Problem 17 - Intermediate Algebra

Question

The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. $$x=y^{2}+6 y+8$$

Step-by-Step Solution

Answer

The vertex of the parabola is \((-1, -3)\).

Step 1: Rewrite the Equation to Standard Form

The given equation is \( x = y^2 + 6y + 8 \). We want to express this in the form \( x = (y - k)^2 + h \). To do this, we will complete the square for the \( y \) terms.

Step 2: Complete the Square

Start by focusing on the expression \( y^2 + 6y \). To complete the square, take the coefficient of \( y \) (which is 6), divide it by 2 to get 3, and then square it to get 9. Add and subtract 9 inside the equation:\[ x = (y^2 + 6y + 9) - 9 + 8 \] This simplifies to:\[ x = (y + 3)^2 - 1 \].

Step 3: Identify the Vertex from Standard Form

The equation \( x = (y + 3)^2 - 1 \) is in the form \( x = (y - k)^2 + h \), where the vertex form identifies the vertex as \((h, -k)\). Here, \( k = -3 \) and \( h = -1 \), so the vertex is at \((-1, -3)\).

Step 4: Graph the Parabola

To graph the parabola, plot the vertex at \((-1, -3)\). Since the equation is in terms of \( x \) as a function of \( y \), the parabola opens to the right. Additional points can be plotted by solving for \( x \) with various \( y \) values to create the complete graph.

Key Concepts

Completing the Square

When you encounter a quadratic expression like \( y^2 + 6y \), completing the square is a vital technique. It's all about reshaping part of the quadratic into a perfect square trinomial. This makes it easier to identify important properties of the parabola, such as its vertex.

The process involves the following steps:

• Take the coefficient of the linear term (\( y \) term here), which is 6.
• Divide it by 2. So, you get 3 in this case.
• Square the result, yielding 9.

Now, incorporate this inside the quadratic expression. You temporarily add and subtract this squared value (9) in the equation.

This transforms \( y^2 + 6y \) into \( (y^2 + 6y + 9) - 9 \). The expression \( (y + 3)^2 \) shows up as a perfect square trinomial, making it simpler to analyze and graph.

Parabola Graphing

Graphing a parabola might seem complex, but it becomes much easier once you've completed the square and pinpointed the vertex. In our example equation, the parabola is represented by the transformation: \( x = (y + 3)^2 - 1 \).

Here’s how you can prepare for graphing:

• Begin with the vertex: Once in vertex form, you can identify the parabola's vertex, which in this case is \((-1, -3)\).
• Determine the orientation: Because our equation is \( x = \) some function of \( y \), the parabola opens horizontally. Specifically, it opens to the right if the leading coefficient (in front of the squared term) is positive.
• Plot additional points: This can be done by choosing various \( y \) values and solving for corresponding \( x \) values based on the equation \( x = (y+3)^2 - 1 \).

Graphing becomes a simple task, with the vertex showing as the critical starting point. You can now connect the dots to visualize the full parabola.

Standard Form of a Quadratic Equation

Quadratic equations can take various forms. The standard form we often encounter is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. However, when dealing with parabolas for graphing purposes, transforming into the vertex form is particularly useful.

The vertex form of a quadratic equation tied to graphing is \( x = (y - k)^2 + h \), indicative of the transformation from the standard quadratic form to pinpoint the vertex. In this form:

• \( h \) and \( k \) represent horizontal and vertical shifts, respectively.
• The form explicitly reveals the vertex at \((h, -k)\).

The transition to vertex form by completing the square allows not only a straightforward way to find the vertex but also an intuitive grasp of the parabola's shift from its standard position. This ease of visualization makes the standard to vertex form conversion invaluable for graphing.