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Chapter 13: Problem 79

Write each equation in standard form, if it is not alreacty 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 is a parabola with vertex at \((-2, -2)\).

Step by step solution

01

Understand the Equation

The given equation is a quadratic in terms of y: \( x = \frac{1}{2}y^2 + 2y \). This form suggests that it is a parabola, opening along the x-axis (since it is quadratic in y).
02

Rearrange the Equation

To express the equation in the standard form of a parabola, rearrange it. The standard form for a parabola opening to the right is \( (y - k)^2 = 4p(x - h) \). Rewrite by completing the square:\( x = \frac{1}{2}y^2 + 2y \) can be rewritten as \( x = \frac{1}{2}(y^2 + 4y) \).Complete the square for \( y^2 + 4y \). First, find \( (\frac{4}{2})^2 = 4 \). Add and subtract 4 inside the parentheses:\( x = \frac{1}{2}((y^2 + 4y + 4) - 4) \).Simplifies to \( x = \frac{1}{2}((y + 2)^2 - 4) \).Distribute the \( \frac{1}{2} \):\( x = \frac{1}{2}(y + 2)^2 - 2 \).This is equivalent to \( (y + 2)^2 = 2(x + 2) \), which is now in standard form.
03

Identify the Vertex

The standard form of the parabola is \( (y + 2)^2 = 2(x + 2) \). Identify the vertex \( (h, k) \) from this form: \( h = -2 \) and \( k = -2 \). Thus, the vertex is at \( (-2, -2) \).
04

Graph the Parabola

Plot the vertex at \((-2, -2)\). Since the equation is of the form \((y - k)^2 = 4p(x - h)\), the parabola opens to the right. The distance \(p\), calculated from \(4p = 2\), gives \(p = \frac{1}{2}\). Mark points to the right side of the vertex, 0.5 units away, to show the direction and opening width.

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Key Concepts

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

Standard Form
The term "standard form" in relation to a parabola is an important concept. It provides a way to express parabolic equations in a manner that makes them easy to analyze and graph. For a parabola that opens horizontally (such as with the variable relationship given by the exercise), the standard form is: \((y - k)^2 = 4p(x - h)\).

In this equation:
  • \(h\) and \(k\) are the coordinates of the vertex of the parabola.
  • \(p\) is the distance from the vertex to the focus, which also determines whether the parabola opens to the right or left.
To convert the equation \(x = \frac{1}{2} y^2 + 2 y\) into standard form, we need to rearrange it by completing the square for the expression in \(y\). This allows us to rewrite the equation as \((y + 2)^2 = 2(x + 2)\), clearly matching the standard form and making it straightforward to identify key features such as the vertex.
Completing the Square
Completing the square is a technique used to convert a quadratic equation into a form that reveals significant properties, like the vertex of a parabola. It is especially handy when we need to rewrite quadratic equations that are not immediately in a standard form.

To start completing the square for the equation \(x = \frac{1}{2} y^2 + 2y\), we focus on the quadratic expression in \(y\): \(y^2 + 4y\). Here's the step-by-step method:
  • First, find the constant that makes \(y^2 + 4y\) a perfect square trinomial. Take half of the linear coefficient \(4\) and square it: \(\left(\frac{4}{2}\right)^2 = 4\).
  • Now, add and subtract this constant inside the expression to form a perfect square trinomial: \((y^2 + 4y + 4 - 4)\).
  • Rewrite \((y^2 + 4y + 4)\) as \((y + 2)^2\) simplifying the form.
  • This manipulation transforms the equation to \(x = \frac{1}{2}((y + 2)^2 - 4)\), which simplifies further to \((y + 2)^2 = 2(x + 2)\).
This method not only eases the path towards standard form but also showcases how algebraic manipulation helps unravel the geometric properties of equations.
Graphing Parabolas
Graphing a parabola becomes straightforward once the equation is in its standard form. For the given problem, our equation is \((y + 2)^2 = 2(x + 2)\), which is in the standard form \((y - k)^2 = 4p(x - h)\).

Let's break down the graphing process:
  • Start by plotting the vertex, which is at the point \((-2, -2)\). This serves as a pivotal feature of the graph.
  • The parameter \(p\) determines how the parabola opens. Since \(4p = 2\), we find \(p = \frac{1}{2}\), indicating a sideways opening (rightward in this case).
  • From the vertex, move \(p\) units to the right to locate the focus, assisting in defining the parabola's orientation and width.
  • The initial width and direction can be showcased by marking additional points at regular \(p\) intervals along the parabolic path, establishing the symmetrical curve.
By following these visual cues and algebraic details, graphing a parabola helps bring the algebraic expressions into a geometric reality, offering insight into their shape and position relative to the Cartesian plane.

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