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Chapter 7: Problem 60

Each equation defines a parabola. Without actually graphing, match the equation in Column I with its description in Column II. A. Vertex \((2,-4) ;\) opens downward B. Vertex \((2,-4) ;\) opens upward C. Vertex \((4,-2) ;\) opens downward D. Vertex \((4,-2) ;\) opens upward E. Vertex \((-2,4) ;\) opens left F. Vertex \((-2,4)\); opens right G. Vertex \((-4,2) ;\) opens left H. Vertex \((-4,2) ;\) opens right $$y=-(x-2)^{2}-4$$

Short Answer

Expert verified
A: Vertex \((2,-4)\); opens downward

Step by step solution

01

Identify the Parabola Form

Recognize that the given equation \( y = -(x-2)^{2}-4 \) is in the vertex form \( y = a(x-h)^{2} + k \). This form shows that the parabola opens upwards if \( a > 0 \) and downwards if \( a < 0 \), with the vertex at \((h, k)\).
02

Extract Vertex and Direction

From the equation \( y = -(x-2)^{2}-4 \), identify \( a = -1 \), \( h = 2 \), and \( k = -4 \). The vertex is therefore \((2, -4)\), and since \( a = -1 \) is negative, the parabola opens downward.
03

Match with Description

Compare the extracted vertex \((2, -4)\) and its upward/downward opening (downward in this case) to the appropriate description. The matching description in Column II is the one that states: "Vertex \((2, -4)\); opens downward", which corresponds to option A.

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Key Concepts

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

Vertex Form
The vertex form of a parabolic equation is key to understanding its characteristics without graphing. It is expressed as \( y = a(x-h)^2 + k \). This form reveals critical aspects of the parabola:
  • \( a \) determines the direction the parabola opens. If \( a > 0 \), it opens upward, and if \( a < 0 \), it opens downward.
  • \( (h, k) \) is the vertex of the parabola. The vertex is the peak or the lowest point, depending on the direction of opening.
By rewriting any quadratic equation in this form, you can easily extract both the vertex and understand the opening direction of the parabola. This allows for quick analysis and matching with theoretical descriptions.
Equation Matching
Matching equations to their descriptions involves identifying components like the vertex and direction of the parabola's opening. Each piece of information in the vertex form points to a specific description:
  • The vertex \((h, k)\) is derived directly from the equation in vertex form.
  • The coefficient \( a \) denotes whether the parabola opens upwards or downwards.
In our exercise, the given equation was \( y = -(x-2)^2 - 4 \). From this, we extract that \( h = 2 \), \( k = -4 \), and \( a = -1 \). Therefore, our matching criteria will lead us to the description "Vertex \((2, -4)\); opens downward", which is a quick process once the vertex form is properly understood.
Graphing Parabolas
Graphing parabolas can be greatly simplified by understanding the vertex form. Each aspect of the equation gives visual cues:
  • \( (h, k) \) is plotted directly on the coordinate grid as the vertex.
  • If \( a > 0 \), the arms of the parabola rise upward from the vertex. If \( a < 0 \), the arms fall downward.
  • The value of \( a \) affects the width of the parabola; larger absolute values of \( a \) narrow the parabola, while smaller values widen it.
These observations allow you to sketch the shape and orientation of the parabola promptly, providing a visual understanding to complement the algebraic interpretation.
Vertex Identification
Identifying the vertex from an equation in vertex form is a straightforward process. The vertex \((h, k)\) is found at the transformation components of the equation \( y = a(x-h)^2 + k \). Specifically:
  • \( h \) is found by observing the shift along the x-axis. If the equation is \( y = a(x-2)^2 + k \), then \( h = 2 \).
  • \( k \) is simply the constant at the end, representing the vertical shift.
Understanding how to extract the vertex provides a clear image of the parabola's highest or lowest point, depending on \( a \). This core concept ensures you can determine essential features of any quadratic function presented in vertex form.

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Most popular questions from this chapter

Find an equation for each hyperbola. \(y\) -intercept \((0,-2)\); center at origin; passing through \((2,3)\)

Path of an Object on a Planet When an object moves under the influence of a gravitational force (without air resistance), its path can be parabolic. This is the path of a ball thrown near the surface of a planet or other celestial object. Suppose two balls are simultaneously thrown upward at a \(45^{\circ}\) angle on two different planets. If their initial velocities are both \(30 \mathrm{mph}\), then their \(x y\) -coordinates in feet can be expressed by the equation $$ y=x-\frac{g}{1922} x^{2} $$ where \(g\) is the acceleration due to gravity. The value of \(g\) will vary with the mass and size of the planet. (Source: Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Saunders College Publishers.) (a) On Earth, \(g=32.2\) and on Mars, \(g=12.6 .\) Find the two equations, and use the same screen of a graphing calculator to graph the paths of the two balls thrown on Earth and Mars. Use the window [0,180] by \([0,120] .\) (Hint: If possible, set the mode on your graphing calculator to simultaneous.) (b) Determine the difference in the horizontal distances traveled by the two balls.

Particle When an alpha particle (a subatomic particle) is moving in a horizontal path along the positive \(x\) -axis and passes between charged plates, it is deflected in a parabolic path. If the plate is charged with 2000 volts and is 0.4 meter long, then an alpha particle's path can be described by the equation \(y=-\frac{k}{2 v_{0}} x^{2}\) where \(k=5 \times 10^{-9}\) is constant and \(v_{0}\) is the initial velocity of the particle. If \(v_{0}=10^{7}\) meters per second, what is the deflection of the alpha particle's path in the \(y\) -direction when \(x=0.4\) meter? (Source: Semat, H. and J. Albright, Introduction to Atomic and Nuclear Physics, Holt, Rinehart and Winston.)

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