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Chapter 5: Problem 44

Identify the graph of the given equation. $$x=-2 y^{2}$$

Short Answer

Expert verified
The graph is a parabola opening to the left with vertex at (0,0).

Step by step solution

01

Rewrite the Equation

Start with the given equation: \(x = -2y^2\). Notice that it is in the form of \(x = ay^2\) where \(a = -2\).
02

Identify the Graph Type

The equation \(x = ay^2\) represents a parabola. If \(a > 0\), the parabola opens to the right, and if \(a < 0\), the parabola opens to the left. In this case, since \(a = -2\), the parabola opens to the left.
03

Confirm Graph Features

The general form \(x = -2y^2\) indicates a parabola with its vertex at the origin (0,0), symmetric about the x-axis, and opening to the left. The coefficient \(-2\) affects the width and direction of opening.

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Key Concepts

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

Conic Sections
Conic sections are curves formed by the intersection of a plane and a double-napped cone. There are several types of conic sections, each with unique properties:
  • Circle: All points are equidistant from the center, forming a perfect round shape.
  • Ellipse: Looks like a stretched circle, where the sum of distances from any point on the curve to two foci is constant.
  • Parabola: Defined by a point called the focus and a line called the directrix. All points on the parabola are equidistant from the focus and directrix.
  • Hyperbola: Consists of two disconnected curves, with the difference in distances from the foci to any point on the curve being constant.
In this context, the parabola is the conic section of interest. Understanding these elements helps in identifying the equation's graph type.
Parabola Characteristics
A parabola is a unique conic section characterized by its vertex, axis of symmetry, focus, and directrix. When given a parabola in the form of an equation, there are key features to recognize:
  • Vertex: The turning point of the parabola. For an equation like \(x = a y^2\), the vertex is at the origin \((0,0)\).
  • Axis of Symmetry: This is a line about which the parabola is symmetric. In the solved equation \(x = -2y^2\), the axis of symmetry is the x-axis.
  • Direction of Opening: Dependent on the coefficient \(a\). If \(a > 0\), the parabola opens to the right, and if \(a < 0\), it opens to the left. Here, since \(a = -2\), the parabola opens to the left.
  • Focus and Directrix: Essential for constructing the graph of a parabola but often not shown in the initial equation.
Understanding these characteristics helps immensely in sketching and analyzing parabolas.
Graph Transformation
Graph transformation is a process of altering the position or shape of a graph on a coordinate plane. For parabolas, transformation usually involves shifting, stretching/compressing, and reflecting:
  • Shifting: Moving the entire graph up, down, left, or right. This changes the position of the vertex.
  • Stretching/Compressing: Altering the width of the parabola. A larger absolute value of \(a\) compresses the parabola making it narrower, while a smaller absolute value stretches it, making it wider.
  • Reflecting: If \(a\) is positive, there's no reflection. If \(a\) is negative, the parabola is reflected across the axis of symmetry.
In the equation \(x = -2y^2\), since \(a = -2\), the graph reflects across the x-axis, maintaining a wide opening due to the absolute value of 2. These transformations affect how the parabola appears without changing its fundamental nature.

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Most popular questions from this chapter

Let \(W\) be a subspace of \(\mathbb{R}^{n}\) and \(\mathbf{v}\) a vector in \(\mathbb{R}^{n}\). Suppose that \(\mathbf{w}\) and \(\mathbf{w}^{\prime}\) are orthogonal vectors with \(\mathbf{w}\) in \(W\) and that \(\mathbf{v}=\mathbf{w}+\mathbf{w}^{\prime} .\) Is it necessarily true that \(\mathbf{w}^{\prime}\) is in \(W^{\perp}\) ? Either prove that it is true or find a counterexample.

Identify the quadric with the given equation and give its equation in standard form. $$\begin{array}{l} 11 x^{2}+11 y^{2}+14 z^{2}+2 x y+8 x z-8 y z-12 x+ \\ 12 y+12 z=6 \end{array}$$

Evaluate the quadratic form \(f(\mathbf{x})=\mathbf{x}^{T} A \mathbf{x}\) for the given A and x. $$A=\left[\begin{array}{rrr} 1 & 0 & -3 \\ 0 & 2 & 1 \\ -3 & 1 & 3 \end{array}\right], \mathbf{x}=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$

If \(A\) and \(B\) are or thogonally diagonalizable and \(A B=\) \(B A,\) show that \(A B\) is orthogonally diagonalizable

Identify the quadric with the given equation and give its equation in standard form. $$16 x^{2}+100 y^{2}+9 z^{2}-24 x z-60 x-80 z=0$$
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