Problem 116 - A Graphical Approach to College Algebra

Question

In Exercises \(109-116\), describe the graph of the equation as either a circle or a parabola with a 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 given viewing window. $$\begin{aligned} &x=-3 y^{2}-6 y+2\\\ &[-9.4,9.4] \text { by }[-6.2,6.2] \end{aligned}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Graph

The given equation is of the form \( x = ay^2 + by + c \), which represents a parabola. Since the variable \( x \) is expressed in terms of \( y \), this indicates a parabola with a horizontal axis of symmetry.

Step 2: Complete the Square

To convert the equation to a form that allows us to define \( y_1 \) and \( y_2 \), we complete the square. Start with the equation \( x = -3y^2 - 6y + 2 \). Factor out the \(-3\) from the \(y\) terms: \( x = -3(y^2 + 2y) + 2 \). Now complete the square inside the parenthesis: \( x = -3((y + 1)^2 - 1) + 2 \). Simplifying gives \( x = -3(y + 1)^2 + 3 + 2 \) or \( x = -3(y + 1)^2 + 5 \).

Step 3: Express the Equation in Vertex Form

The equation \( x = -3(y + 1)^2 + 5 \) reveals that the vertex form of the parabola is \( x = a(y-b)^2 + c \) with vertex at \((5, -1)\), opening leftward because \(a = -3\) is negative.

Step 4: Define Functions \( y_1 \) and \( y_2 \)

We solve for \( y \) to get functions \( y_1 \) and \( y_2 \). From \( x = -3(y + 1)^2 + 5 \), rearrange to find \( y \): \( -3(y+1)^2 = x - 5 \), which becomes \( (y+1)^2 = \frac{5-x}{3} \). Solving for \( y \) gives two functions: \( y_1 = -1 + \sqrt{\frac{5-x}{3}} \) and \( y_2 = -1 - \sqrt{\frac{5-x}{3}} \).

Step 5: Graph the Functions

Plot \( y_1 \) and \( y_2 \) within the specified viewing window \([-9.4,9.4] \text{ by } [-6.2,6.2]\). The union of these two functions represents the full parabola. The graph should show a parabola opening to the left with a vertex at \( (5, -1) \).

Key Concepts

Completing the Square

Completing the square is an essential technique used to rewrite quadratic expressions in a more manageable form. It simplifies the process of identifying key features of the graph, such as the vertex. Here’s how it works for our specific example where the equation is given as \( x = -3y^2 - 6y + 2 \).
Steps:

• Factor out the coefficient: Start by factoring out \(-3\) from the terms involving \(y\). This gives \( x = -3(y^2 + 2y) + 2 \).
• Complete the square inside the parenthesis: To transform \( y^2 + 2y \), we add and subtract \(1\) inside the parentheses, getting \( (y + 1)^2 - 1 \).
• Simplify the expression: Substitute back into the equation, resulting in \( x = -3((y + 1)^2 - 1) + 2 \). After simplifying, we have \( x = -3(y + 1)^2 + 5 \).

This form makes it easier to find the vertex and other attributes of the parabola.

Vertex Form

The vertex form of a quadratic equation delivers straightforward information about the parabola, specifically its vertex, which is a crucial point. The general vertex form is \( x = a(y-b)^2 + c \).
For our equation, we have \( x = -3(y+1)^2 + 5 \). From this, we can see:

• Vertex: The vertex of the parabola is \((5, -1)\), indicating where it "turns." This tells us that the highest or lowest point along the axis of symmetry is at \( x = 5 \).
• Direction of opening: Since \( a = -3 \) is negative, the parabola opens to the left.

Using vertex form clarifies the shape and position of the parabola, making it a powerful tool for graphing.

Functions y1 and y2

In this context, functions \(y_1\) and \(y_2\) represent solutions to our equation when rearranged and solved for \(y\). These functions outline the complete graph of the parabola.
Finding y1 and y2:

• Rearranging the equation \( x = -3(y+1)^2 + 5 \), we solve for \(y\) to get \( (y+1)^2 = \frac{5-x}{3} \).
• The functions become \( y_1 = -1 + \sqrt{\frac{5-x}{3}} \) and \( y_2 = -1 - \sqrt{\frac{5-x}{3}} \).

These two functions together depict the full parabola on the graph.
Graphing:

• Using the given window \([-9.4, 9.4] \text{ by } [-6.2, 6.2]\), plot both \(y_1\) and \(y_2\) to show the complete parabola.
• The parabola should be seen opening to the left, with the vertex positioned at \((5, -1)\).

Thus, combining \(y_1\) and \(y_2\) helps visualize the full geometric representation.