Problem 11 - Algebra A Combined Function

Question

The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. See Examples 1 through 4. $$ x=y^{2}+6 y+8 $$

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Equation in Standard Form

The given equation is \( x = y^2 + 6y + 8 \). To find the vertex, we need the equation in the form \( y = a(x-h)^2 + k \) or \( x = a(y-k)^2 + h \). Let's complete the square for the \( y \) terms.

Step 2: Complete the Square

To complete the square for the \( y \) terms in the equation \( x = y^2 + 6y + 8 \), start by focusing on \( y^2 + 6y \). We need to add and subtract the square of half of the coefficient of \( y \), which is \( \left(\frac{6}{2}\right)^2 = 9 \). This changes the equation to \( x = (y^2 + 6y + 9) - 9 + 8 \).

Step 3: Simplify the Equation

Now, \( y^2 + 6y + 9 \) can be written as \( (y+3)^2 \). Substitute back into the equation to get \( x = (y+3)^2 - 1 \). This puts the equation in the vertex form \( x = a(y-k)^2 + h \), where \( a = 1 \), \( k = -3 \), and \( h = -1 \).

Step 4: Identify the Vertex

The equation \( x = (y+3)^2 - 1 \) is now in vertex form \( x = a(y-k)^2 + h \). In this format, the vertex of the parabola is at the point \( (h, k) \). For our equation, this means the vertex is \( (-1, -3) \).

Step 5: Graph the Parabola Using the Vertex

Using the vertex \( (-1, -3) \), plot this point on the graph. Since the parabola opens to the right (as seen from \( x = (y+3)^2 - 1 \)), sketch the curve of the parabola. Make sure it is symmetric across the horizontal line through the vertex.

Key Concepts

Completing the Square

Completing the square is an essential technique when dealing with quadratic expressions. It helps in transforming a quadratic equation into a perfectly squared form, making it easier to identify key features such as the vertex of a parabola. To complete the square for the equation \( x = y^2 + 6y + 8 \), we focus on the quadratic term \( y^2 + 6y \). We take half of the coefficient of \( y \), which is 6, divide it by 2 to get 3, and then square it to get 9. By adding and subtracting 9 within the expression, we change the equation part to \( (y^2 + 6y + 9) - 9 \). This process enables us to write the expression as \((y+3)^2 - 1\).

Mastering the skill of completing the square can greatly simplify the process of graphing parabolas and deducing their properties. It's a method applicable not just to solving equations but to many areas of algebraic manipulation.

Vertex Form of a Parabola

The vertex form of a parabola is central to transforming equations into a format that readily reveals the vertex. It reads \( y = a(x-h)^2 + k \) or \( x = a(y-k)^2 + h \) for parabolas that open horizontally. Converting the given equation \( x = y^2 + 6y + 8 \) to vertex form involved rewriting it as \( x = (y+3)^2 - 1 \). Here, the vertex form made it straightforward to identify the vertex as \((-1, -3)\).

By expressing a quadratic in vertex form, it becomes much easier to determine whether the parabola opens upwards, downwards, to the left, or to the right, depending on the axis of symmetry. This form offers clear insight into the location and nature of the shape within a coordinate plane.

Graphing Parabolas

Graphing parabolas accurately requires understanding their structure and key features, such as the vertex and the direction in which the parabola opens. Once we have the equation in vertex form \( x = (y+3)^2 - 1 \), plotting starts with marking the vertex at \((-1, -3)\).

The next step involves detecting the direction of the parabola. With an \( x \) as the dependent variable and the parabola taking the form \((y+3)^2\), we know it opens to the right.

• Begin by plotting the vertex.
• Draw the parabola symmetrically around the horizontal line through the vertex.
• Try a few other points to ensure accurate curvature.

Graphing by hand reinforces the dispersed relationship between the equation and its visual representation. Additionally, graphing tools can aid in checking your work against expected outcomes.

Parabola Properties

Parabolas have intrinsic properties that make them unique among other conics. Understanding these properties helps in their applications and graphing. A parabola is symmetric about its axis, which, for horizontally oriented parabolas like ours, runs along a vertical line passing through the vertex.

Key properties to remember include:

• The vertex, which serves as the turning point.
• The axis of symmetry, here it's parallel to the y-axis as \( x \) is the dependent variable.
• The direction of opening, determined by the orientation of the squared term.\((y+3)^2\) indicates a horizontal opening.

In practical applications, these features help in tasks such as optimizing paths and predicting movements modeled by parabolic motion. Recognizing these traits assists not only in academic examinations but in real-world problem-solving scenarios.