Problem 43 - Precalculus with Limits

Question

In Exercises 33-46, find the vertex, focus, and directrix of the parabola, and sketch its graph. \(y=\frac{1}{4}(x^2 - 2x + 5)\)

Step-by-Step Solution

Answer

The vertex of the parabola is at (1, 1), the focus is at (1, 2), and the directrix is y=0.

Step 1: Rewrite the equation in vertex form

The given equation is \(y=\frac{1}{4}(x^2 - 2x + 5)\). It needs to be rewritten in vertex form which is \(y=a(x-h)^2 + k\), where (h,k) is the vertex of the parabola. We can complete the square to rewrite the given equation to its vertex form. Completing the square of \(x^2 - 2x + 5\) gives \( (x-1)^2 + 4 \). Therefore, the equation in vertex form is \(y=\frac{1}{4}(x-1)^2 + 1\).

Step 2: Find the vertex

From the vertex form \(y=\frac{1}{4}(x-1)^2 + 1\), we can clearly see the vertex of the parabola is at (1, 1).

Step 3: Find the focus

The value of a in the vertex form is 1/4. Since the parabola is upwards as a is positive, use the formula for the focus of an upward parabola, which is (h, k + 1/4a). Plugging in the values, we get for the focus (1, 1 + 1/(4*1/4)) = (1, 2).

Step 4: Find the directrix

The formula for the directrix of an upward parabola is y = k - 1/4a. Plugging in the values, we get for the directrix y = 1 - 1/(4*1/4) = 0.

Step 5: Sketch the graph

Plot the vertex at point (1, 1). The focus is located at point (1, 2) which is 1 unit above the vertex. Draw a horizontal line y=0 representing the directrix which is 1 unit below the vertex. The parabola graphs as a U shape opening upwards with vertex at (1,1), focus at (1,2) and directrix at y=0.

Key Concepts

Parabola

A parabola is a U-shaped curve commonly found in algebra and geometry. This elegant curve has some unique properties. It's a symmetric shape that can open upward or downward, depending on the orientation of its leading term. In our example, the equation is given as \[y = \frac{1}{4}(x^2 - 2x + 5)\] which simplifies to the U-shaped parabola that opens upwards.

Parabolas have several essential parts:

• Axis of Symmetry: This is the vertical line that runs through the vertex, and it balances the parabola evenly on both sides.
• Vertex: This is the lowest point of the parabola when facing upwards or the highest point when facing downwards.
• Focus: A fixed point inside the curve that represents a constant distance from every other point on the parabola to the directrix line.
• Directrix: A horizontal line outside the curve, helping to define the parabola's shape alongside the focus.

When you understand these parts, identifying components on a graph becomes more straightforward! Each vertex, focus, and directrix connects to the others, creating a perfectly balanced curve.

Vertex Form

The vertex form of a parabola's equation is a neat way to express the curve, as it clearly demonstrates the vertex's position. This version is written as \[y = a(x-h)^2 + k\] where

• (h, k) represents the vertex of the parabola.
• a signifies if the parabola opens upwards (\(a > 0\)) or downwards (\(a < 0\)), as well as the "width" of the parabola.

In the original equation \[y = \frac{1}{4}(x^2 - 2x + 5)\], converting by completing the square leads to \[y = \frac{1}{4}(x-1)^2 + 1\]. Here, the vertex is at (1, 1) and value \(a = \frac{1}{4}\), indicating the parabola opens upwards and is relatively wide due to the smaller size of \(a\).

Focus and Directrix

The focus and directrix give a parabola its unique shape and mathematical symmetry. Imagine them as guiding points that define where the curve extends.

For our parabola, the vertex is at (1, 1), and the form is \(y=\frac{1}{4}(x-1)^2 + 1\):

• The Focus: Located at (1, 2). The formula used here for an upward parabola is \((h, k + \frac{1}{4a})\), giving insight into the curve's direction and curvature.
• The Directrix: The line \(y = 0\), a horizontal guideline placed symmetrically against the focus. The formula, \(y = k - \frac{1}{4a}\) gives the perfect complement, balancing the focus on the graph.

These elements focus and directrix, coincide with the vertex, portraying the ideal shape and maintaining the parabola's distinctive U form!