Problem 74 - Algebra and Trigonometry

Question

73–76 ? Sketch the graph of the equation. $$ 4 x^{2}+4 y^{2}+2 x=0 $$

Step-by-Step Solution

Answer

The graph is a circle centered at \((-\frac{1}{4},0)\) with a radius of \(\frac{1}{4}\).

Step 1: Rewrite the Equation

Start by rewriting the given equation in a more standard form. We have the equation: \( 4x^2 + 4y^2 + 2x = 0 \).First, we need to factor out a 4 from the \(x\) terms: \( 4(x^2 + \frac{1}{2}x) + 4y^2 = 0 \). This simplifies to \( x^2 + \frac{1}{2}x + y^2 = 0 \).

Step 2: Complete the Square

To simplify \( x^2 + \frac{1}{2}x \) and make it easier to graph, we should complete the square.For the term \( x^2 + \frac{1}{2}x \), take half of the coefficient of \(x\), which is \(\frac{1}{4}\), and square it, resulting in \(\frac{1}{16}\). Add and subtract \(\frac{1}{16}\) inside the equation: \((x + \frac{1}{4})^2 - \frac{1}{16} + y^2 = 0 \).

Step 3: Adjust the Equation

Add \(\frac{1}{16}\) to both sides to balance the equation: \((x + \frac{1}{4})^2 + y^2 = \frac{1}{16} \).This is now in the form of a circle equation, \((x - h)^2 + (y - k)^2 = r^2\). Here, \(h = -\frac{1}{4}\), \(k = 0\), and \(r = \frac{1}{4}\).

Step 4: Sketch the Graph

Plot the circle with the center at \((-\frac{1}{4}, 0)\) and a radius of \(\frac{1}{4}\).First, identify the center on the coordinate plane. The radius means the circle will extend \(\frac{1}{4}\) unit in all directions from this point. Since the circle's radius is quite small, it would closely encircle the center point.

Key Concepts

Completing the Square

Completing the square is a method used to simplify quadratic equations, and it plays an essential role in rewriting the equation of a circle in standard form. Let’s break down this technique with our equation from the exercise. We have the expression:

• Start with: \(x^2 + \frac{1}{2}x\)
• Take the coefficient of \(x\), which is \(\frac{1}{2}\), divide it by \(2\), giving us \(\frac{1}{4}\).
• Square \(\frac{1}{4}\) to get \(\frac{1}{16}\).

To complete the square, we add and subtract \(\frac{1}{16}\) within the equation:

• \((x + \frac{1}{4})^2 - \frac{1}{16}\)

This transforms the quadratic expression into a perfect square trinomial, making it easier to understand and graph. Completing the square is a pivotal step in converting quadratic forms into more manageable expressions, especially useful when dealing with equations of circles.

Standard Form of a Circle

The standard form of a circle is an equation that allows us to identify the circle’s key features easily: its center and radius. This form looks like:
\((x - h)^2 + (y - k)^2 = r^2\)
Where \((h,k)\) represents the center of the circle, and \(r\) is the radius.
By using the completing the square process, the original problem was transformed from:
\[4x^2 + 4y^2 + 2x = 0\]into the standard circle form:

• \((x + \frac{1}{4})^2 + y^2 = \frac{1}{16}\)

In this form, it is clear that the circle has:

• A center at \(( -\frac{1}{4}, 0)\)
• And a radius of \(\frac{1}{4}\)

This easy-to-read format lets us quickly deduce these features, making sketching the graph more intuitive.

Graphing Circles

Graphing circles correctly requires a solid understanding of their properties, particularly focusing on the circle's center and radius. Now that we have the circle in standard form, plotting becomes straightforward. Our exercise gives the circle expressed as:

• \((x + \frac{1}{4})^2 + y^2 = \frac{1}{16}\)

To graph this:

• Identify the circle’s center, which in this case is at \(( -\frac{1}{4}, 0)\).
• Determine the radius, here being \(\frac{1}{4}\).
• Locate the center on the coordinate plane.
• Extend the radius \(\frac{1}{4}\) unit in all directions around the center, outlining the small circle.

This process ensures the circle is accurately represented. Graphing not only visualizes equations but also aids in understanding the spatial geometry of circles, important in various fields of mathematics.