Problem 36 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 35-38: Find an equation for the indicated half of the parabola. $$ \text { Upper half of }(y-2)^{2}=x-4 $$

Step-by-Step Solution

Answer

The equation for the upper half is \(y = \sqrt{x-4} + 2\).

Step 1: Identify the Form of the Equation

The given equation is \((y-2)^2 = x-4\). This is a variation of the parabola equation \((y-k)^2 = 4p(x-h)\), which represents a parabola that opens to the left or right, with the vertex at \((h, k)\).

Step 2: Determine the Direction of the Parabola

The form \((y-2)^2 = x-4\) suggests the parabola opens sideways (left or right). Specifically, it opens rightward because \(x\) is positive on the right side of the equation.

Step 3: Find the Vertex

The equation \((y-2)^2 = x-4\) can be seen as \((y-k)^2 = (x-h)\) where \(h = 4\) and \(k = 2\). Thus, the vertex of the parabola is \((4, 2)\).

Step 4: Choose the Half of the Parabola

The problem specifies finding the equation for the upper half of this side-opening parabola. Since the complete set of values for \(y\) would be \(y\geq 0\) and \(y\leq 4\), for the upper half, we consider \(y \geq 2\).

Step 5: Solve for the Upper Half

To express \(y\) in terms of \(x\) for the upper half, solve for \(y\) in the equation: \((y-2)^2 = x - 4\). Take the square root of both sides: \(y - 2 = \sqrt{x - 4}\). So, \(y = \sqrt{x-4} + 2\). For the upper half, we take the positive square root.

Key Concepts

Vertex Form

Understanding the vertex form of a parabolic equation is essential for manipulating and analyzing parabolas. In vertex form, a parabola opening sideways is often expressed as \[(y-k)^2 = 4p(x-h)\].Here:

• \( (h, k) \) is the vertex of the parabola, providing its central point. This gives you a clear idea of the parabola's position on the graph.
• \( p \) indicates how far the focus is from the vertex, affecting the parabola's width.

In the given equation, \[(y-2)^2 = (x-4),\] we can match it to the template by identifying \( h = 4 \) and \( k = 2 \). Thus, the vertex is at \((4, 2)\). The vertex is a crucial starting point when graphing a parabola, especially for ones which open sideways.

Side-Opening Parabola

A side-opening parabola is one that opens horizontally, either to the left or right, rather than vertically. The equation \((y-k)^2 = 4p(x-h)\) characterizes such parabolas. In our case, \((y-2)^2 = x-4\) indicates a rightward-opening parabola. The left-hand side is a square of a \(y\) term, making \(y\) the variable with squared term and \(x\) on the right, resulting in horizontal opening.

• Rightward-opening means \ x \ is positive; this directs the parabola to the right.
• Leftward-opening is indicated by \ (y-k)^2 = -(x-h) \ which would suggest a negative sign influencing the x variable.

Recognizing whether a parabola opens sideways or vertically is essential. It affects how we solve certain equations, particularly when isolating terms related to \ x \ or \ y \.

Solving for y

Once we understand the equation \((y-2)^2 = x-4\), we can solve for \(y\). This requires taking the square root of both sides of the equation. This step separates our variable of interest, simplifying the problem.Steps:

• Start with: \((y-2)^2 = x-4\). This sets the stage for extraction.
• Apply the square root to both sides: \(y - 2 = \pm \sqrt{x - 4}\).
• Solve the equation for \(y\): \(y = \pm \sqrt{x - 4} + 2\).

In this context, since we are considering the upper half of the parabola, we choose the positive value: \ y = \sqrt{x - 4} + 2. \ This solution reflects the correct portion we need for the analysis.

Upper Half of Parabola

When examining a side-opening parabola, selecting which 'half' you need to represent is a common task. For the upper half of a parabola opening sideways, consider the values of \ y \ which are greater than the vertex's \ y \ ordinate. In our example:

• Vertex is \(4, 2\)\.
• Upper half is \ y \geq 2 \.

Any result for \ y \ must align with these expectations:

• Use \( y = \sqrt{x-4} + 2\) for the upper section, affirming that \ y \ values exceed the vertex point.
• This ensures we're capturing the correct graph portion, consistent with the problem's requirement.

Focusing on the upper half helps us zero in on the boundary, excluding the lower section for clearer, directed calculations.