Problem 19
Question
Graph each inequality. $$ y \leq x^{2}+x-2 $$
Step-by-Step Solution
Verified Answer
Graph the parabola for \( y = x^2 + x - 2 \) and shade the region below it.
1Step 1: Understand the Inequality
The inequality given is \( y \leq x^{2} + x - 2 \). This means we want to graph the region where the value of \( y \) is less than or equal to the quadratic function \( x^{2} + x - 2 \). Our task will involve finding the boundary line (or curve), which is \( y = x^{2} + x - 2 \), and shading the appropriate region on the graph.
2Step 2: Graph the Parabola
Since the inequality involves a quadratic expression, the graph of the equation \( y = x^{2} + x - 2 \) is a parabola. To graph it, find the vertex and key points on the parabola. The vertex of a parabola in the form \( ax^2 + bx + c \) can be found at \( x = -\frac{b}{2a} \). For this parabola, \( a = 1 \), \( b = 1 \), and \( x = -\frac{1}{2} \). Compute \( y \) at \( x = -\frac{1}{2} \) to complete the vertex and plot essential points.
3Step 3: Calculate and Plot Points
To further plot the parabola, calculate a few points. Substitute some x-values into the equation to get corresponding y-values. For example, let \( x = 0 \), then \( y = 0^{2} + 0 - 2 = -2 \); when \( x = 1 \), \( y = 1^{2} + 1 - 2 = 0 \); and when \( x = -1 \), \( y = (-1)^{2} - 1 - 2 = -2 \). By plotting these points and others, you can shape the graph of the parabola.
4Step 4: Draw the Parabola
Connect the points plotted from Step 3 smoothly to illustrate the parabolic shape of the equation \( y = x^{2} + x - 2 \). The parabola opens upwards due to the positive coefficient of \( x^2 \). Ensure the vertex and plotted points are accurate to maintain precision in the graph.
5Step 5: Shade the Region
The inequality is \( y \leq x^{2} + x - 2 \), indicating the shaded region should be the area below the parabola. Since the inequality sign is \( \leq \), the line \( y = x^{2} + x - 2 \) itself is included in the solution set. Shade the region beneath the entire parabola to represent all values of \( y \) that satisfy the inequality.
6Step 6: Verify Solution
Choose a point below the parabola to test if it satisfies the inequality. For example, the origin \( (0,0) \) lies below the curve. Substitute this back into the inequality: \( 0 \leq 0^{2}+0-2 \), or \( 0 \leq -2 \), which is false. Try a point like \( (0,-3) \); it should satisfy the inequality \( -3 \leq 0^{2} + 0 - 2 \), or \( -3 \leq -2 \), which is true, verifying our shaded region.
Key Concepts
Quadratic InequalitiesParabolaGraphing FunctionsInequality Solution
Quadratic Inequalities
Quadratic inequalities involve expressions with a squared variable, like \( x^2 \), and require us to solve for a range of values rather than a single solution. In the context of graphing, an inequality like \( y \leq x^2 + x - 2 \) represents a region of the coordinate plane rather than just a line or a curve. Here, the focus is on determining where the function \( y = x^2 + x - 2 \) provides a boundary and which region lies below it. The usage of "\( \leq \)" means that not only do we include the solutions on the line (or curve), but also every possible solution beneath it. Understanding this ensures that when you graph, the shaded area correctly represents the solution set.
Parabola
A parabola is the U-shaped curve that occurs when we graph quadratic equations like \( y = x^2 + x - 2 \). To plot a parabola accurately, you need to identify its vertex, intercepts, and direction. The vertex can be calculated using the formula \( x = -\frac{b}{2a} \) for equations in the form \( ax^2 + bx + c \). By substituting \( x = -\frac{1}{2} \) into the equation, you can find the corresponding \( y \)-value, completing the vertex point \((-\frac{1}{2}, -\frac{9}{4})\). This point provides a crucial reference position from which the parabola curves upward, due to the positive coefficient of \( x^2 \). Graphing additional points around the vertex helps in showing the accuracy and the exact shape of the parabola on the coordinate plane.
Graphing Functions
When graphing functions, especially quadratics, your goal is to depict the entire relationship implied by the equation visually. Start by plotting points from calculated values as in the given inequality. For our equation, values such as \( x = 0 \) yielding \( y = -2 \), \( x = 1 \) with \( y = 0 \), and others help to outline the shape. Connect these points smoothly, ensuring the parabola emerges with an upward opening because the leading coefficient is positive. This visually demonstrates the quadratic's expressions. Besides its form, evaluate if it correctly maps onto the established axes with accurately plotted points to confirm the graph's precision. These steps are vital as they reflect the inequality solution meaning for students to use and understand further math concepts.
Inequality Solution
Solving an inequality like \( y \leq x^2 + x - 2 \) involves deciding the area of the plane that satisfies this condition. Begin by establishing the boundary with the function's graph, here a parabola. The solution then incorporates shading the region below, including the parabola itself, as given by "\( \leq \)." This indicates areas where \( y \) is less, visually denoted by shading beneath the curve. Checking the accuracy of this region involves testing points; any point under the parabola bound \( (-\frac{1}{2}, -\frac{9}{4}) \) should hold true for the inequality. For instance, \( (0,-3) \) satisfies \( -3 \leq -2 \), making it a valid point within the solution. This verification helps comprehend inequality solutions not just as a graph, but as a key concept in understanding quadratic functions and their relationships.
Other exercises in this chapter
Problem 19
Graph each hyperbola. $$ 16 y^{2}-x^{2}=16 $$
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The graph of each equation is a circle. Find the center and the radius, and then graph the circle. See Examples 5 through 7. $$ x^{2}+(y-2)^{2}=1 $$
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Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} 2 x^{2}+3 y^{2}=14 \\ -x^{2}+y^{2}=3 \end{array}\right. $$
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Graph each hyperbola. $$ 4 y^{2}-25 x^{2}=100 $$
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