Problem 117 - 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. $$x=2 y^{2}+8 y+1$$

Step-by-Step Solution

Answer

The graph is a parabola with a horizontal axis of symmetry. The functions are \(y_1 = -2 + \sqrt{\frac{x + 7}{2}}\) and \(y_2 = -2 - \sqrt{\frac{x + 7}{2}}\).

Step 1: Identify the Type of Graph

The given equation is \(x = 2y^2 + 8y + 1\). This equation is quadratic in terms of \(y\), and all terms are expressed in powers of \(y\). Hence, the equation represents a parabola. Furthermore, since the variable \(x\) is isolated and the quadratic term is associated with \(y\), the parabola has a horizontal axis of symmetry.

Step 2: Complete the Square

To make it easier to determine \(y_1\) and \(y_2\), we need to complete the square for the quadratic expression in \(y\). Starting from the equation:\[x = 2(y^2 + 4y) + 1\]First factor out 2 inside the parentheses:\[x = 2((y^2 + 4y + 4) - 4) + 1\]Complete the square within the parentheses:\[x = 2((y + 2)^2 - 4) + 1 = 2(y+2)^2 - 8 + 1\]Simplify to obtain:\[x = 2(y+2)^2 - 7\]

Step 3: Solve for y in Terms of x

Having expressed the equation as \(x = 2(y+2)^2 - 7\), we rearrange to solve for \(y\):\[x + 7 = 2(y+2)^2\]\[\frac{x + 7}{2} = (y+2)^2\]Taking the square root on both sides, we get:\[y + 2 = \pm \sqrt{\frac{x + 7}{2}}\] Thus:\[y_1 = -2 + \sqrt{\frac{x + 7}{2}}\] \[y_2 = -2 - \sqrt{\frac{x + 7}{2}}\]

Step 4: Graph y1 and y2

The functions \(y_1 = -2 + \sqrt{\frac{x + 7}{2}}\) and \(y_2 = -2 - \sqrt{\frac{x + 7}{2}}\) should be graphed in the same viewing rectangle. These represent the upper and lower halves of the parabola, respectively. For each \(x\), calculate \(y_1\) and \(y_2\) and plot the points to visualize the entire parabola.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation of the second degree, typically written in the standard form as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The special characteristic of quadratic equations is the presence of the square of the variable, which defines the equation as quadratic.
Unlike linear equations, which produce straight-line graphs, quadratic equations result in parabolas on a graph. The direction of the parabola (opening upwards or downwards) depends on the sign of the leading coefficient: if \( a > 0 \), the parabola opens upwards; if \( a < 0 \), it opens downwards. In our exercise, the equation is quadratic in terms of \( y \), making it a horizontal parabola.
This means it opens either to the left or right, depending on the terms involved. Recognizing quadratic equations and their forms is crucial for graph interpretation and solving systems.

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial, making it easier to solve or graph. This process is particularly helpful in graphing functions, as it allows us to easily find the vertex of the parabola.
In our problem, the equation \( x = 2y^2 + 8y + 1 \) is modified by completing the square. Begin by factoring out any coefficients from the quadratic and linear terms in \( y \):

• First, factor out 2 from \( (y^2 + 4y) \), giving \( 2(y^2 + 4y) \).
• Add and subtract the same value inside the parentheses to form a perfect square. Here, add 4 and subtract 4 inside the parentheses.
• This transforms into \( 2((y+2)^2 - 4) \).

Simplifying gives us \( x = 2(y+2)^2 - 7 \). Completing the square simplifies solving equations, as it changes the form into something that is easier to manipulate mathematically and understand graphically.

Graphing Functions

Graphing functions involves plotting points on a coordinate plane to visualize the shape of the function. For quadratic functions, this often results in a parabola, as seen in our exercise. The task at hand is to use the completed square form to plot and visualize the equation's graph.
By rearranging \( x = 2(y+2)^2 - 7 \), we express \( y \) in terms of \( x \). After completing the square, we solve for \( y \) to get:

• \( y_1 = -2 + \sqrt{\frac{x + 7}{2}} \)
• \( y_2 = -2 - \sqrt{\frac{x + 7}{2}} \)

These two mathematical expressions represent the upper and lower parts of the parabola, which when plotted, give us the full picture of the function. By calculating and plotting multiple values of \( x \) to find corresponding \( y_1 \) and \( y_2 \), we sketch the true shape of this horizontal parabola.
Graphing helps in understanding how the function behaves across different values of the variables, showing symmetry, direction, and key points like the vertex.