Problem 118 - A Graphical Approach to Precalculus with Limits

Question

Describe the graph of the equation as either a circle or a parabola with horizontal axis of symmetry. Then determine two functions, designated by \(y_{1}\) and \(y_{2},\) such that their union will give the graph of the given equation. Finally, graph \(y_{1}\) and \(y_{2}\) in the same viewing rectangle. $$x=-3 y^{2}-6 y+2$$

Step-by-Step Solution

Answer

The graph is a parabola with a horizontal axis of symmetry. Functions: \( y_1 = -1 + \sqrt{\frac{5 - x}{3}} \) and \( y_2 = -1 - \sqrt{\frac{5 - x}{3}} \).

Step 1: Identifying the Type of Graph

The given equation is of the form \( x = ay^2 + by + c \), which is characteristic of a parabola. Since it is expressed with the variable \( x \) as a function of \( y \), it indicates that the parabola has a horizontal axis of symmetry.

Step 2: Rearranging the Equation

To better understand the graph, we need to express the equation in a more familiar quadratic form. Start by rearranging the terms: \( x = -3(y^2 + 2y) + 2 \).

Step 3: Completing the Square

Complete the square for the expression in the parentheses: 1. Take the coefficient of \( y \), which is 2, divide by 2 to get 1, and square it to obtain 1.2. Rewrite the equation as \( x = -3((y + 1)^2 - 1) + 2 \). 3. Simplify to get \( x = -3(y + 1)^2 + 3 + 2 \), resulting in \( x = -3(y + 1)^2 + 5 \).

Step 4: Isolating the Square and Solving for y

Isolate \( (y + 1)^2 \) by following these steps:- Add and subtract to get \( -3(y + 1)^2 = x - 5 \).- Divide by \(-3\): \( (y + 1)^2 = \frac{5 - x}{3} \).

Step 5: Solving for y and Identifying Functions

Take the square root of both sides to solve for \( y \): \[ y + 1 = \pm \sqrt{\frac{5 - x}{3}} \]This results in two functions:1. \( y_1 = -1 + \sqrt{\frac{5 - x}{3}} \)2. \( y_2 = -1 - \sqrt{\frac{5 - x}{3}} \)

Step 6: Graphing the Functions

Both functions, \( y_1 \) and \( y_2 \), should be graphed in the same viewing rectangle to display the parabola. These functions represent the upper and lower halves of the parabola with horizontal symmetry around the line \( y = -1 \).

Key Concepts

Horizontal Axis of Symmetry

When a parabola has a horizontal axis of symmetry, it means the parabola opens to the left or right instead of up or down. In mathematical terms, we say that the axis of symmetry is parallel to the x-axis. For the equation \( x = ay^2 + by + c \), \( y \) is the variable, and \( x \) is described in terms of \( y \). This arrangement suggests that the "U" shape of the parabola is lying on its side.
Understanding the axis of symmetry is important because:

• It divides the parabola into two mirror-image halves.
• All points on the parabola are symmetric with respect to this line.

In the given problem, the axis of symmetry is determined by the equation \( y = -1 \), which can be found by completing the square.

Quadratic Equation

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). For our exercise, the equation is expressed as \( x = -3y^2 - 6y + 2 \), which is a quadratic expression in terms of \( y \). This indicates that:

• The parabola's direction depends on the sign of \( a \); if \( a \) is negative, it opens to the left, as in our example.
• Re-writing the equation helps reveal properties like the axis of symmetry and the vertex.

To gain insights about the parabola, we rearrange the terms, eventually leading us to complete the square.

Completing the Square

Completing the square is a valuable algebraic method for solving quadratic equations and is essential for rewriting them in vertex form: \( a(y + h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.Here’s a step-by-step guide:

• Start with the equation: \( x = -3(y^2 + 2y) + 2 \).
• Focus on \( y^2 + 2y \): Take half of 2 (the coefficient of \( y \)), square it to get 1. Add and subtract this square within the parentheses to complete the square.
• This gives: \( x = -3((y + 1)^2 - 1) + 2 \).
• Simplify to find \( x = -3(y + 1)^2 + 5 \).

By completing the square, we transform the quadratic into a form that demonstrates the parabola's vertex, providing a clear path to graph the functions \( y_1 \) and \( y_2 \).