Problem 29 - Precalculus with Limits

Question

In Exercises 19-36, determine whether the equation represents \(y\) as a function of \(x\). \(y = \sqrt{16-x^2}\)

Step-by-Step Solution

Answer

Yes, the equation \(y = \sqrt{16-x^2}\) represents \(y\) as a function of \(x\).

Step 1: Understanding the Function

The equation \(y = \sqrt{16-x^2}\) belongs to the family of equations for a circle centered at the origin, and is split into two halves: the upper half and the lower half. However, as per the definition, the primary square root function \(y = \sqrt{x}\) only outputs nonnegative values (i.e. it only represents the upper half of the circle).

Step 2: Determine whether y is a function of x

Given that the square root function only outputs nonnegative values, the equation \(y = \sqrt{16-x^2}\) will also only output nonnegative values of \(y\) for any value of \(x\), which means for every input \(x\), there is exactly one output \(y\).

Key Concepts

Circle Equations

Circle equations are fascinating geometrical expressions. They define the shape and location of a circle within a coordinate plane. The most common form is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius.
In this equation, any point \((x, y)\) that satisfies the equation lies on the circle's perimeter.
For instance, when we look at the equation \(y = \sqrt{16-x^2}\), we are observing a special part of a circle's equation.
It represents a semicircle because it accounts only for the nonnegative values of \(y\). This semicircle is centered on the origin \((0, 0)\) with a radius of 4, as we can rewrite it to match our standard form: \(x^2 + y^2 = 16\).

• The semicircle results because the square root function calculates only the nonnegative half, effectively slicing the circle horizontally in half at \(x\).
• In simpler terms, it is like having a pizza and only being able to see the top half after cutting it horizontally straight through the middle.

Understanding these properties helps predict the graphical representation of the equation.

Square Root Function

The square root function, often written as \(y = \sqrt{x}\), is one of the most fundamental functions in mathematics. It transforms a nonnegative input \(x\) into a nonnegative output \(y\). The output is effectively the principal square root of the input.
When we apply this concept to \(y = \sqrt{16-x^2}\), it operates similarly by taking the result of \(16-x^2\) and giving us its principal square root.
This is crucial as it restricts \(y\) to values from 0 upwards, excluding any negative values.

• This restriction is what defines the graph of this equation as a semicircle rather than a full circle, focusing only on the upper half.
• Furthermore, understanding that square roots give nonnegative results helps prevent errors when analyzing or graphing such functions.

Through this lens, the square root function assists in interpreting and predicting results in various mathematical scenarios.

Nonnegative Values

In mathematics, nonnegative values refer to numbers that are either positive or zero. This concept is essential when dealing with functions like the square root, because it ensures that calculations do not produce undefined outputs.
Specifically, for the equation \(y = \sqrt{16-x^2}\), the nonnegative condition means that the value inside the square root, \(16-x^2\), must itself be nonnegative.
Thus, \(-4 \leq x \leq 4\).

• This restriction is because if \(16-x^2\) were negative, we wouldn't be able to calculate a real square root, as square roots of negative numbers are not real values.
• Therefore, the domain of \(x\) is limited to values that keep \(16-x^2\) nonnegative.

Understanding nonnegative values ensures that the solution to any function is both real and meaningful, stabilizing its interpretation and practical application.