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Chapter 10: Problem 79

Write each equation in standard form, if it is not already so, and graph it. If the graph is a circle, give the coordinates of its center and its radius. If the graph is a parabola, give the coordinates of its vertex. $$ x=\frac{1}{2} y^{2}+2 y $$

Short Answer

Expert verified
The equation represents a parabola with vertex at (-2, -2).

Step by step solution

01

Recognize the Type of Equation

The given equation is \( x = \frac{1}{2} y^2 + 2y \). Notice that the equation involves \( y^2 \), which indicates that it might be a parabola since it is quadratic in terms of \( y \). We aim to rewrite it in the standard form of a parabola.
02

Rearrange the Equation

Rearrange the equation to express it in the standard form for a parabola: \( x = a(y - k)^2 + h \). The form we have is \( x = \frac{1}{2}y^2 + 2y \). This isn't quite in standard form yet. We can complete the square to transform it.
03

Complete the Square

Start with the equation \( x = \frac{1}{2}y^2 + 2y \). Factor out the coefficient of \( y^2 \), which is \( \frac{1}{2} \): \( x = \frac{1}{2}(y^2 + 4y) \). To complete the square inside the parentheses, take half of the linear coefficient (4), square it to get 4, and add and subtract it inside to maintain equality: \( x = \frac{1}{2}((y^2 + 4y + 4) - 4) \). This becomes \( x = \frac{1}{2}((y + 2)^2 - 4) \).
04

Simplify to Standard Form

Distribute the \( \frac{1}{2} \) across the terms: \( x = \frac{1}{2}(y + 2)^2 - \frac{1}{2} \cdot 4 \), simplifying to \( x = \frac{1}{2}(y + 2)^2 - 2 \). This is now in the standard form \( x = a(y - k)^2 + h \) for a parabola, where \( a = \frac{1}{2}, \ h = -2, \) and \( k = -2 \).
05

Identify the Vertex and Graph

The standard form \( x = \frac{1}{2}(y + 2)^2 - 2 \) indicates a parabola opening horizontally. The vertex, given by \( (h, k) \), is \((-2, -2)\). Plot this as the vertex. Since \( a = \frac{1}{2} \), the parabola opens to the right with a wider shape compared to a parabola that opens in the \( y \)-direction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Form of a Parabola
The standard form of a parabola simplifies the process of identifying key components like the vertex and direction of the parabola's opening. Parabolas can have equations either in terms of x or y, and for visualization:
  • When in the form \( y = a(x - h)^2 + k \), the parabola opens vertically.
  • When in the form \( x = a(y - k)^2 + h \), the parabola opens horizontally.
In the context of the exercise, the equation \( x = \frac{1}{2}(y + 2)^2 - 2 \) indicates a horizontally opening parabola, which isn't the common form for parabolas opening upward or downward. This form is useful in graphing by clearly identifying the vertex, \((h, k)\), and understanding the shift in position from the origin.
Completing the Square
Completing the square is a critical technique for transforming quadratic expressions into a more manageable form. This technique helps in converting an equation to its standard form. Here's a brief on how it works:
  • You start with a quadratic expression, such as \( y^2 + 4y \).
  • Take half of the coefficient of the linear term (which is 4 in our example), becoming 2, and square it to get 4.
  • You add and subtract this square within the equation to complete the square effectively: \( y^2 + 4y + 4 - 4 \).
  • Thus, the expression \( y^2 + 4y \) becomes \((y + 2)^2 - 4\).
This method turns the equation into a perfect square trinomial, making it easier to rewrite in the standard form of a parabola. This transformation is essential for graphing and identifying key features like the direction and vertex of the parabola.
Graphing Parabolas
Graphing parabolas involves several steps to ensure accurate plotting on a coordinate plane. Let's take the example equation \( x = \frac{1}{2}(y + 2)^2 - 2 \):
  • Vertex Identification: The vertex is crucial as it serves as the central point. For the given equation, the vertex is at \((-2, -2)\).
  • Direction of Opening: Since \(a = \frac{1}{2}\) and is positive, the parabola opens to the right, signifying a wider shape compared to if it opened upwards.
  • Effect of "a": The value of \(a\) affects how "wide" or "narrow" a parabola appears. A smaller \( |a| \) leads to a wider opening.
To graph:- Start at the vertex \((-2, -2)\) on the coordinate plane.- Use the direction of opening and vertex to sketch the parabola, ensuring the shape reflects the specified orientation and width determined by \(a\).- Check additional points if necessary by substitifying values to ensure that the plot aligns correctly with calculations.

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Most popular questions from this chapter

Find each value of \(x .\) $$ \log _{8} x=2 $$

Solve each system of equations by substitution for real values of \(x\) and \(y.\) See Examples 2 and 3. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=20 \\ y=x^{2} \end{array}\right. $$

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Find each value of \(x .\) $$ \log _{6} 216=x $$

Solve the system for real solutions: \(\left\\{\begin{array}{l}\frac{1}{x}+\frac{2}{y}=1 \\\ \frac{2}{x}-\frac{1}{y}=\frac{1}{3}\end{array}\right.\)
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