Problem 113 - 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. $$\begin{aligned} &x=y^{2}+6 y+9\\\ &[-10,10] \text { by }[-10,10] \end{aligned}$$

Step-by-Step Solution

Answer

The graph is a parabola with a horizontal axis of symmetry. The functions are \(y_1 = \sqrt{x} - 3\) and \(y_2 = -\sqrt{x} - 3\).

Step 1: Recognize the Type of Conic Section

Look at the given equation \(x = y^2 + 6y + 9\). Notice that the equation is of the form \(x = ay^2 + by + c\), which suggests it is a parabola because there is only a squared term in \(y\) and no squared term in \(x\). This confirms that the graph is a parabola.

Step 2: Identify the Axis of Symmetry

Since the equation is in terms of \(x\) and \(y\), and it's solved for \(x\), this parabola has a horizontal axis of symmetry. Parabolas of the form \(x = ay^2 + by + c\) have horizontal axes.

Step 3: Rewrite the Equation in Standard Form

The equation \(x = y^2 + 6y + 9\) can be rewritten by completing the square on the right-hand side. \(y^2 + 6y + 9\) can be factored as \((y + 3)^2\). Therefore, the equation becomes \(x = (y + 3)^2\).

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

The original equation is \(x = (y + 3)^2\), so solving for \(y\) gives two functions: \(y_{1} = \sqrt{x} - 3\) and \(y_{2} = -\sqrt{x} - 3\). These two functions together will form the parabola.

Step 5: Graph the Functions

Plot \(y_1\) and \(y_2\) on a coordinate plane with the given window \([-10,10] \) by \([-10,10]\). For each \(x\) greater than or equal to 0, compute \(y_1\) and \(y_2\) to obtain corresponding points and create the parabola.

Key Concepts

Parabola

A parabola is a beautiful and simple curve that you encounter frequently in mathematics, especially in the study of conic sections. Parabolas have a few distinct characteristics:

• They have a symmetrical shape, meaning one side is a mirror image of the other.
• They can open either vertically or horizontally.
• The direction they open depends on the variable that is squared in their equation.

In the given exercise, we have the equation \(x = y^2 + 6y + 9\). Here, the term \(y^2\) ensures this is a parabola since only one variable, \(y\), is squared. This indicates the parabola opens horizontally, unlike the standard form \(y = ax^2 + bx + c\), which opens vertically.

Axis of Symmetry

The axis of symmetry is a crucial part of understanding parabolas. It is the line that divides the parabola into two equal halves. For a parabola that opens horizontally, like in our equation, this axis of symmetry is a horizontal line.
Just as a parabola that opens vertically has a vertical axis of symmetry, a parabola that opens horizontally, with an equation structured in terms of \(x\) like \(x = ay^2 + by + c\), will have a horizontal axis. This helps us quickly identify how the parabola will appear when graphed. In this exercise, the axis of symmetry signifies the balance line where all points on the parabola are mirrored.

Completing the Square

Completing the square is an algebraic technique used to rewrite quadratic equations in a way that easily reveals important properties about their graph. In our case, we start with \(x = y^2 + 6y + 9\).
To complete the square:

• First, we focus on the quadratic expression \(y^2 + 6y + 9\).
• The expression can be rewritten as \((y + 3)^2\), which is a perfect square trinomial.

Why is this useful? Expressing the quadratic in this form shows us the turning point of the parabola, thus simplifying both solving and graphing. Consequently, we get \(x = (y + 3)^2\), indicating a shift of the standard parabola along the \(y\)-axis.

Graphing Functions

Graphing functions lets us bring equations to life on the coordinate plane. For the parabola given by the equation \(x = (y + 3)^2\), we need to find two functions, \(y_1\) and \(y_2\), which represent the positive and negative solutions for \(y\).

• Solving for \(y\) yields: \(y_1 = \sqrt{x} - 3\) and \(y_2 = -\sqrt{x} - 3\).
• These two functions will form the upper and lower parts of the parabola, respectively.

To graph these functions within the specified window ([-10, 10] by [-10, 10]), we plot points for various values of \(x\) (ensuring \(x \geq 0\) due to the square root) and then sketch the parabola. This visual representation helps in understanding the symmetry and position of the parabola, as well as confirming the accuracy of previous algebraic manipulations.