Problem 14 - Precalculus: Mathematical for Calculus

Question

Graph the inequality. $$x^{2}+(y-1)^{2} \leq 1$$

Step-by-Step Solution

Answer

Shade the circle centered at \((0, 1)\) with radius 1.

Step 1: Identify the Type of Graph

The inequality \( x^2 + (y-1)^2 \leq 1 \) represents a circle. The standard form of a circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) is the radius. In this inequality, \(h = 0\), \(k = 1\), and \(r = 1\).

Step 2: Determine the Circle's Center and Radius

From the equation \( x^2 + (y-1)^2 = 1 \), we deduce that the circle's center is at \((0, 1)\) with a radius of 1 unit.

Step 3: Graph the Circle

Plot the center of the circle at \((0, 1)\) on the coordinate plane. Then, use the radius to draw the circle. Since the radius is 1, measure 1 unit in all directions from the center and draw the circle through these points.

Step 4: Shade the Interior

Since the inequality is \( \leq \), include the area inside the circle in the graph. Shade the entire area enclosed by the circle to represent solutions satisfying the inequality.

Key Concepts

Circle Equations

A circle equation in coordinate geometry helps us define the set of all points that lie the same distance from a fixed point called the center. The standard circle equation is \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h, k)\) is the center of the circle, and \(r\) is the radius. This form is derived from the distance formula used to find the distance between two points. The equation is essentially saying the distance from any point \((x, y)\) on the circle to the center \((h, k)\) is constant and equal to the radius \(r\). To better understand:

• \((x-h)^2 + (y-k)^2 = r^2\) is the fundamental definition of a circle.
• You might need to rearrange equations into this form to identify the circle's center and radius easily.
• The given exercise involves a circle centered at \((0, 1)\) with a radius of 1, which fits nicely in the standard form.

By examining the given inequality \(x^2 + (y-1)^2 \leq 1\), you can deduce that any solution to this inequality lies on or within the circle defined by \(x^2 + (y-1)^2 = 1\). This makes it essential to properly grasp circle equations, so you can easily visualize geometric problems on the coordinate plane.

Graphing Inequalities

Graphing inequalities revolves around the area of a graph where these inequalities hold true. In this case, we have the inequality \(x^2 + (y-1)^2 \leq 1\), which requires us to graph within a circle. The enclosed area within the circle (including its boundary) signifies all points \((x, y)\) that satisfy this inequality. The process to graph inequalities like this involves:

• Determining the boundary, which for \(x^2 + (y-1)^2 = 1\) is a circle centered at \((0,1)\).
• Checking the inequality sign; \(\leq\) indicates that points on the boundary are solutions, requiring the entire circle to be drawn solid.
• Since the inequality responds to \(\leq\), you need to shade the area inside the circle that includes all solutions.

Observing these crucial steps allows one to accurately represent inequality solutions on a graph, showing both the boundaries and valid points inside them. Understanding the key difference between different inequality signs (like \(>\), \(<\), \(\geq\), and \(\leq\)) is crucial for proper graph interpretation.

Coordinate Geometry

Coordinate geometry, sometimes known as analytic geometry, involves plotting and interpreting shapes, lines, and curves on a coordinate plane using algebra. This form of geometry utilizes equations to describe geometric figures, which can then be translated into visual graphs.For the inequality \(x^2 + (y-1)^2 \leq 1\), coordinate geometry helps us:

• Visualize the circle by plotting its center at \((0, 1)\) and using the radius to establish the boundary.
• Locate relevant points and verify which coordinates satisfy the inequality conditions.
• Draw conclusions about the space regions, like shading areas inside circles for given inequalities.

Mastering coordinate geometry is essential for solving such problems since it directly links algebraic expressions of inequalities or equations with their geometric counterparts on the graph. The power of visual interpretation allows students to better understand complex relationships and solutions in mathematics.