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Chapter 10: Problem 24

Use a graphing device to graph the parabola. $$x-2 y^{2}=0$$

Short Answer

Expert verified
The parabola \(x = 2y^2\) opens to the right with its vertex at the origin \((0, 0)\).

Step by step solution

01

Write the Equation in the Form of a Parabola

The given equation is \(x - 2y^2 = 0\). Let's write it in the form \(x = ay^2\), where \(a\) is a constant. Add \(2y^2\) to both sides to get \(x = 2y^2\). This equation represents a parabola that opens to the right.
02

Identify the Vertex and Orientation

For the equation \(x = 2y^2\), the vertex is at the origin \((0, 0)\) because there are no added terms to shift it up/down or right/left. The parabola opens to the right as the coefficient of \(y^2\) is positive.
03

Sketch the Parabola

You know that the vertex is at \((0,0)\) and the parabola opens to the right. To sketch it, plot the vertex and choose values for \(y\) to find corresponding \(x\) values. For example, if \(y = -1\), then \(x = 2(-1)^2 = 2\). If \(y = 1\), then \(x = 2(1)^2 = 2\). Plot these points \((2, -1)\) and \((2, 1)\). Draw a symmetric curve through these points and the vertex.
04

Use a Graphing Device

To accurately graph \(x = 2y^2\), input the equation into a graphing calculator or software. This will create a precise representation of the parabola, giving a visual confirmation of its shape and orientation.

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Key Concepts

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

Vertex of a Parabola
The vertex is a crucial point on a parabola. It is where the graph changes direction. For the equation \(x = 2y^2\), the vertex lies at \((0,0)\). This is because there are no additional terms that alter its position from the origin.

Understanding the vertex's location is important because it serves as a reference point when drawing or analyzing the parabola. It鈥檚 like the starting spot! Always check if the equation includes any terms that shift this point right or left, up or down. For instance, had the equation been \(x = 2(y - b)^2 + a\), the vertex would shift to \((a, b)\).
Parabola Orientation
Orientation tells us which way a parabola opens. Does it open up, down, left, or right? For a parabola in the form \(x = ay^2\), we determine orientation with the sign of \(a\).
  • If \(a > 0\), like in \(x = 2y^2\), the parabola opens to the right.
  • If \(a < 0\), it opens to the left.
For parabolas in the form \(y = ax^2\), you check the direction similarly:
  • \(a > 0\) means the parabola opens upwards.
  • \(a < 0\) means it opens downwards.
Understanding orientation helps in correctly sketching and analyzing the graph.
Graphing Calculators
Graphing calculators are handy tools for visualizing equations like parabolas. They plot the graph by processing the equation you input, showing you its shape and key points automatically.

To use a graphing calculator for \(x = 2y^2\), simply input the equation, and let the device calculate it for you. The calculator will display the parabola, including its vertex at \((0,0)\) and show the rightward orientation.

This tool is particularly useful for checking your hand-drawn sketches. It confirms your understanding and highlights any possible mistakes in the manual process, allowing you to correct and learn effectively.

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Most popular questions from this chapter

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. $$x=\cos ^{3} t, \quad y=\sin ^{3} t, \quad 0 \leq t \leq 2 \pi$$

(a)Use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola. (b) Use a rotation of axes to eliminate the \(x y\) -term. (c) Sketch the graph. $$52 x^{2}+72 x y+73 y^{2}=40 x-30 y+75$$

(a) Use the discriminant to identify the conic. (b) Confirm your answer by graphing the conic using a graphing device. $$9 x^{2}-6 x y+y^{2}+6 x-2 y=0$$

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. $$x=\sec t, \quad y=\tan ^{2} t, \quad 0 \leq t<\pi / 2$$

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. $$3 x^{2}+4 y^{2}-6 x-24 y+39=0$$
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