknown values in equations. It's particularly useful when working alongside tools like calculators, as it verifies results from graphical analysis. Here鈥檚 how it helps:
  • Directly calculate exact y-values by substituting known x-values into the equation.
  • Offer a way to double-check results obtained from graphical trace outputs, ensuring accuracy.
  • Enhance understanding of algebraic manipulations and their impact on variables.
Use substitution by taking your given equation and plugging your x-value straight into it. Solve the resulting expression to find the corresponding y-value. This method complements graphical solutions, providing a full picture of solving equations algorithmically and visually.
Algebraic Expressions
Algebraic expressions form the backbone of modeling real-world scenarios through equations. Mastery of these expressions is essential for solving problems and facilitating calculations with tools like calculators. Understanding them helps:
  • Recognize and manipulate different forms of expressions such as linear, quadratic, or polynomial.
  • Translate practical situations into mathematical models to calculate unknowns logically.
  • Develop strategies for solving equations, whether through graphical, substitution, or other algebraic methods.
In any given problem, identify the variables and constants, determine their relationships, and construct equations that represent these relationships clearly. Breaking down complex expressions into simpler parts can aid in understanding and solving intricate problems effectively.

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Most popular questions from this chapter

APPLICATION Manny has a part-time job as a waiter. He makes \(\$ 45\) per day plus tips. He has calculated that his average tip is \(12 \%\) of the total amount his customers spend on food and beverages. a. Define variables and write an equation in intercept form to represent Manny's daily income in terms of the amount his customers spend on food and beverages. b. Graph this relationship for food and beverage amounts between \(\$ 0\) and \(\$ 900\). c. Write and evaluate an expression to find how much Manny makes in one day if his customers spend \(\$ 312\) on food and beverages. d. What amounts spent on food and beverages will give him a daily income between \(\$ 105\) and \(\$ 120\) ?

Draw a scatter plot on your paper picturing (time, distance) at 1 s intervals if you start timing Carol's walk as she walks toward her partner starting at a distance of \(5.9 \mathrm{~m}\) and moving at a constart speed of \(0.6 \mathrm{~m} / \mathrm{s}\).

A bicyclist, \(1 \mathrm{mi}(5280 \mathrm{ft})\) away, pedals toward you at a rate of \(600 \mathrm{ft} / \mathrm{min}\) for \(3 \mathrm{~min}\). The bicyclist then pedals at a rate of \(1000 \mathrm{ft} / \mathrm{min}\) for the next \(5 \mathrm{~min}\). a. Describe what you think the plot of (time, distance from you) will look like. (a) b. Graph the data using 1 min intervals for your plot. (a) c. Invent a question about the situation, and use your graph to answer the question.

Draw and label a coordinate plane with each axis scaled from \(-10\) to 10 . a. Represent each point named with a dot, and label it using its letter name. \(\begin{array}{llll}A(3,-2) & B(-8,1.5) & C(9,0) & D(-9.5,-3) \\ F(1,-1) & G(0,-6.5) & H(2.5,3) & I(-6,7.5)\end{array}\) b. List the points in Quadrant I, Quadrant II, Quadrant III, and Quadrant IV. Which points are on the \(x\)-axis? Which points are on the \(y\)-axis? c. Explain how to tell which quadrant a point will be in by looking at the coordinates. Explain how to tell if a point lies on one of the axes.

These tables show the changing depths of two submarines as they come to the surface. $$ \begin{aligned} &\begin{array}{|l|r|r|r|r|r|r|c|} \hline \text { Times (s) } & 0 & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline \text { Depth (ft) } & -38 & -31 & -24 & -17 & -10 & -3 & 4 \\ \hline \end{array}\\\ &\text { USS Dallas }\\\ &\begin{array}{|l|r|r|r|r|r|r|r|} \hline \text { Times (s) } & 0 & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline \text { Depth (ft) } & -48 & -40 & -32 & -24 & -16 & -8 & 0 \\ \hline \hline \end{array} \end{aligned} $$ a. Graph the data from both tables on the same set of coordinate axes. b. Describe what you found by graphing the data. How are the graphs the same? How are they different? c. Does it make sense to draw a line through each set of points? Explain what these lines mean. d. What is the real-world meaning of the point \((30,4)\) for the USS Alabama?
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