Problem 26 - Algebra and Trigonometry

Question

\(11-36\) Make a table of values and sketch the graph of the equation. Find the \(x\) - and \(y\) -intercepts. $$ x^{2}+y^{2}=9 $$

Step-by-Step Solution

Answer

The circle centered at the origin has x-intercepts at (-3,0) and (3,0), and y-intercepts at (0,-3) and (0,3).

Step 1: Understand the Equation Type

The given equation is \(x^2 + y^2 = 9\), which represents a circle with center at the origin \((0,0)\) and radius 3, because the equation is in the standard form of a circle which is \((x-h)^2 + (y-k)^2 = r^2\). Here, \(h = 0\), \(k = 0\), and \(r = 3\).

Step 2: Create a Table of Values

To create a table of values, substitute several values of \(x\) into the equation and solve for \(y\). For example, if \(x = 0\), \(y^2 = 9\), so \(y = \pm 3\). Similarly, if \(x = 3\), \(y^2 = 0\), so \(y = 0\). Fill in more values for \(x\) to see the corresponding \(y\) values.

Step 3: Determine x-intercepts

To find the \(x\)-intercepts, set \(y = 0\) in the equation. This gives \(x^2 + 0^2 = 9\), which simplifies to \(x^2 = 9\). Thus, the \(x\)-intercepts are \(x = \pm 3\).

Step 4: Determine y-intercepts

To find the \(y\)-intercepts, set \(x = 0\) in the equation. This gives \(0^2 + y^2 = 9\), which simplifies to \(y^2 = 9\). Thus, the \(y\)-intercepts are \(y = \pm 3\).

Step 5: Sketch the Graph

Plot the points from the table of values onto a coordinate plane. Since the equation represents a circle, connect these points to form a circle. Ensure the circle is centered at the origin \((0,0)\) with a consistent radius of 3.

Key Concepts

Graphing Circles

When graphing a circle, you begin by identifying its key features from the equation. The equation \( x^2 + y^2 = 9 \) represents a circle in the standard form. This form is \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) is the center and \( r \) is the radius of the circle. In this case, \( h = 0 \) and \( k = 0 \), so the center is at the origin \((0,0)\). The radius \( r \) is 3, derived from \( r^2 = 9 \).

When sketching the circle, use the center and radius to ensure that the circle spans correctly. Start from the center at \((0,0)\) and measure 3 units in all directions, forming a symmetric shape. By plotting this, you ensure that your sketch accurately represents the circle defined by the equation.

x-intercepts

Finding the \(x\)-intercepts involves setting \( y = 0 \) in the circle's equation and solving for \( x \). For the equation \( x^2 + y^2 = 9 \), replace \( y \) with 0, leading to \( x^2 = 9 \). This simplifies to \( x = \pm 3 \).

The intercepts are the points on the graph where the circle crosses the \( x \)-axis. For this circle, they occur at \( (3, 0) \) and \( (-3, 0) \). These points are critical for plotting the circle, as they mark the furthest points on the \( x \)-axis that the circle touches.

y-intercepts

Similarly, to find the \(y\)-intercepts, set \( x = 0 \) and solve for \( y \). Using the circle's equation \( x^2 + y^2 = 9 \), substituting \( x \) with 0 converts it to \( y^2 = 9 \), which simplifies to \( y = \pm 3 \).

These intercepts are where the circle crosses the \( y \)-axis, located at \( (0, 3) \) and \( (0, -3) \). Like the \( x \)-intercepts, these points are essential for constructing the circle's graph. They highlight the circle's extent in the vertical direction.

Table of Values

Creating a table of values helps in understanding and plotting the circle, especially when confirming the shape and size. Start by selecting various \( x \) values, then substitute them into the equation to find corresponding \( y \) values. For example:

• \( x = 0 \) leads to \( y = \pm 3 \)
• \( x = 3 \) gives \( y = 0 \)
• \( x = 2 \) results in \( y = \pm \sqrt{5} \)

Once you have several coordinate pairs, such as \((0, 3), (3, 0), (2, \sqrt{5})\), plot these on a graph to begin seeing the circular shape. The table ensures accurate positioning of points, providing a check against the radius and symmetry inherent to circles.