Problem 15 - Algebra and Trigonometry

Question

11–18 ? Find the x- and y-intercepts of the graph of the equation. $$ x^{2}+y^{2}=4 $$

Step-by-Step Solution

Answer

X-intercepts: (2, 0), (-2, 0); Y-intercepts: (0, 2), (0, -2).

Step 1: Understand Intercepts

Intercepts are points where the graph intersects the axes. The x-intercept is where the graph crosses the x-axis, meaning y = 0, while the y-intercept is where the graph crosses the y-axis, meaning x = 0.

Step 2: Finding the x-intercepts

To find x-intercepts, set y = 0 and solve for x in the equation. Start with the equation: \(x^2 + y^2 = 4\). Substitute \(y = 0\): \(x^2 + 0^2 = 4\), which simplifies to \(x^2 = 4\). Then solve for x: \(x = \pm 2\). Thus, the x-intercepts are (2, 0) and (-2, 0).

Step 3: Finding the y-intercepts

To find y-intercepts, set x = 0 and solve for y in the equation. Start with the equation: \(x^2 + y^2 = 4\). Substitute \(x = 0\): \(0^2 + y^2 = 4\), which simplifies to \(y^2 = 4\). Then solve for y: \(y = \pm 2\). Thus, the y-intercepts are (0, 2) and (0, -2).

Key Concepts

equation of a circle

The equation of a circle in the standard form is expressed as \(x^2 + y^2 = r^2\), where \(r\) represents the radius of the circle, and the center is located at the origin, (0,0). This general form allows us to quickly identify key properties of the circle.

• Radius: The radius \(r\) is the square root of the constant on the right side of the equation. For example, in \(x^2 + y^2 = 4\), the radius is \(\sqrt{4} = 2\).
• Center: The center of this circle is at (0,0) since there is no \(h\) or \(k\) term.

The simplicity of this equation makes it easy to visualize and plot the circle. Each point \((x, y)\) that satisfies this equation is a point on the circumference of the circle.

finding intercepts

Finding intercepts involves determining where the figure crosses the axes. These are special points on the graph:

• x-intercepts: Located where the graph touches or crosses the x-axis. To find them, set \(y = 0\) in the equation and solve for \(x\).
• y-intercepts: Located where the graph touches or crosses the y-axis. To find these, set \(x = 0\) in the equation and solve for \(y\).

These intercepts are particularly useful as they provide clear and simple starting points to graph the circle. In our example with the circle equation \(x^2 + y^2 = 4\):

• For the x-intercepts, substituting \(y = 0\) into the equation gives the solutions \(x = \pm 2\), producing intercept points (2, 0) and (-2, 0).
• Similarly, for the y-intercepts, substituting \(x = 0\) gives \(y = \pm 2\), producing intercept points (0, 2) and (0, -2).

solving quadratic equations

Solving quadratic equations is a fundamental skill in mathematics, encountered frequently when identifying intercepts for circles and other curves. A quadratic equation typically takes the form \(ax^2 + bx + c = 0\). Here's how to go about solving it:

• Factoring: If the equation can be factored easily, solve for roots by setting each factor equal to zero.
• Square Root Method: Used when the equation is in the form \(x^2 = d\). Directly take the square root on both sides to solve for \(x\).
• Quadratic Formula: Apply when the equation does not easily factor: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

For the equation \(x^2 + 0 = 4\), the simplest method is the square root method. This yields \(x = \pm 2\), providing the solutions for x-intercepts directly and fast.