Problem 49 - Algebra 2 and Trigonometry

Question

In \(44-51 :\) a. Graph the given inequality. b. Determine if the given point is in the solution set. $$ -4(x+2)^{2}-5 \leq y ;\left(1, \frac{2}{3}\right) $$

Step-by-Step Solution

Answer

Graph the parabola opening downwards centered at (-2,-5), the point (1, 2/3) is in the solution set.

Step 1: Rewrite the Inequality

The given inequality is \(-4(x+2)^2 - 5 \leq y\). To simplify, understand that the equation describes a quadratic curve. Rewrite it as \(y \geq -4(x+2)^2 - 5\). This represents a parabola opening upwards, shifted left by 2 units and down by 5 units.

Step 2: Vertex Form Identification

Identify the vertex for the parabola \(y = -4(x+2)^2 - 5\). The vertex is at \((-2, -5)\) because the standard form \(-(x-h)^2 + k\) shows a shift of \(h = -2\) left and \(k = -5\) down.

Step 3: Graph the Inequality

Plot the vertex at \((-2, -5)\). Determine the direction of opening. The coefficient \(-4\) indicates the parabola opens downwards. The inequality \(\leq\) signifies that the region to shade is below and includes the parabola line.

Step 4: Test the Given Point

Substitute the point \((1, \frac{2}{3})\) into the inequality to check if it belongs to the solution set: \(-4(1+2)^2 - 5 \leq \frac{2}{3}\). Calculate the left side: \(-4(9) - 5 = -36 - 5 = -41\). Since \(-41 \leq \frac{2}{3}\), the point is within the solution set.

Key Concepts

Parabola

A parabola is a specific type of curve on a graph, generated by a quadratic equation. It can open upwards or downwards, depending on the sign of the coefficient in front of the squared term.
In the case of \(y = -4(x+2)^2 - 5\), the parabola opens downwards because the coefficient of the squared term is negative (-4).
This particular parabola:

• Has a vertex at (-2, -5), indicating its lowest point on the graph.
• Opens downwards, representing a concave shape.
• Is centered horizontally around x = -2.

Understanding parabolas is crucial for solving quadratic inequalities, as they determine the shape and location of the possible solutions.

Graphing Inequalities

Graphing quadratic inequalities involves more than just sketching a parabola. Instead of a single line or curve, inequalities involve shading a region that satisfies the given condition.
For this exercise\( -4(x+2)^2 - 5 \leq y \), the inequality sign \( \leq \) signifies that we include the entire area on and below the parabola in the solution set. Here’s how to graphically represent it:

• First, plot the parabola, ensuring to mark the vertex and other symmetrical points.
• Shaded area should cover the region below the curve since it represents values that make the inequality true.
• A dashed line signifies a strict inequality (without "or equal to"), while a solid line is used here to include the boundary as it is \( \leq \).

Graphing helps visualize which areas satisfy the inequality, providing a comprehensive understanding of the solutions.

Vertex Form

The vertex form of a quadratic equation is essential for quickly identifying the characteristics of a parabola. It is usually expressed as \(y = a(x-h)^2 + k\), where:

• \(a\) determines the parabola's direction and width.
• \(h\) and \(k\) represent the vertex's coordinates (h, k).

For the inequality \(y = -4(x+2)^2 - 5\), the vertex form reveals:

• Vertex at (-2, -5), indicating horizontal and vertical shifts from the origin.
• A negative \(a = -4\) implies a downward opening parabola.

This quick identification helps in graphing and analyzing the solution set appropriately without recalculating.

Solution Set Testing

Testing a specific point in the context of a quadratic inequality allows us to verify if it lies within the solution region. For the inequality \(-4(x+2)^2 - 5 \leq y\), testing the point \((1, rac{2}{3})\), one must:

• Substitute x = 1 and y = \(\frac{2}{3}\) into the inequality.
• Calculate the left side: \(-4(1+2)^2 - 5 = -41\).
• Compare \(-41\) with \(\frac{2}{3}\).

Since \(-41 \leq \frac{2}{3}\), the point is indeed part of the solution set.
Solution set testing is essential to not only confirm particular solutions but also to reinforce understanding of the graph's shaded region.