Problem 11 - Precalculus Mathematics for Calculus

Question

Find the vertex, focus, and directrix of the parabola. Then sketch the graph. $$-4\left(x+\frac{1}{2}\right)^{2}=y$$

Step-by-Step Solution

Answer

Vertex: \((-\frac{1}{2}, 0)\), Focus: \((-\frac{1}{2}, -1)\), Directrix: \(y = 1\).

Step 1: Identify the parabola form

The given equation is \(-4\left(x + \frac{1}{2}\right)^{2} = y\). This is a parabola, and it resembles the standard form \((x-h)^2 = 4p(y-k)\), which is a vertical parabola opening either up or down. From this, we can compare directly to find \(p\), \(h\), and \(k\).

Step 2: Rewrite and determine vertex

Rewrite the equation as \((x + \frac{1}{2})^2 = -\frac{1}{4}y\). By comparing with the standard form, \(h = -\frac{1}{2}\), \(k = 0\), and \(p = -1\). Hence, the vertex is \((-\frac{1}{2}, 0)\).

Step 3: Calculate focus

The focus of the parabola is calculated using \((h, k + p)\). Substituting the known values, the focus is \((-\frac{1}{2}, 0 + (-1))\) or \((-\frac{1}{2}, -1)\).

Step 4: Determine directrix

The directrix of the parabola is found using the equation \(y = k - p\). Substituting the known values gives the directrix as \(y = 0 - (-1)\), which is \(y = 1\).

Step 5: Sketch the graph

Plot the vertex \((-\frac{1}{2}, 0)\), the focus \((-\frac{1}{2}, -1)\), and draw the directrix line at \(y = 1\). Since \(p\) is negative, the parabola opens downward. Sketch the parabola opening down, using the focus and directrix as guides.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is the point where the parabola changes direction. You can think of it as the "tip" of the curve.
For vertical parabolas, this point is either the top or the bottom of the curve, while for horizontal parabolas, it's on the left or right.
In mathematical terms, the vertex of a parabola given by \(y = a(x - h)^2 + k\) is \((h, k)\).
In this exercise, the vertex was found by transforming the given equation into its standard form.

• The original equation is \( -4(x + \frac{1}{2})^2 = y \).
• After manipulation, we identified \(-\frac{1}{2}\) as \(h\) and 0 as \(k\).

Thus, the vertex is \((-\frac{1}{2}, 0)\). This point is critical when drawing the parabola because it helps set the baseline from which the curve opens either upwards or downwards.

Focus and Directrix

The focus and directrix are essential components of a parabola that help define its shape and position. The focus is a fixed point inside the parabola that helps direct its curve.
The directrix is a line that is perpendicular to the axis of symmetry of the parabola.
Together, they establish the parabola's unique property that any point on the curve is equidistant from the focus and the directrix.
For a parabola in the format \((x-h)^2 = 4p(y-k)\), we calculate:

• **Focus** using the formula \( (h, k + p) \).
• **Directrix** as \( y = k - p \).

In this example,
- **Focus**: Substituting \(h = -\frac{1}{2}, k = 0, p = -1\), the focus is \((-\frac{1}{2}, -1)\).
- **Directrix**: The equation \(y = k - p\) gives the line \(y = 1\).
These elements, the focus and directrix, help guide the sketching of the parabola and ensure it follows its intended path.

Graphing Parabolas

Graphing parabolas requires an understanding of their key features like the vertex, focus, and directrix. These elements guide how the parabola opens and in which direction.
Once you have calculated the vertex, focus, and directrix:

• **Plot the vertex**, which in this case is \((-\frac{1}{2}, 0)\).
• **Mark the focus** \((-\frac{1}{2}, -1)\) on the graph, providing an inside point.
• **Draw the directrix** as a horizontal line at \(y=1\).
• Since **\(p\) is negative**, the parabola opens downwards.

Use the vertex and focus to - identify how sharply or widely the parabola curves and- ensure that its path is equidistant from the focus and directrix at any point.This methodical approach to graphing ensures that each parabola you draw is accurate and mathematically sound.

Standard Form of a Parabola

The standard form of a parabola is crucial for understanding and plotting its graph.
For a vertical parabola, it is expressed as \((x-h)^2 = 4p(y-k)\), where:

• \( (h, k) \) is the vertex.
• \( p \) dictates how far the focus is from the vertex.

This format makes it easy to identify core elements such as the vertex, focus, and directrix.
In the given exercise, we converted the equation \(-4(x+\frac{1}{2})^2 = y\) into a form that matched the standard formula:
- \(h = -\frac{1}{2}\), \(k = 0\), \(p = -1\), guiding us to locate the vertex, focus, and directrix.
Using the standard form provides a clear, organized method of dissecting and understanding the structure of any parabola.