Problem 107 - A Graphical Approach to Precalculus with Limits

Question

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$x=\frac{2}{3} y^{2}-4 y+8$$

Step-by-Step Solution

Answer

Vertex: (2, 3); Axis: y=3; Domain: [2, ∞); Range: (-∞, ∞).

Step 1: Identify the Standard Form

The given equation is \( x = \frac{2}{3} y^2 - 4y + 8 \). This is a parabolic equation in terms of \( y \). For parabolas oriented horizontally, the standard form is \( x = a(y - k)^2 + h \). We need to manipulate the equation to achieve this form.

Step 2: Complete the Square

We start with \( x = \frac{2}{3} y^2 - 4y + 8 \). Focus on the \( y \) terms: \( \frac{2}{3} (y^2 - 6y) + 8 \). To complete the square: \( (y - 3)^2 = y^2 - 6y + 9 \). Replacing gives: \( x = \frac{2}{3}[(y-3)^2 - 9] + 8 \). Expanding: \( x = \frac{2}{3}(y-3)^2 - 6 + 8 \), hence, \( x = \frac{2}{3}(y-3)^2 + 2 \).

Step 3: Identify the Vertex and Axis

In the completed square form \( x = \frac{2}{3}(y-3)^2 + 2 \), the vertex is at \((h, k) = (2, 3)\). The axis of symmetry is the line \( y = 3 \), since this is where \( y \) is centered around in the square.

Step 4: Determine the Domain and Range

Since the parabola opens horizontally to the right (positive \( \frac{2}{3} \)), the domain starts from the x-value of the vertex, \( [2, \infty) \). The range is all real numbers, \( (-\infty, \infty) \), because the parabola can take on any \( y \) value as it extends horizontally.

Key Concepts

Vertex

The vertex of a parabola is a crucial point that can be thought of as the peak or trough, depending on how the parabola opens. In the case of the parabola oriented horizontally along the equation \( x = \frac{2}{3}(y-3)^2 + 2 \), the vertex represents the starting point from which the parabola's width extends. The vertex in this form is denoted as \((h, k)\), and for our given equation, the vertex is located at \((2, 3)\). This means that the parabola is shifted 2 units horizontally and 3 units vertically in the plane.
Understanding the vertex helps us quickly pinpoint the parabola's position and guide us in plotting the rest of the shape accurately.

Axis of symmetry

The axis of symmetry in a parabola is a vertical line that cuts the parabola into two mirror-image halves. For a horizontally oriented parabola, the axis of symmetry focuses on the \( y \) coordinate of the vertex. In the formula \( x = a(y-k)^2 + h \), the axis of symmetry will correspond to the line \( y = k \).
In our example, the equation \( x = \frac{2}{3}(y-3)^2 + 2 \) reveals the axis of symmetry as the line \( y = 3 \). This vertical line is crucial because it helps in sketching the parabola accurately, ensuring it is balanced around this central line.

• Use the axis of symmetry to check symmetry when drawing the graph manually.

Domain and range

When discussing the domain and range of a parabola, it's essential to understand its orientation. For a horizontally opening parabola like ours, the domain and range definitions shift from the typical vertical parabolas.
The domain, which is the set of possible \( x \) values the function can take, begins at the x-coordinate of the vertex. Here, since the parabola opens to the right and \( x = \frac{2}{3}(y-3)^2 + 2 \), the domain is \([2, \infty)\).
The range, in contrast, includes all real numbers for \( y \) because the parabola can stretch infinitely in both the positive and negative \( y \) directions: \((-\infty, \infty)\).

• The domain indicates restrictions for plotting x-values.
• The range guides you on possible y-values.

Completing the square

Completing the square is a useful algebraic technique that turns quadratic expressions into a format that easily reveals properties like the vertex. In the equation \( x = \frac{2}{3} y^2 - 4y + 8 \), our goal is to rearrange it to standard form using this process.
Start by focusing on \( y^2 - 6y \), derived from factoring out \( \frac{2}{3} \). To complete the square, you need to balance the equation by finding a perfect square trinomial. We adjust by adding and subtracting 9 inside the square to maintain equality: \( (y - 3)^2 = y^2 - 6y + 9 \).
Substituting back, we get \( \frac{2}{3}[(y-3)^2 - 9] \) which upon expansion results in \( \frac{2}{3}(y-3)^2 + 2 \). This transformation into vertex form uncovers the vertex and makes graph sketching and other calculations straightforward.

• This technique simplifies solving quadratic equations.
• Helps in graphing parabolas by highlighting key features.