Problem 19 - A Graphical Approach to College Algebra

Question

Graph each inequality. Do not use a calculator. $$x^{2}+(y+3)^{2} \leq 16$$

Step-by-Step Solution

Answer

Shade the region of a circle centered at (0, -3) with radius 4.

Step 1: Recognize the Inequality as a Circle Equation

The given inequality \( x^2 + (y + 3)^2 \leq 16 \) looks similar to the equation of a circle in the form \( (x - h)^2 + (y - k)^2 = r^2 \). Here, \( h = 0 \), \( k = -3 \), and \( r^2 = 16 \), thus \( r = 4 \). This represents a circle centered at \((0, -3)\) with radius 4.

Step 2: Understand What the Inequality Represents

The inequality \( \leq \) indicates all points inside or on the circle. If it were \( < \), it would include only the points strictly inside the circle.

Step 3: Plot the Center of the Circle

Begin by marking the center of the circle at the point \((0, -3)\) on a coordinate plane. This point is crucial as all other points of the circle are equidistant from it.

Step 4: Draw the Circle

Using the radius of 4, sketch a circle around the center \((0, -3)\). Ensure that the circle passes through points that are exactly 4 units away from the center in all directions.

Step 5: Shade the Region Inside the Circle

Since the inequality is \( \leq \), shade the entire region inside this circle, as well as the boundary of the circle itself, to indicate that these points satisfy the inequality.

Key Concepts

Circle Equation

When we encounter an inequality like \( x^2 + (y + 3)^2 \leq 16 \), it hints at the concept of a circle equation. A typical circle equation is represented as \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center of the circle and \(r\) is the radius. In this exercise, comparing with the standard form helps us identify:

• Center: \((0, -3)\), since there is no horizontal shift \((x - 0)^2\) and \((y + 3)^2\) means shifting down by 3 units.
• Radius: \(4\), found as the square root of \(16\) (from \(r^2\)).

The inequality sign \(\leq\) suggests the influence of the circle extends to all points on and within this boundary, indicating more than just the circumference is involved. It allows us to consider a filled-in circle, rather than just a perimeter, assisting us to visualize combined regions when graphing.

Coordinate Plane

Graphing on the coordinate plane is fundamental for visualizing mathematical concepts. When we plot inequalities, understanding the plane's basics ensures accurate representation. The coordinate plane consists of two perpendicular lines:

• x-axis: Horizontal line where \(y = 0\).
• y-axis: Vertical line where \(x = 0\).

In our case, the circle's center at \((0, -3)\) lies on the y-axis, down 3 units from the origin \((0,0)\). Using this plane helps illustrate spatial relationships, such as the distance from the center to any point on the circumference, leveraging the grid to measure precisely. Place tick marks at consistent intervals to aid in drawing circles and lines accurately. With the center and radius identified, this grid facilitates plotting not just shapes, but solutions like inequalities too.

Plotting Inequalities

Translating inequalities onto the graph involves specific steps to encompass the entire solution set. Let's break it down using the example inequality, \( x^2 + (y+3)^2 \leq 16 \):

• Plot the center: Start by locating \((0, -3)\).
• Draw the circle: Utilize the radius (4 units) to sketch a circle. The boundary includes any point four units away from the center, in any direction.
• Determine shading: The inequality symbol \(\leq\) signifies that points on and inside the circle meet the inequality conditions. Thus, shade the entire circular region, highlighting that all contained points are valid solutions.

This technique not only aids in graphical depiction but also enhances our comprehension of connectivity between algebraic conditions and their geometrical forms.