Problem 13 - Precalculus with Calculus Previews

Question

Find the vertex, focus, directrix, and axis of the given parabola. Graph the parabola. \(y^{2}+12 y-4 x+16=0\)

Step-by-Step Solution

Answer

Vertex: (-5, -6); Focus: (-4, -6); Directrix: x = -6; Axis: y = -6.

Step 1: Rewrite the Equation

First, rewrite the given equation in a format that will make it easier to find the vertex, focus, and directrix. The original equation is \(y^2 + 12y - 4x + 16 = 0\). Rearrange it to isolate terms with \(y\) on one side: \(y^2 + 12y = 4x - 16\).

Step 2: Complete the Square

To convert the equation into standard form, complete the square for the \(y\) terms. Take half of the coefficient of \(y\), which is 6, square it to get 36, and add and subtract this inside the equation: \((y^2 + 12y + 36) - 36 = 4x - 16\). This results in \((y+6)^2 = 4x + 20\).

Step 3: Simplify the Equation

Simplify the equation to get it closer to standard form. Rewrite the right side as \((y+6)^2 = 4(x + 5)\). Now the equation is in the form of \((y-k)^2 = 4p(x-h)\), which is the standard form of a parabola that opens horizontally.

Step 4: Identify the Vertex

In the standard form \((y-k)^2 = 4p(x-h)\), the vertex \((h, k)\) can be identified directly. Here, \(h = -5\) and \(k = -6\), so the vertex of the parabola is \((-5, -6)\).

Step 5: Identify the Focus and Directrix

From the standard equation \((y+6)^2 = 4(x+5)\), \(4p = 4\) gives \(p = 1\). - The parabola opens in the positive x-direction, so the focus is \((h+p, k) = (-5+1, -6) = (-4, -6)\).- The directrix is the line \(x = h-p = -5-1 = -6\).

Step 6: Determine the Axis of Symmetry

The axis of symmetry for a horizontally-oriented parabola is the horizontal line through \(y = k\). For this parabola, the axis of symmetry is the line \(y = -6\).

Step 7: Sketch the Graph

Plot the vertex \((-5, -6)\), focus \((-4, -6)\), and draw the vertical directrix line \(x = -6\). Sketch the parabola opening to the right with the vertex at the indicated coordinate and the axis of symmetry \(y = -6\).

Key Concepts

Completing the Square

Completing the square is a method often used to manipulate quadratic equations. This technique is especially helpful when the goal is to convert a quadratic equation into a form that makes it easy to identify specific properties, such as the vertex of a parabola.
To complete the square, you focus on the quadratic and linear terms involving one of the variables. In the example provided, the equation involves the variable \( y \): \( y^2 + 12y \).

• First, take half of the coefficient of \( y \), which is 12. Half of 12 is 6.
• Next, square that number to get 36.
• Add and subtract this squared number within the equation: \((y^2 + 12y + 36) - 36 = 4x - 16\).

This process allows you to rewrite the quadratic part as a perfect square trinomial. Hence, \( (y+6)^2 \). By using this technique, you simplify the parabola's equation, making it easier to match with the vertex form.

Vertex Form of a Parabola

The vertex form of a parabola is a particularly useful way to express the equation of a parabola, as it clearly reveals the vertex, which is the point where the parabola turns or changes direction.
The standard vertex form is given by \( (y-k)^2 = 4p(x-h) \) for a parabola that opens sideways. Here, \((h, k)\) is the vertex.
In the exercise's equation, once the square was completed, we obtained \( (y+6)^2 = 4(x+5) \).

• This corresponds to the vertex form \( (y-k)^2 = 4p(x-h) \), where \( h = -5 \) and \( k = -6 \).
• Thus, the vertex of the parabola is at \((-5, -6)\).

The values of \( h \) and \( k \) are directly read from the vertex form, making it quite intuitive to find the vertex.

Focus of a Parabola

The focus of a parabola is a specific point from which distances to points on the parabola are defined. When manipulating and graphing parabolas, the focus is especially important as it helps to determine the parabola's shape.
In the standard form of a horizontally opening parabola \( (y-k)^2 = 4p(x-h) \), the focus is located at \((h+p, k)\).

• For the equation \( (y+6)^2 = 4(x+5) \), calculate \( p \): since \( 4p = 4 \), we have \( p = 1 \).
• Adding \( p \) to \( h \), the focus becomes \( (-5+1, -6) \) or \((-4, -6)\).

This focus affects the "width" of the parabola's opening and orientation, providing a valuable piece to fully sketch the parabola accurately.

Directrix of a Parabola

The directrix of a parabola is a line that serves as a reference to determine distances to the curve, similar to the focus. The parabola is defined as the set of all points equidistant from the focus and the directrix.
Like the focus, the directrix plays an essential role in determining the parabola's orientation and position. For a parabola given by the equation \( (y-k)^2 = 4p(x-h) \), with \( p \) representing how far the focus is from the vertex:

• The directrix, in this horizontal orientation, will be a vertical line.
• Its equation is \( x = h - p \).

Given \( h = -5 \) and \( p = 1 \), the equation of the directrix becomes \( x = -5 - 1 \) or \( x = -6 \).
By knowing both the focus and directrix, you can fully understand the parabola's geometric properties and accurately graph its shape.