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Chapter 5: Problem 53

Use a graphing utility to determine whether the system of equations has one solution, two solutions, or no solution. $$\left\\{\begin{array}{l}y=x^{2}+2 x-1 \\ y=2 x+5\end{array}\right.$$

Short Answer

Expert verified
The number of solutions for the system of equations depends on the number of intersection points of the graphs of the functions \( y = x^{2} + 2x - 1 \) and \( y = 2x + 5 \)

Step by step solution

01

Graph the First Equation

Using a graphing utility, input the equation for the first function \( y = x^{2} + 2x - 1 \). Here, the graph will have a parabolic shape due to the \( x^{2} \) term.
02

Graph the Second Equation

Next, overlay the graph of the second function \( y = 2x + 5 \) on the same plot. This function will be represented by a straight line due to the linear form of the equation.
03

Analyze the Intersection Points

Identify intersection points of two graphs (the points where the graphs of both functions meet). The number of intersection points indicates the number of solutions to the system of equations.\n
04

Verify

To confirm the number of solutions, click on the intersection points (if any) using the cursor in the graphing utility. The tool will provide the coordinates of the intersection points. These coordinates are the solutions to the system of equations.

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Key Concepts

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

Parabolic Graphs
A parabolic graph is the graphical representation of a quadratic function. The standard form of a quadratic equation is \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In our exercise, the equation \( y = x^2 + 2x - 1 \) is a quadratic equation, thus forming a parabola when graphed. This type of curve is symmetrical and often has a distinct "U" shape which can either open upwards or downwards depending on the sign of \( a \).

Key features of a parabola include:
  • Vertex: The peak or the lowest point of the parabola.
  • Axis of Symmetry: A vertical line through the vertex dividing the parabola into mirror images.
  • Direction: The parabola opens upwards if \( a > 0 \) and downwards if \( a < 0 \).
Understanding these features helps in predicting the shape and position of the graph.
Linear Equations
Linear equations produce graphs that are straight lines. A linear equation generally takes the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In our context, \( y = 2x + 5 \) represents a linear equation.

The slope \( m \) determines the steepness and direction of the line, with positive values indicating an upward slope and negative values a downward slope. The y-intercept \( b \) is the point where the line crosses the y-axis.

Essential characteristics of linear equations are:
  • Slope: Influences the angle of the line; larger slopes lead to steeper lines.
  • Y-intercept: Helps in determining starting point on the graph.
  • Graph as a Straight Line: Contrasts sharply with the curves of non-linear equations.
Grasping these concepts is crucial for understanding how linear equations behave and appear on a graph.
Intersection Points
Intersection points are the locations where graphs of different equations meet on a plot. When graphing systems of equations, these points are crucial because they represent solutions to the system. For the system given, finding where the parabola \( y = x^2 + 2x - 1 \) intersects with the line \( y = 2x + 5 \) provides the solutions to the system of equations.

To identify intersection points:
  • Graph both equations on the same set of axes using a graphing utility.
  • Look for points where the line and the parabola cross each other.
  • The x-coordinates at these crossing points are the solutions for \( x \), and the y-coordinates confirm the values for both equations being equal.
Accurately spotting these points leads you to the solution(s) of the system.
Graphing Utilities
Graphing utilities are digital tools that aid in graphing mathematical functions accurately and efficiently. They allow users to enter equations and instantly see their graphical representation. These tools are exceptionally helpful for visualizing functions such as parabolas and lines.

Benefits of using graphing utilities include:
  • Accuracy: Precise plotting of functions with reduced manual errors.
  • Efficiency: Quick generation of graphs that would take longer by hand.
  • Interactivity: Explore features like zooming, rotating, and identifying key points or intersections with a cursor.
When dealing with complex systems of equations, graphing utilities simplify the process of determining solutions through visual representation.

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Most popular questions from this chapter

Investments An investor has up to $$\$ 250,000$$ to invest in two types of investments. Type A investments pay \(7 \%\) annually and type \(\mathrm{B}\) pay \(12 \%\) annually. To have a well-balanced portfolio, the investor imposes the following conditions. At least one-fourth of the total portfolio is to be allocated to type A investments and at least one-fourth is to be allocated to type \(\mathrm{B}\) investments. What is the optimal amount that should be invested in each type of investment? What is the optimal return?

Sketch the region determined by the constraints. Then find the minimum anc maximum values of the objective function and where they occur, subject to the indicated constraints. Objective function: $$ z=x $$ Constraints: $$ \begin{array}{r} x \geq 0 \\ y \geq 0 \\ 2 x+3 y \leq 60 \\ 2 x+y \leq 28 \\ 4 x+y \leq 48 \end{array} $$

MAKE A DECISION: STOPPING DISTANCE In testing of the new braking system of an automobile, the speed (in miles per hour) and the stopping distance (in feet) were recorded in the table below. $$ \begin{array}{|c|c|} \hline \text { Speed, } x & \text { Stopping distance, } y \\ \hline 30 & 54 \\ \hline 40 & 116 \\ \hline 50 & 203 \\ \hline 60 & 315 \\ \hline 70 & 452 \\ \hline \end{array} $$ (a) Find the least squares regression parabola \(y=a x^{2}+b x+c\) for the data by solving the following system. \(\left\\{\begin{array}{r}5 c+250 b+13,500 a=1140 \\ 250 c+13,500 b+775,000 a=66,950 \\ 13,500 c+775,000 b+46,590,000 a=4,090,500\end{array}\right.\) (b) Use the regression feature of a graphing utility to check your answer to part (a). (c) A car design specification requires the car to stop within 520 feet when traveling 75 miles per hour. Does the new braking system meet this specification?

The given linear programming problem has an unusual characteristic. Sketch a graph of the solution region for the problem and describe the unusual characteristic. Find the maximum value of the objective function and where it occurs. Objective function: \(z=x+2 y\) Constraints: $$ \begin{aligned} x & \geq 0 \\ y & \geq 0 \\ x+2 y & \leq 4 \\ 2 x+y & \leq 4 \end{aligned} $$

Maximize the objective function subject to the constraints \(3 x+y \leq 15,4 x+3 y \leq 30\) \(x \geq 0\), and \(y \geq 0\) $$z=5 x+y$$
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