Problem 42 - Algebra 2

Question

For Exercises \(40-43,\) use the following information. since a circle is not the graph of a function, you cannot enter its equation directly into a graphing calculator. Instead, you must solve the equation for \(y .\) The result will contain a \pm symbol, so you will have two functions. Graph \((x+3)^{2}+y^{2}=16\) on a graphing calculator.

Step-by-Step Solution

Answer

Graph separately \(y = \sqrt{16 - (x+3)^2}\) and \(y = -\sqrt{16 - (x+3)^2}\) on the calculator.

Step 1: Understand the Circle Equation

The given equation \((x+3)^2 + y^2 = 16\) represents a circle with center at \((-3, 0)\) and radius \(4\). This is because the equation is of the form \((x-h)^2 + (y-k)^2 = r^2\).

Step 2: Isolate the \(y^2\) Term

Rewrite the equation to solve for \(y^2\):\[(x+3)^2 + y^2 = 16\]Subtract \((x+3)^2\) from both sides:\[y^2 = 16 - (x+3)^2\]

Step 3: Solve for \(y\) using Square Root

Take the square root of both sides to solve for \(y\):\[y = \pm \sqrt{16 - (x+3)^2}\]This gives us two separate equations to graph on a calculator:1. \(y = \sqrt{16 - (x+3)^2}\)2. \(y = -\sqrt{16 - (x+3)^2}\)

Step 4: Graph the Functions

Enter the two equations from Step 3 into a graphing calculator as two separate functions:- \(y_1 = \sqrt{16 - (x+3)^2}\)- \(y_2 = -\sqrt{16 - (x+3)^2}\)Adjust the viewing window to ensure the full circle is visible, typically setting both \(x\) and \(y\) limits to range from \(-8\) to \(8\) will work.

Key Concepts

Equation of a Circle

Understanding the equation of a circle is fundamental in graphing it correctly. The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\). Here, the point \((h, k)\) represents the circle's center, and \(r\) is its radius.

In the given exercise, we have \((x+3)^2 + y^2 = 16\). By comparing this with the standard form, it becomes clear that the center of the circle is at \((-3, 0)\), because \((x+3)\) can be rewritten as \((x-(-3))\). The radius \(r\) is found by taking the square root of 16, which gives us a radius of 4.

Knowing the center and radius allows us to visualize the circle. The circle is centered at \((-3, 0)\) on the Cartesian plane and it extends in all directions from the center up to a distance of 4 units.

Solving for \(y\)

To graph a circle on a calculator that only accepts functions, you need to solve the circle's equation for \(y\). This involves isolating \(y\) on one side of the equation.

Starting with \((x+3)^2 + y^2 = 16\), subtract \((x+3)^2\) from both sides to isolate \(y^2\):

• \(y^2 = 16 - (x+3)^2\)

Next, solve for \(y\) by taking the square root of both sides:

• \(y = \pm \sqrt{16 - (x+3)^2}\)

This gives us two equations, \(y = \sqrt{16 - (x+3)^2}\) and \(y = -\sqrt{16 - (x+3)^2}\), which represent the top and bottom halves of the circle, respectively.

Each part of these equations can be graphed as separate functions on a graphing calculator, effectively creating the circle's complete shape.

Graphing Calculator Usage

To graph these equations on a calculator, you need to enter them as two separate functions. Many graphing calculators do not handle the \(\pm\) symbol directly, so you will enter the positive and negative cases individually.

Here's how you can input them into a typical graphing calculator:

• Function 1: \(y_1 = \sqrt{16 - (x+3)^2}\)
• Function 2: \(y_2 = -\sqrt{16 - (x+3)^2}\)

Ensure your calculator's viewing window is set correctly to visualize the circle fully. A recommended window setting might be to extend both \(x\) and \(y\) axes from \(-8\) to \(8\). This ensures the complete circle is visible given the circle's center and radius.

By graphing these functions, you'll see the top half of the circle drawn by \(y_1\) and the bottom half by \(y_2\), which together form the whole circle on your calculator's screen.