Chapter 10: Problem 17
The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. $$x=y^{2}+6 y+8$$
Short Answer
Expert verified
The vertex of the parabola is \((-1, -3)\).
Step by step solution
01
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.
02
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:\[ x = (y + 3)^2 - 1 \].
03
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 \((-1, -3)\).
04
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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.
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:
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 \( x = (y+3)^2 - 1 \).
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:
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 \((h, -k)\).