Problem 44 - Algebra A Combined Function

Question

$$ \begin{aligned} &\text { Graph the system: }\\\ &\left\\{\begin{array}{l} x \geq 0 \\ y \geq 0 \\ y \geq x^{2}+1 \\ y \leq 4-x \end{array}\right. \end{aligned} $$

Step-by-Step Solution

Answer

The feasible region is a section in the first quadrant bounded by the parabola and the line, starting from the point (1, 3).

Step 1: Identify Constraints

The system is comprised of four inequalities: \( x \geq 0 \), \( y \geq 0 \), \( y \geq x^2 + 1 \), and \( y \leq 4 - x \). These constraints define the feasible region in the coordinate plane.

Step 2: Graphing the Non-Negativity Constraints

First, graph the lines \( x = 0 \) and \( y = 0 \) which correspond to the x-axis and y-axis, respectively. The region defined by \( x \geq 0 \) and \( y \geq 0 \) is the first quadrant of the coordinate plane.

Step 3: Graphing the Parabola

Graph the parabola defined by \( y = x^2 + 1 \). Since we have \( y \geq x^2 + 1 \), shade the region above the parabola. The parabola opens upwards and intersects the y-axis at \((0, 1)\).

Step 4: Graphing the Line

Graph the line defined by \( y = 4 - x \). Since we have \( y \leq 4 - x \), shade the region below this line. The line has a y-intercept at \((0, 4)\) and a slope of -1.

Step 5: Finding the Intersection

Find the point where the line \( y = 4 - x \) intersects the parabola \( y = x^2 + 1 \) by solving the equation: \( x^2 + 1 = 4 - x \). This simplifies to \( x^2 + x - 3 = 0 \). Factor or use the quadratic formula to find its solutions, \( x = 1 \) and \( x = -3 \). The relevant intersection point is \((1, 3)\).

Step 6: Identify Feasible Region

Considering all inequalities, the feasible region is bounded by the intersection of the line and parabola, and extends indefinitely in the positive quadrant. The region lies above the parabola starting from \((1, 3)\), under the line \( y = 4 - x \), and in the first quadrant.

Key Concepts

Quadratic Inequalities

Quadratic inequalities involve expressions where a quadratic polynomial is set in relation to an inequality symbol. In our context, the inequality is represented by the quadratic equation \( y \geq x^2 + 1 \). This means that for any value of \( x \) within our constraints, \( y \) must fall on or above the parabola described by \( y = x^2 + 1 \).
Understanding quadratic inequalities is crucial because it allows us to determine which sections of a parabola are included or excluded from a solution set. The key is to appreciate that the inequality sign dictates the portion of the region if you should shade above or below the curve.

• "\( y \geq x^2 + 1 \)" means accepting values of \( y\) that are equal to or larger than the parabola.
• The parabola \( y = x^2 + 1 \) opens upwards and its vertex is at point \((0, 1)\).

Recognizing this involves identifying the region where all these conditions are satisfied. By plotting the parabola on the graph, you can visually see the space above it becoming part of the possible solutions.

Feasible Region

The feasible region is at the heart of solving systems of inequalities. It is where all conditions or constraints overlap, producing the area containing all conceivable solutions. For the system given, the feasible region is where all four inequalities overlap on the coordinate plane.
The feasible region in this system is constrained by:

• Both non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \), meaning the solution must be in the first quadrant.
• The parabola \( y = x^2 + 1 \), dictating that all solutions lie on or above it.
• The line \( y = 4 - x \) where solutions are on or below it.

All these conditions narrow down to an area that stretches upward from the intersection point \((1, 3)\) above the parabola and below the line. It's essential when working on these problems to correctly plot each boundary and understand its impact on possible solutions.

Coordinate Plane Graphing

Graphing on the coordinate plane involves plotting points, lines, or regions to visually solve inequalities or systems of equations. This process helps identify the feasible region where solution sets lie.
To graph the system provided, we start with the axes. The x-axis \( (x = 0) \) and y-axis \( (y = 0) \) outline the quadrant where real solutions can exist according to \( x \geq 0 \) and \( y \geq 0 \).
The parabola \( y = x^2 + 1 \) is especially important because its curve influences the feasible region shape. You plot it by selecting various \( x \) values and determining their corresponding \( y \) values. Ensure plotting is precise since the shading direction—either above or below—is derived from this graph.

• After graphing the parabola, shade the area above it.
• Graph the line \( y = 4 - x \) with precision, regarding its slope and intercept, and shade the region below.

These steps are crucial for the visual verification of solutions and provide a clear picture of where the feasible region exists on the plane.