Problem 25 - Precalculus with Limits

Question

In Exercises 19-36, determine whether the equation represents \(y\) as a function of \(x\). \((x+2)^2 + (y-1)^2 = 25\)

Step-by-Step Solution

Answer

The equation \((x+2)^2 + (y-1)^2 = 25\) does not represent \(y\) as a function of \(x\) because for each value of \(x\), there are two corresponding values of \(y\).

Step 1: Isolate y

To begin the solution, the given equation \((x+2)^2 + (y-1)^2 = 25\) needs to be solved for \(y\). So, subtract \((x+2)^2\) from both sides to isolate \((y-1)^2\) on one side. The equation now becomes \((y-1)^2 = 25 - (x+2)^2\).

Step 2: Square-Root both sides

Taking the square root of both sides in order to further isolate y, two solutions will be obtained \(y-1 = \sqrt{25-(x+2)^2}\) and \(y-1 = -\sqrt{25-(x+2)^2}\). So, the two equations become \(y = 1 + \sqrt{25-(x+2)^2}\) and \(y = 1 - \sqrt{25-(x+2)^2}\). Each has a different value of \(y\) for every value of \(x\), so both \(y\) values exist for each \(x\) value.

Step 3: Determine the conclusion

When y was isolated in terms of x, two corresponding y-values were found for each x-value, which implies that each x-value does not map to a unique y-value. Therefore, the original equation does not represent \(y\) as a function of \(x\).

Key Concepts

Isolating Variables

To solve an equation and understand the relationship between two variables, isolating one variable is often a crucial first step. Let's focus on how this applies when dealing with circle equations. Imagine we start with an equation \((x+2)^2 + (y-1)^2 = 25\). Our goal is to express \(y\) in terms of \(x\).
To achieve this, we need to restructure the equation, isolating the term \((y-1)^2\). Begin by performing simple algebraic operations: subtract \((x+2)^2\) from both sides. As a result, you get:

• \((y-1)^2 = 25 - (x+2)^2\)

This step is straightforward yet vital. By isolating \(y\), we pave the way for further simplification. Now, \(y\) is on one side of the equation, and the rest is expressed in terms of \(x\). This manipulated form makes it simpler to solve for \(y\) by using additional techniques like taking square roots later.

Square Root Method

Once we've isolated \((y-1)^2\), the next logical step is to solve for \(y\) itself. Here is where the square root method comes in handy.
If you take the square root of both sides of the equation \((y-1)^2 = 25 - (x+2)^2\), you acquire two possible results for \(y\):

• \(y-1 = \sqrt{25 - (x+2)^2}\)
• \(y-1 = -\sqrt{25 - (x+2)^2}\)

This duality arises because both positive and negative numbers squared yield the same result. Therefore, from these equations, we have:

• \(y = 1 + \sqrt{25 - (x+2)^2}\)
• \(y = 1 - \sqrt{25 - (x+2)^2}\)

As you can see, the square root method reveals that each \(x\)-value corresponds to two different \(y\)-values, explaining why the original equation doesn't represent \(y\) as a function of \(x\).

Circle Equations

The initial equation you encountered represents a circle on a plane: \((x+2)^2 + (y-1)^2 = 25\). This is a standard form of a circle equation, which reads as follows:

• \((x-h)^2 + (y-k)^2 = r^2\)

In this context:

• \(h = -2\)
• \(k = 1\)
• \(r = 5\)

where \((h, k)\) is the center of the circle and \(r\) is its radius.
This representation is useful because it directly shows the geometric features of the circle:

• The circle is centered at the point \((-2, 1)\) on the Cartesian plane
• With a radius of 5

Understanding this can help identify whether equations like these represent functions. A circle does not meet the criterion for a function because it can have multiple \(y\) values for a single \(x\) value, which we verified when finding \(y\). This circle equation exemplifies why not every equation involving \(x\) and \(y\) qualifies as a function.