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Chapter 7: Problem 69

Write the given equation either in the form \((y-k)^{2}=a(x-h)\) or in the form \((x-h)^{2}=a(y-k)\). $$ x=2 y^{2}+4 y-1 $$

Short Answer

Expert verified
The equation becomes \((y+1)^2 = \frac{1}{2}(x + 3)\).

Step by step solution

01

Identify the Equation Type

The provided equation is in the form of a quadratic in terms of \(y\): \(x = 2y^2 + 4y - 1\). This type indicates that we'll aim to express it in the form \((x-h)^2 = a(y-k)\).
02

Complete the Square

First, identify the quadratic and linear terms: \(2y^2 + 4y\). Factoring out the 2 gives \(2(y^2 + 2y)\). To complete the square inside the parentheses, take half of 2 (the coefficient of \(y\)), which is 1, and then square it to get 1. Add and subtract this value inside the parentheses: \(2(y^2 + 2y + 1 - 1)\).
03

Simplify and Rearrange

The expression becomes: \(2((y+1)^2 - 1) = 2(y+1)^2 - 2\). Substitute back into the original equation: \(x = 2(y+1)^2 - 2 - 1\). Simplify the constants: \(x = 2(y+1)^2 - 3\).
04

Equation in Vertex Form

The equation is now \(x = 2(y+1)^2 - 3\), which is of the form \((x-h)^2 = a(y-k)\). To explicitly write it in the requested form, we can note the parameters directly: \((y+1)^2 = \frac{1}{2}(x + 3)\). Therefore, it is in the form \((y-k)^2 = a(x-h)\) where \(k = -1\), \(a = \frac{1}{2}\), and \(h = -3\).

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Key Concepts

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

Completing the Square
Completing the square is a method used to simplify quadratic equations, making it easier to identify their conic sections. The goal of this approach is to transform a quadratic equation into a perfect square trinomial, which highlights its essential features.
  1. First, we look at the quadratic term and the linear term. For example, in the expression \(2y^2 + 4y\), we first factor out the leading coefficient, in this case, 2, giving us \(2(y^2 + 2y)\).
  2. Next, take the linear coefficient (here it's 2), halve it, and then square the result. This results in 1, as \((\frac{2}{2})^2 = 1\).
  3. Add and subtract this squared value within the parentheses: \(2(y^2 + 2y + 1 - 1)\).
  4. Rewriting gives us \(2((y + 1)^2 - 1)\), which simplifies further to express the original equation more cleanly.
Completing the square is a crucial step to transform quadratics into a form that reveals their "vertex" properties and prepares them for identifying conic sections.
Conic Sections
Conic sections refer to the curves obtained by intersecting a plane with a cone. Such curves include circles, ellipses, parabolas, and hyperbolas. Each has particular equations that describe their shape and properties.
  • Parabolas: Typically described by equations of the form \((x-h)^2 = a(y-k)\) or \((y-k)^2 = a(x-h)\). Here, the parabola opens either horizontally or vertically.
  • Ellipses and Circles: These are expressed with equations like \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\) for ellipses, which simplifies to a circle when \(a=b\).
  • Hyperbolas: With equations resembling \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), hyperbolas have two separate curves called branches.
In our exercise, the equation \(x = 2(y+1)^2 - 3\) is formatted into the standard parabola form, indicating it represents a parabola with its vertex at a specific point on the Cartesian plane. Recognizing this helps students understand and interpret the graph of the equation.
Vertex Form
The vertex form of a quadratic equation is useful because it readily displays the maximum or minimum point of the parabola. In our example, we achieved \(x = 2(y+1)^2 - 3\), illustrating a quadratic in vertex form.The vertex form is typically \((y-k)^2 = a(x-h)\) or \((x-h)^2 = a(y-k)\), where:
  • \((h, k)\) represents the "vertex" of the parabola, marking a significant point such as the highest or lowest point on the graph.
  • The term \(a\) affects the width and direction of the parabola; a larger \(|a|\) makes it "narrower," while a negative \(a\) flips the parabola upside down or to the opposite direction.
For our solution, the converted form \((y+1)^2 = \frac{1}{2}(x + 3)\) clearly displays the vertex at \((-3, -1)\) and the coefficient \(\frac{1}{2}\) suggests the parabola's orientation and size. Understanding how to manipulate a quadratic equation into vertex form allows students to graph the shape efficiently and comprehend its maximum or minimum behaviors.

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Most popular questions from this chapter

Shade the solutions set to the system. $$ \begin{array}{r} x^{2}+y^{2} \leq 4 \\ x^{2}+(y-2)^{2} \leq 4 \end{array} $$

Find an equation of a parabola that satisfies the given conditions. Focus \((-1,3)\) and directrix \(y=7\)

Perimeter of an Ellipse The perimeter \(P\) of an ellipse can be approximated by $$ P \approx 2 \pi \sqrt{\frac{a^{2}+b^{2}}{2}} $$ (a) Approximate the distance in miles that Mercury travels in one orbit of the sun if \(a=36.0, b=35.2\) and the units are in millions of miles. (b) If a planet has a circular orbit, does this formula give the exact perimeter? Explain.

Write the equation in standard form for an ellipse centered at (h, k). Identify the center and the vertices. $$ 16 x^{2}+48 x+4 y^{2}-20 y+57=0 $$

Solve each system. $$ \begin{aligned} x^{2} &=-3 y \\ -x^{2} &=2 y-2 \end{aligned} $$
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