Problem 45 - Algebra and Trigonometry

Question

\(41-48\) 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) and (-2, 0); y-intercepts: (0, 2) and (0, -2).

Step 1: Understanding Intercepts

The intercepts are the points where the graph of the equation crosses the x-axis and y-axis. At the x-intercept(s), the y-value is 0, and at the y-intercept(s), the x-value is 0.

Step 2: Finding the x-intercepts

To find the x-intercepts, set \(y = 0\) in the equation. This gives \(x^2 + 0^2 = 4\), simplifying to \(x^2 = 4\). Solving for \(x\), we get \(x = \pm 2\). Thus, the x-intercepts are \((2, 0)\) and \((-2, 0)\).

Step 3: Finding the y-intercepts

To find the y-intercepts, set \(x = 0\) in the equation. This gives \(0^2 + y^2 = 4\), simplifying to \(y^2 = 4\). Solving for \(y\), we get \(y = \pm 2\). Thus, the y-intercepts are \((0, 2)\) and \((0, -2)\).

Step 4: Summary of Intercepts

The x-intercepts are the points \((2, 0)\) and \((-2, 0)\), while the y-intercepts are the points \((0, 2)\) and \((0, -2)\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest degree of the variable is squared. They are usually written in the form of \(ax^2 + bx + c = 0\). However, in this exercise, we encounter a slightly different form: \(x^2 + y^2 = 4\), which is a special type of equation representing a circle centered at the origin with a radius of 2. The key to solving problems involving quadratic equations is identifying the standard form and recognizing its graph's shape. Quadratic equations can represent different geometric shapes depending on their coefficients and terms:

• If there's one squared variable, it usually indicates a parabola, either opening upwards or downwards.
• If both variables are squared and with certain symmetry, like in this case, it indicates a circle or other conic sections.

When dealing with quadratic equations with both \(x\) and \(y\), it's essential to know whether you're working with a parabola, circle, ellipse, or hyperbola by examining the terms.

X-Intercepts

The x-intercepts of a graph are the points where the graph crosses the x-axis. This means that at these points, the value of \(y\) is zero. To find the x-intercepts, substitute \(y = 0\) into the equation. For the equation \(x^2 + y^2 = 4\):- Set \(y = 0\) giving us \(x^2 + 0^2 = 4\).- This simplifies to \(x^2 = 4\).- Solving for \(x\) gives us \(x = \pm 2\).- Therefore, the x-intercepts are at \((2, 0)\) and \((-2, 0)\).The significance of x-intercepts lies in their ability to provide critical information about the graph's behavior relative to the x-axis. They tell us where the graph touches or crosses the axis and helps in visualizing the symmetry of the geometric shape represented.

Y-Intercepts

Y-intercepts are the points where the equation's graph crosses the y-axis, meaning the value of \(x\) at these points is zero. To determine the y-intercepts for the equation \(x^2 + y^2 = 4\):- Set \(x = 0\) which translates the equation into \(0^2 + y^2 = 4\).- This simplifies to \(y^2 = 4\).- Solving for \(y\) gives us \(y = \pm 2\).- Hence, the y-intercepts are \((0, 2)\) and \((0, -2)\).Y-intercepts are crucial in understanding the vertical extent of the graph. They indicate how high or low the graph reaches along the y-axis and are essential for sketching the graph accurately. This understanding helps in grasping the symmetry and range of the graph of a circle with a radius centered at the origin in the coordinate plane.