Problem 115 - 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. $$\begin{aligned} &x=2 y^{2}+8 y+1\\\ &[-10,10] \text { by }[-10,10] \end{aligned}$$

Step-by-Step Solution

Answer

Parabola; \(y_1 = -2 + \sqrt{\frac{x + 7}{2}}\), \(y_2 = -2 - \sqrt{\frac{x + 7}{2}}\).

Step 1: Identify the Type of Graph

Examine the given equation \(x = 2y^2 + 8y + 1\). This is a quadratic equation in terms of \(y\), indicating that the graph is a parabola with a horizontal axis of symmetry.

Step 2: Rewrite in Standard Form

Complete the square for the expression in \(y\): \(x = 2 (y^2 + 4y) + 1\). Factor out the 2 from the quadratic to rewrite as \(x = 2 ((y + 2)^2 - 4) + 1\). This becomes \(x = 2(y + 2)^2 - 8 + 1 = 2(y + 2)^2 - 7\). Now, it's in the form \(x = a(y - k)^2 + h\), which describes a parabola.

Step 3: Determine Functions for Graph

The standard quadratic form \(x = a(y - k)^2 + h\) represents a parabola with a horizontal axis. To solve for \(y\), rearrange \(x = 2(y + 2)^2 - 7\) to find \(y\): \1. Solve \(x + 7 = 2(y + 2)^2\). \2. Divide by 2: \(\frac{x + 7}{2} = (y + 2)^2\). \3. Take the square root: \(y + 2 = \pm \sqrt{\frac{x + 7}{2}}\). \4. Solve for \(y\) yields: \(y = -2 \pm \sqrt{\frac{x + 7}{2}}\).

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

From the solutions in Step 3, we define \(y_1\) and \(y_2\): \\(y_1 = -2 + \sqrt{\frac{x + 7}{2}}\) \\(y_2 = -2 - \sqrt{\frac{x + 7}{2}}\). These two functions, when graphed, will provide the complete graph of the original equation.

Step 5: Graph the Functions

On a coordinate plane with the provided window \([-10,10]\) by \([-10,10]\), graph \(y_1 = -2 + \sqrt{\frac{x + 7}{2}}\) and \(y_2 = -2 - \sqrt{\frac{x + 7}{2}}\). They should form a parabola with its vertex at \((-7, -2)\) opening horizontally.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that can be written in the general form:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In our exercise, we see the equation expressed differently:

• \( x = 2y^2 + 8y + 1 \).

This is still a quadratic equation, but it's structured with respect to the variable \( y \) instead of \( x \). This arrangement means the orientation of the parabola is different—in this case, opening horizontally. Quadratic equations like this one can describe a parabola, which is a U-shaped graph. Depending on how the equation is structured, the parabola can open upwards, downwards, or in our example, to the left or right.

Axis of Symmetry

The axis of symmetry is a critical concept in understanding the graph of a parabola. It is an imaginary line that divides the parabola into two mirror-image halves. For a quadratic equation of the form \( x = a(y - k)^2 + h \), as identified in the step-by-step solution, the parabola has a horizontal axis of symmetry. This means the line dividing the parabola runs parallel to the x-axis.
In our example, the symmetry occurs at \( y = -2 \), so every point on the parabola reflects over this horizontal line. This feature makes the parabola have an identical shape on either side of its axis of symmetry. Understanding this axis not only helps in graphing but also in analyzing the properties of the parabola, such as its vertex and focus.

Completing the Square

Completing the square is a technique used to transform a quadratic equation into a simpler form, making it easier to graph parabola shapes. This involves writing the equation in a way that clearly shows its geometric properties, like the vertex form.
To complete the square, you need to manipulate a quadratic expression. Take the equation from the example:

• \( x = 2(y^2 + 4y) + 1 \).

First, factor out the coefficient of \( y^2 \), which is 2, to get:

• \( x = 2((y + 2)^2 - 4) + 1 \).

This transitions the expression into vertex form:

• \( x = 2(y + 2)^2 - 7 \),

where the parabola's vertex is at the point \( (-7, -2) \). The act of completing the square reveals this important feature of the parabola effortlessly, thereby simplifying complex quadratic problems.