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Chapter 5: Problem 36

Graph the solution set. \(y \geq-(x+1)^{2}-2\)

Short Answer

Expert verified
Shade the region above and on the parabola opening downward with vertex at \((-1, -2)\).

Step by step solution

01

Identify the inequality and the quadratic equation

The given inequality is \(y \geq -(x + 1)^{2} - 2\) . The corresponding quadratic equation is \(y = -(x + 1)^{2} - 2\).
02

Vertex of the parabola

The quadratic equation is in the form \(y = a(x - h)^{2} + k\). Here, \(a = -1\), \(h = -1\), and \(k = - 2\). So, the vertex of the parabola is \((-1, -2)\).
03

Determine if the parabola opens up or down

Since the coefficient \(a = -1\) is negative, the parabola opens downward.
04

Plot the vertex and the parabola

Plot the vertex \((-1, -2)\) on the coordinate plane. Since the parabola opens downward, sketch the graph by drawing a downward curve starting from the vertex.
05

Shading the solution set

Because the inequality is \(y \geq -(x + 1)^{2} - 2\), shade the region above and on the parabola. This represents all the points where \(y\) is greater than or equal to \( -(x+1)^2 -2 \).

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Key Concepts

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

quadratic equations
Quadratic equations are polynomial equations of the second degree and can be written in the general form ax^{2} + bx + c = 0 , where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. These equations have a characteristic U-shaped graph known as a parabola. Parabolas can open upwards or downwards depending on the coefficient 'a'. If a > 0 , the parabola opens upwards, and if a < 0 , it opens downwards. The solutions to quadratic equations are also known as the roots or zeros of the equation, representing the points where the parabola intersects the x-axis.
vertex of a parabola
Understanding the vertex of a parabola is crucial for graphing quadratic equations and inequalities. The vertex represents the highest or lowest point on the parabola, depending on its orientation. The vertex form of a quadratic equation is y = a(x - h)^{2} + k , where (h, k) is the vertex of the parabola. In the exercise given, y = -(x + 1)^{2} - 2 , the values of h and k are -1 and -2 respectively, giving us the vertex (-1, -2) . This point is where the parabola changes direction on the coordinate plane.
shading solution sets
Shading solution sets is a visual way to represent all possible solutions to an inequality. For the inequality y 鈮� -(x+1)^{2} - 2, the solution set includes all points where y is greater than or equal to the value on the right side of the inequality. First, graph the corresponding quadratic equation y = -(x + 1)^{2} - 2 as a parabola. Then, because the inequality symbol is 'greater than or equal to,' we shade the region above and on the parabola. This shaded area represents all the points that satisfy the inequality, indicating where y meets the criteria.
parabola orientation
The orientation of a parabola depends on the sign of the leading coefficient 'a' in the quadratic equation y = ax^{2} + bx + c . If 'a' is positive, the parabola opens upwards, resembling a U shape. If 'a' is negative, the parabola opens downwards, resembling an inverted U shape. In our given quadratic inequality y 鈮� -(x + 1)^{2} - 2, the coefficient 'a' is - 1, so the parabola opens downwards. Identifying the orientation is essential for correctly graphing the parabola and determining the shading for the solution set of the inequality.

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Most popular questions from this chapter

A coordinate system is placed at the center of a town with the positive \(x\) -axis pointing east, and the positive \(y\) -axis pointing north. A cell tower is located \(4 \mathrm{mi}\) west and \(5 \mathrm{mi}\) north of the origin. a. If the tower has a 8 -mi range, write an inequality that represents the points on the map serviced by this tower. b. Can a resident 5 mi east of the center of town get a signal from this tower?

Solve the system of equations by using the substitution method. (See Example 2\()\) $$ \begin{array}{l} 3 x=2 y-11 \\ 6+5 x=1 \end{array} $$

A plant nursery sells two sizes of oak trees to landscapers. Large trees cost the nursery \(\$ 120\) from the grower. Small trees cost the nursery \(\$ 80\). The profit for each large tree sold is \(\$ 35\) and the profit for each small tree sold is \(\$ 30 .\) The monthly demand is at most 400 oak trees. Furthermore, the nursery does not want to allocate more than \(\$ 43,200\) each month on inventory for oak trees. a. Determine the number of large oak trees and the number of small oak trees that the nursery should have in its inventory each month to maximize profit. (Assume that all trees in inventory are sold.) b. What is the maximum profit? c. If the profit on large trees were \(\$ 50\), and the profit on small trees remained the same, then how many of each should the nursery have to maximize profit?

Solve the system using any method. $$ \begin{array}{l} 5(2 x+y)=y-x-8 \\ x-\frac{3}{2} y=\frac{5}{2} \end{array} $$

Determine if the ordered pair is a solution to the system of equations. (See Example 1\()\) \(y=-\frac{1}{5} x+2\) \(2 x+10 y=10\) a. (5,1) b. (-10,4)
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