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Magazine Chapter 9: Problem 31
Graph each system of inequalities or indicate that the system has no solution.
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Short Answer
Expert verified
The solution is the region below the line \( y = x+1 \) and below/on the parabola \( y \leq x^2 \).
Step by step solution
01
Graph the First Inequality
The first inequality is \( y < x + 1 \). Start by graphing the line \( y = x + 1 \), which is a straight line with a slope of 1 and a y-intercept at (0, 1). Since it is a strict inequality (\(<\) rather than \(\leq\)), use a dashed line to indicate that points on the line are not included in the solution. Shade the region below the line, as we are interested in points where \( y \) is less than (not equal to) \( x + 1 \).
02
Graph the Second Inequality
The second inequality is \( y \leq x^2 \). First, graph the curve \( y = x^2 \), which is a parabola with its vertex at the origin (0, 0) that opens upwards. Since it is a \( \leq \) inequality, include the boundary by drawing a solid line for the parabola. Shade the region below or on the curve where \( y \) is less than or equal to \( x^2 \).
03
Find the Intersection of the Regions
The solution to the system of inequalities is the set of points that satisfy both inequalities simultaneously. Locate the region where the shaded area from Step 1 (below the line \( y = x + 1 \)) and the shaded area from Step 2 (below or on the parabola \( y \leq x^2 \)) overlap. This region will be bounded above by the dashed line and below by the solid parabola.
04
Check for Feasibility
Verify that the region found in Step 3 does exist. The parabola \( y = x^2 \) intersects the line \( y = x + 1 \) at points that satisfy both equations simultaneously. Solve for these points: set \( x^2 = x + 1 \) or \( x^2 - x - 1 = 0 \). Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), to find the intersection points of the curves. Calculate \( x = \frac{1 \pm \sqrt{5}}{2} \). For these \( x \)-values, ensure the overlapping region indeed exists.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Inequalities
When tackling systems of inequalities, graphing each inequality helps in visualizing the solutions, and finding where these solutions overlap. Here's how to graph two types of inequalities typically found in these systems:
- Linear Inequalities: Given an inequality such as \( y < x + 1 \), begin by graphing the line \( y = x + 1 \). For this specific inequality, use a dashed line, because the "less than" sign \( < \) means we do not include the line itself in the solution. Shade the region below this line to represent all the \( y \)-values that are less than \( x + 1 \).
- Quadratic Inequalities: For an inequality like \( y \leq x^2 \), graph the parabola \( y = x^2 \). Since the inequality is "less than or equal to," the boundary line here will be solid, including the parabola itself in the solution. Shade below this curve, representing points where \( y \) is less than or equal to \( x^2 \).
Parabolas
Parabolas are common graphs in systems of inequalities, especially when dealing with quadratic functions. A standard parabola has the equation form \( y = ax^2 + bx + c \). For the inequality \( y \leq x^2 \), here are the core features:
- Vertex: This is the turning point of the parabola, and for \( y = x^2 \), it is at the origin (0,0).
- Opening: The parabola \( y = x^2 \) opens upwards because the coefficient of \( x^2 \) is positive. This means it U-shapes upwards.
- Boundary: In graphing \( y \leq x^2 \), draw a solid boundary line, since solutions on the parabola are valid points for the inequality.
Straight Lines
Straight lines form the backbone of linear equations and inequalities, with the general equation \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Let's break down the line \( y = x + 1 \) in the first inequality \( y < x + 1 \):
- Slope: For \( y = x + 1 \), the slope \( m \) is 1, meaning the line rises one unit up for every unit right.
- Y-intercept: This is the value of \( y \) when \( x \) is zero, and here it is 1. Thus, the line crosses the y-axis at (0,1).
- Boundary Representation: Use a dashed line for \( y < x + 1 \), indicating that points on the line itself are not solutions to the inequality. The shaded region below the line represents all possible solutions where \( y \) is indeed less than \( x + 1 \).
