/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Use a graphing device to graph t... [FREE SOLUTION] | 91Ó°ÊÓ

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Use a graphing device to graph the parabola. $$8 y^{2}=x$$

Short Answer

Expert verified
Graph the parabola \( 8y^2 = x \) on a device; it opens right from the origin.

Step by step solution

01

Understanding the Equation

The given parabola is represented by the equation \( 8y^2 = x \), which is in the form of \( y^2 = \frac{x}{8} \). This shows that the parabola opens to the right because the \( y^2 \) term is positive.
02

Identify Key Components

In the equation \( y^2 = \frac{1}{8}x \), the vertex of the parabola is at the origin \((0, 0)\). The coefficient of \( x \) is \( \frac{1}{8} \), which will affect the width of the parabola.
03

Graphing the Parabola

Using a graphing device, input the equation \( y^2 = \frac{1}{8}x \). The graph should appear as a parabola situated on the y-axis, with its vertex at the origin \((0, 0)\) and opening to the right. The parabolic arms will extend symmetrically on either side of the x-axis.

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Key Concepts

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

vertex of parabola
In a parabola, the vertex is a key point where the curve changes direction. It acts like the 'nose' from where the parabola opens out. For the equation \(8 y^2 = x\), the vertex is located at the origin \((0, 0)\). Why? Because in the equation \(y^2 = \frac{1}{8}x \), you can see there are no additional constants added or subtracted from the \(x\) or \(y\). This indicates that the highest or lowest point is at \(0, 0\), which is typical for simple forms of parabolas.
Understanding the vertex is crucial because it helps in determining the parabola's position and direction on a graph. In this case, the parabola opens to the right from the vertex, as there's no \(x^2\) term restricting the opening to vertical direction. This is consistent with equations where \(y^2\) is a function of \(x\).
equation of parabola
The equation of a parabola in general can vary but has certain forms. With \(8y^2 = x\), we see that it resembles the form \(y^2 = 4px\), where \(p\) indicates how wide or narrow it is. The form helps us understand key attributes of the parabola, like its direction and how it opens.
In this equation, rearranging gives \(y^2 = \frac{1}{8}x\). Notice \(\frac{1}{8}\) here. It's a small fraction implying the parabola will open wider. The opening direction is determined by which variable the square term is attached to. If the equation were \(x^2 = 4py\), it would open vertically rather than horizontally. Recognizing these parts in an equation helps predict what the parabola will look like even before graphing.
graphing device usage
Graphing devices, like graphing calculators or software, are invaluable for visualizing equations like \(8y^2 = x\). To graph this, input it as \(y^2 = \frac{1}{8}x\), which is equivalent by rearrangement. Most graphing devices allow you to plot this directly, showing you the shape practically customize your view by zooming to see different sections.
These devices handle the heavy lifting, drawing complex graphs effortlessly and accurately. This facilitates better understanding, especially when equations become more complicated. They allow instant verification of manually plotted points or when checking characteristic points like vertexes or intersection with axes. Using graphing devices efficiently complements manual calculation skills.

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Most popular questions from this chapter

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$x^{2}-y^{2}=10(x-y)+1$$

A pair of parametric equations is given. (a) Sketch the curve represented by the parametric equations. (b) Find a rectangular-coordinate equation for the curve by eliminating the parameter. $$x=\sin ^{2} t, \quad y=\cos t$$

Sketch the curve given by the parametric equations. $$x=\cot t, \quad y=2 \sin ^{2} t, \quad 0

This exercise deals with confocal parabolas, that is, families of parabolas that have the same focus. (a) Draw graphs of the family of parabolas $$x^{2}=4 p(y+p)$$ for \(p=-2,-\frac{3}{2},-1,-\frac{1}{2}, \frac{1}{2}, 1, \frac{3}{2}, 2\). (b) Show that each parabola in this family has its focus at the origin. (c) Describe the effect on the graph of moving the vertex closer to the origin.

In this section we stated that parametric equations contain more information than just the shape of a curve. Write a short paragraph explaining this statement. Use the following example and your answers to parts (a) and (b) below in your explanation. The position of a particle is given by the parametric equations $$x=\sin t \quad y=\cos t$$ where \(t\) represents time. We know that the shape of the path of the particle is a circle. (a) How long does it take the particle to go once around the circle? Find parametric equations if the particle moves twice as fast around the circle. (b) Does the particle travel clockwise or counterclockwise around the circle? Find parametric equations if the particle moves in the opposite direction around the circle.

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