Problem 36 - Precalculus with Limits

Question

In Exercises 33-46, sketch the graph (and label the vertices) of the solution set of the system of inequalities. $$\left\\{\begin{aligned} 4 x^{2}+y & \geq 2 \\ x & \leq 1 \\ y & \leq 1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution set of the given system of inequalities is a bounded region right to the y-axis, below the line y=1, left to the line x=1, and above the parabola \(4x^{2}+y = 2\).

Step 1: Graph the Linear Inequalities

Start by graphing the linear inequalities \(x\leq1\) and \(y\leq1\). These are straight lines parallel to the axes, and the area of solution is to the left of the line \(x=1\) and below the line \(y=1\).

Step 2: Graph the Quadratic Inequality

To graph the quadratic inequality \(4x^{2}+y \geq 2\), first set it as an equation \(4x^{2}+y = 2\) which is a parabola. The graph of the parabola opens upwards with its vertex at the origin (0,2). For the inequality \(4x^{2}+y \geq 2\) the solution set is above the graph of the parabola.

Step 3: Combine the Inequalities

The overall solution set is the region where all three inequalities overlap. Exactly, considering the bounds imposed by \(x \leq 1\) and \(y \leq 1\) and the overall solution area is a bounded region in the first quadrant.

Key Concepts

Linear Inequalities

Linear inequalities represent a linear relationship that is not equated but rather expressed as greater than, less than, or equal to. In our exercise, the inequalities involved are \(x \leq 1\) and \(y \leq 1\). This means that we are looking for all the points that lie to the left of the line you get if \(x\) always equals 1, as well as all points below the line where \(y\) is always 1. To graph these, you draw straight lines along the respective values on their axes.

• These inequalities divide the coordinate plane into two parts.
• The line for \(x \leq 1\) is vertical.
• The line for \(y \leq 1\) is horizontal.

The region representing the solution is to the left and below these lines. These lines act as borders or edges which constrain the solution set on the graph.

Quadratic Inequalities

Quadratic inequalities, unlike linear ones, involve an equation where the highest-degree term is squared, as in \(4x^2 + y \geq 2\). You start by dealing with the equality portion, \(4x^2 + y = 2\), which forms a parabola.

• This parabola opens upwards as the term is \(x^2\).
• The vertex, the highest or lowest point depending on its direction, is at \((0, 2)\).

In this inequality, we are interested in the region above this parabola because of the \(\geq\) sign. Therefore, when graphing, identify the areas above the curve to indicate solutions to \(4x^2 + y \geq 2\).

Solution Set

The solution set is essentially the subset of points or coordinates that satisfy all the given inequalities at once. For the current system, it’s where the areas of \(x \leq 1\), \(y \leq 1\), and \(4x^2 + y \geq 2\) intersect.

• To find this, observe the intersections of the shaded regions from each inequality on a graph.
• Start by shading the areas for the linear inequalities.
• Then, shade the area above the parabola derived from the quadratic inequality.

The overlap of all these shaded regions provides the complete solution set. This intersection is typically found in the bounds of all conditions, resulting in a distinct geometric shape in the plane, usually a bounded region.

Vertices of Inequalities

Vertices in the context of inequalities refer to the corner points of the solution set. These are critical in identifying the actual extent and area of the solution space in graphical representations. For this system, you need to identify where the boundary lines intersect each other or the parabola.

• The point \((1, 1)\) is formed by the intersection of the lines \(x \leq 1\) and \(y \leq 1\).
• Check intersections between \(4x^2+y=2\) and \(x=1\), and \(y=1\), if there are any.

These vertices are essential since they define the limit of the shaded region. They mark transitions between boundary conditions and help ensure that the graph reflects all constraints accurately.