Problem 70 - College Algebra with Modeling and Visualization

Question

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

Step-by-Step Solution

Answer

y is not a function of x, as it can take two values for each x.

Step 1: Identify Equation Type

The given equation \((x+2)^{2}+(y+1)^{2}=1\) represents a circle because it is in the standard form of a circle \((x-h)^{2} + (y-k)^{2} = r^{2}\). Here, \(h = -2\), \(k = -1\), and \(r^{2} = 1\), so the radius \(r = 1\).

Step 2: Solve for y Explicitly

Rearrange to express \(y\) in terms of \(x\). Start with the original equation:\[(x+2)^{2} + (y+1)^{2} = 1\]Rearrange terms to isolate the \((y+1)^{2}\) term:\[(y+1)^{2} = 1 - (x+2)^{2}\]Take the square root of both sides to solve for \(y+1\):\[y + 1 = \pm \sqrt{1 - (x+2)^{2}}\]Finally, solve for \(y\):\[y = -1 \pm \sqrt{1 - (x+2)^{2}}\]

Step 3: Determine If y is a Function of x

For \(y\) to be a function of \(x\), each \(x\)-value must be associated with exactly one \(y\)-value. The expression \[y = -1 \pm \sqrt{1 - (x+2)^{2}}\] suggests two values of \(y\) for each \(x\), except where the square root equals zero. This means \(y\) is not a function of \(x\), as each input does not map to exactly one output.

Key Concepts

Equations of a Circle

Understanding that an equation like \((x+2)^{2} + (y+1)^{2} = 1\) describes a circle is a crucial first step in algebra. A circle's equation is generally written as \((x-h)^{2} + (y-k)^{2} = r^{2}\), where \((h, k)\) is the center of the circle and \(r\) is the radius. Here, \(h = -2\) and \(k = -1\), moving the circle's center to the point \((-2, -1)\). This specific equation also tells us that the radius of the circle is 1, because \(r^{2} = 1\) means \(r = \sqrt{1} = 1\).
Centrally, the circle equation visually represents all points \((x, y)\) such that their distance from the center \((-2, -1)\) is exactly 1. The symmetry of this circle around its center helps illustrate why, when solving for one variable, both positive and negative roots might exist, as the shape is perfectly even in all directions.

Function Definition

Determining whether the variable \(y\) can be considered a function of \(x\) involves checking if each \(x\) maps to exactly one \(y\). A key aspect of a function is its ability to provide a unique output for every specific input. In our equation context, each substitution of \(x\) could yield a \(y\) because of the \(\pm\) in the solved expression \(y = -1 \pm \sqrt{1 - (x+2)^{2}}\).

• If for a specific \(x\), two values of \(y\) emerge, then \(y\) is not a valid function of \(x\).
• In geometrical terms, if a vertical line can cross more than one point on the graph of the relationship, then it's not a function. This is known as the vertical line test.

In our circumstance, what this double root tells us is that not every input relates to one unique output, thus breaking the definition of a function.

Solving Algebraic Equations

Solving algebraic equations like \((x+2)^{2} + (y+1)^{2} = 1\) typically means simplifying the equation to explicitly isolate one variable in terms of the other. Here, we aim to solve explicitly for \(y\).

• First, isolate where the \(y\) occurs: \((y+1)^{2} = 1 - (x+2)^{2}\).
• Next, the crucial step is extracting \(y\) by taking the square root, remembering that the square root of a function can have both positive and negative results.
• This step results in \(y+1 = \pm\sqrt{1 - (x+2)^{2}}\).
• Lastly, simplify to find \(y = -1 \pm \sqrt{1 - (x+2)^{2}}\).

The presence of the \(\pm\) symbol signifies that there are two potential values of \(y\) for each \(x\), except in situations where the inner expression equals zero. Understanding this step by step not only helps in solving the circle equation but also in recognizing when a relationship doesn't define a function.