Problem 15 - Precalculus with Calculus Previews

Question

Find the vertex, focus, directrix, and axis of the given parabola. Graph the parabola. \(x^{2}+5 x-\frac{1}{4} y+6=0\)

Step-by-Step Solution

Answer

Vertex: \((-\frac{5}{2}, -1)\), Focus: \((-\frac{5}{2}, 0)\), Directrix: \(y = -2\), Axis: \(x = -\frac{5}{2}\).

Step 1: Rewrite the Equation in Standard Form

The given equation is \(x^{2} + 5x - \frac{1}{4}y + 6 = 0\). First, we need to convert it into the standard form of a parabola equation, which is \((x - h)^{2} = 4p(y - k)\) for a vertical parabola. Rearrange to isolate the \(y\) terms: \(\frac{1}{4}y = x^{2} + 5x + 6\). Multiply the entire equation by \(4\) to get rid of the fraction: \(y = 4x^{2} + 20x + 24\). Next, complete the square for the \(x\) terms.

Step 2: Complete the Square

Focus on the terms \(4x^{2} + 20x\). Factor out the \(4\): \(y = 4(x^{2} + 5x) + 24\). Now, complete the square inside the parenthesis. For \(x^{2} + 5x\), take half of \(5\), which is \(\frac{5}{2}\), and square it to get \(\left(\frac{5}{2}\right)^{2} = \frac{25}{4}\). Add and subtract \(\frac{25}{4}\) inside the parenthesis: \(y = 4(x^{2} + 5x + \frac{25}{4} - \frac{25}{4}) + 24\), which simplifies to \(y = 4((x + \frac{5}{2})^{2} - \frac{25}{4}) + 24\). Distribute the \(4\): \(y = 4(x + \frac{5}{2})^{2} - 25 + 24\) or \(y = 4(x + \frac{5}{2})^{2} - 1\).

Step 3: Identify the Vertex

From the equation \(y = 4(x + \frac{5}{2})^{2} - 1\), the vertex form \((x - h)^{2} = 4p(y - k)\) reveals that the vertex \( (h, k) \) is \((-\frac{5}{2}, -1)\).

Step 4: Determine the Focus and Directrix

The value of \(4p\) is \(4\), so \(p = 1\). Since the parabola opens upwards (\(4p > 0\)), the focus is \(1\) unit above the vertex. Thus, the focus is \((-\frac{5}{2}, 0)\). The directrix is \(1\) unit below the vertex at \(y = -2\).

Step 5: Axis of Symmetry

The axis of symmetry for a parabola in the form \((x - h)^{2} = 4p(y - k)\) is \(x = h\). For this parabola, the axis of symmetry is \(x = -\frac{5}{2}\).

Step 6: Graph the Parabola

Plot the vertex at \((-\frac{5}{2}, -1)\), the focus at \((-\frac{5}{2}, 0)\), and draw the directrix line at \(y = -2\). The axis of symmetry is \(x = -\frac{5}{2}\). Sketch the parabola opening upwards.

Key Concepts

Vertex

The vertex of a parabola is a crucial point where the curve changes direction. In mathematical terms, it represents the lowest point (for a parabola that opens upwards) or the highest point (for a parabola that opens downwards). In the case of our exercise, we have determined the vertex from the standard form equation:

• The parabola is expressed as \( y = 4(x + \frac{5}{2})^{2} - 1 \).
• This is written in the form \((x - h)^{2} = 4p(y - k)\), where \((h, k)\) is the vertex.
• Therefore, the vertex is located at the coordinates \((-\frac{5}{2}, -1)\).

At this point, you can imagine the parabola opening upwards from \((-\frac{5}{2}, -1)\). This vertex acts as the point of equilibrium for our parabolic curve, and understanding it helps in graphing the parabola accurately.
Grasping the concept of the vertex allows you to predict the shape and direction in which the parabola opens.

Focus

The focus of a parabola is a special point inside the curve. It plays a significant role in defining the shape and position of the parabola. This point is always located along the axis of symmetry and acts as a key characteristic that, together with the directrix, defines the parabola. Let's go through its characteristics:

• The distance from the vertex to the focus, denoted by \(p\), was determined to be 1 in our solution.
• This means the focus is 1 unit in the direction the parabola opens from the vertex.
• Since our parabola opens upwards, the focus lies 1 unit above the vertex, resulting in a focus at \((-\frac{5}{2}, 0)\).

Understanding where the focus is helps in understanding how light reflects off the parabola. In the case of satellite dishes or headlights, this point is critical for concentrating signals or light. The elegance of a parabola lies in how every point on it is equidistant from the focus and directrix.

Directrix

The directrix of a parabola is a fixed line used to define the curve in conjunction with the focus. Together with the vertex and focus, it assists in shaping the parabola and giving it binary balance:

• The directrix is parallel to the axis of symmetry and located \(-p\) units in the opposite direction to the opening of the parabola from the vertex.
• In this particular exercise, since \(p = 1\), the directrix is 1 unit below the vertex as the parabola opens upwards.
• This placement of the directrix makes it at \(y = -2\).

While it doesn't appear directly on the parabolic curve, the directrix holds geometrical significance. Every point on the parabola is equidistant from the focus and the directrix, ensuring the perfect symmetric shape of the curve.

Axis of Symmetry

An essential feature of any parabola is its axis of symmetry. This is a vertical line that passes through the vertex and divides the parabola into two mirror-image halves. It helps in understanding and plotting the equation with greater accuracy. For our specific parabola:

• The axis of symmetry can be found directly from the vertex form of the equation \((x - h)^{2} = 4p(y - k)\).
• It essentially runs through the vertex coordinate \((h, k)\).
• As a result, for our equation, the axis of symmetry is \(x = -\frac{5}{2}\).

Recognizing the axis of symmetry helps in constructing the graph symmetrically. If you were to draw a perpendicular line along this axis, you should find equal parts of the parabola on both sides. This axis acts as the backbone or framework that ensures the overall balance and aesthetics of the parabola.