Problem 88 - College Algebra with Modeling and Visualization

Question

Determine if \(y\) is a function of \(x\). $$ (x-1)^{2}+y^{2}=1 $$

Step-by-Step Solution

Answer

No, \(y\) is not a function of \(x\).

Step 1: Identify the Equation Type

The given equation is \((x-1)^2 + y^2 = 1\). This equation is in the form of a circle centered at \((1,0)\) with a radius of 1.

Step 2: Analyze for Function Properties

A function from \(x\) to \(y\) requires that for every \(x\)-value, there is exactly one corresponding \(y\)-value, fulfilling the vertical line test.

Step 3: Apply the Vertical Line Test Conceptually

For the equation of a circle, particularly \((x-1)^2 + y^2 = 1\), each \(x\)-value within the domain can map to two different \(y\)-values (one positive and one negative), except at the endpoints of the horizontal diameter. Thus, it fails the vertical line test.

Key Concepts

Vertical Line Test

The vertical line test is a simple yet effective way to determine if a graph represents a function. For a set of points to be a function, every vertical line drawn through the graph should intersect it at most once. This ensures that for each input (or each value of \(x\)), there is exactly one output (or \(y\)).

In our exercise, we have the equation of a circle \((x-1)^2 + y^2 = 1\). When we draw vertical lines through this circle on the graph, there are places where a line intersects the circle twice: once at a positive \(y\)-value and once at a negative \(y\)-value. This shows that certain \(x\)-values have more than one corresponding \(y\)-value, causing the test to fail.

Therefore, the given equation does not represent \(y\) as a function of \(x\), because it doesn't pass the vertical line test. Keep in mind that passing this test is crucial for determining a true function between the two variables.

Circles in Coordinate Geometry

In coordinate geometry, a circle is defined by its center and radius. The general equation for a circle centered at \((h,k)\) with radius \(r\) is given by \((x-h)^2 + (y-k)^2 = r^2\).

For our equation, \((x-1)^2 + y^2 = 1\), we understand that:

• The circle is centered at \((1, 0)\), indicating it's moved 1 unit away from the origin on the x-axis.
• The radius is 1, since \(r^2 = 1\).

The circular shape implies that for most \(x\)-values within its reach, there are multiple corresponding \(y\)-values, due to the top and bottom halves of the circle.

Thus, circles typically do not yield functions when each \(x\) point maps to more than one \(y\) point unless their equations are adjusted or restricted.

Domain and Range of Functions

Understanding the domain and range of functions is pivotal in determining their properties. The domain refers to all the possible input values (or \(x\)-values) a function can accept, while the range is the set of all possible output values (or \(y\)-values) that the function can produce.

In our exercise here, since we are dealing with a circle, let's consider an ideal function to illustrate these concepts. Suppose we consider half of this circle as our function, we will have:

• Domain: If we restrict the circle in our function to only allow 1 output per input, the altered domain might be \([-1, 3]\), accounting for some conventionally permissible parts of the circle.
• Range: Correspondingly, the range might then be restricted from 0 to 1 if we use only the upper half.

By analyzing the domain and range, you can better comprehend how the function behaves and why certain inputs or outputs might not be feasible given the graph’s inherent restrictions.