Problem 20 - A Graphical Approach to Precalculus with Limits

Question

The equation of a conic section is given in a familiar form. Identify the type of graph (if any) that each equation has, without actually graphing. See the summary chart in this section. Do not use a calculator. $$x-4 y^{2}-8 y=0$$

Step-by-Step Solution

Answer

The equation represents a parabola.

Step 1: Rewrite the Equation

To identify the type of conic section, we first need to rewrite the given equation in a standard form. Start by isolating the terms with the variable \(y\). The equation is: \(x - 4y^2 - 8y = 0\). This can be rewritten as \(x = 4y^2 + 8y\).

Step 2: Complete the Square

To transform \(4y^2+8y\) into a perfect square, first factor out \(4\): \(4(y^2 + 2y)\). Next, complete the square inside the parentheses by adding and subtracting \(1\) (half of \(2\) squared, \((\frac{2}{2})^2=1\)): \(4(y^2 + 2y + 1 - 1) = 4((y+1)^2 - 1)\). Thus, \(4y^2 + 8y\) becomes \(4((y+1)^2 - 1) = 4(y+1)^2 - 4\).

Step 3: Identify the Conic Section

Substitute the completed square back into the equation. It becomes \(x = 4(y+1)^2 - 4\). Further simplifying gives \(x + 4 = 4(y+1)^2\), or \(x + 4 = 4(y+1)^2\). This equation is in the form \(x = a(y-k)^2 + h\), indicating it is a sideways-opening parabola.

Step 4: Determine the Vertex

The standard form of a parabola \(x = a(y-k)^2 + h\) indicates the vertex is \((h, k)\). In our equation, \(x + 4 = 4(y+1)^2\) corresponds to \(h = -4\) and \(k = -1\), so the vertex is \((-4, -1)\).

Key Concepts

Graphing

Graphing conic sections starts with rewriting equations in their particular standard forms. This helps clearly identify the features of the graph, such as vertices, axes, and orientation.
When we deal with an equation like \(x = 4(y+1)^2 - 4\), we see it resembles the form of a sideways opening parabola.
For such graphs, finding the vertex and understanding its orientation is crucial.

• Ensure the equation is simplified and rewritten in one of the standard forms.
• Identify constants in the formula which will guide you in mapping the vertex and the curve’s spread.
• A positive \(a\) value results in a right-opening parabola, while a negative opens left.

Taking the time to set up the equation correctly aids in graphing with confidence.

Parabola

The parabola is a basic conic section that is easily identifiable by its distinct U-shape or curve formed as it opens to the side. Typically, you'll spot them in algebraic equations when either the \(x\) or \(y\) variable is squared but not both.
When graphing, remember:

• The standard form \(y = ax^2 + bx + c\) represents a vertical parabola, while \(x = a(y-k)^2 + h\) is a sideways parabola.
• Sideways parabolas have equations of the form \((x = a(y-k)^2 + h)\), indicating they open horizontally.
• Key features to note include the vertex, parabola direction, and axis of symmetry.

This specific equation \(x = 4(y+1)^2 - 4\) is a sideways-opening parabola, meaning it opens horizontally rather than vertically.

Completing the Square

Completing the square is a valuable technique used to rewrite quadratic equations. This method helps to identify parabolas easily when dealing with conic sections by transforming the quadratic equation into a perfect square trinomial.
Let's say we have a quadratic term \(+\) a linear term like \(4y^2 + 8y\). Here’s how to complete the square:

• Factor out the coefficient of \(y^2\), which is \(4\), giving \(4(y^2 + 2y)\).
• Take half of the linear coefficient within the parentheses, square it, and both add and subtract it within the bracket.
• Here, with \(2\) being in \(y^2 + 2y\), half is \(1\), hence, \((1)^2 = 1\).

This results in restructuring the equation into \(4((y+1)^2 - 1)\), simplifying further to form \(4(y+1)^2 - 4\). Completing the square allows the easier translation of the equation into the standard form, clearly showing that it represents a sideways-opening parabola.