/*! 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 42 Sketch the graph of each equatio... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the graph of each equation. If the graph is a parabola, find irs vertex. If the graph is a circle, find its center and radius. $$x=-2(y+5)^{2}$$

Short Answer

Expert verified
The graph is a parabola; vertex: (0, -5), opens left.

Step by step solution

01

Identify the Equation Type

The given equation is \( x = -2(y+5)^2 \). It is in the form \( x = a(y-k)^2 + h \), which represents a parabola. Parabolas that open horizontally rather than vertically are of this form.
02

Find the Vertex of the Parabola

For the parabola in the form \( x = a(y-k)^2 + h \), the vertex is located at \((h, k)\). In the equation \( x = -2(y+5)^2 \), comparing to the standard form gives \( h = 0 \) and \( k = -5 \). Therefore, the vertex of this parabola is at \((0, -5)\).
03

Determine the Orientation and Open Direction

Since the coefficient of \((y+5)^2\) is negative (\(-2\)), the parabola opens horizontally to the left.
04

Sketch the Graph

Begin by plotting the vertex at \((0, -5)\). Since the parabola opens to the left, draw the parabola curving leftward from the vertex, indicating a wide, symmetric U-shape.

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Key Concepts

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

Parabola Vertex
The vertex of a parabola is a crucial point that gives us valuable insight into its graph. It can be seen as the peak or the lowest point of the parabola, depending on its orientation. To find the vertex of a parabola represented in the form \( x = a(y-k)^2 + h \), we can use the coordinates \( (h, k) \). This formula is quite different from the more common vertical parabola.
In our example, the equation \( x = -2(y+5)^2 \) is already in this vertex form. By comparing it with the standard form \( x = a(y-k)^2 + h \), we can plainly observe that \( h = 0 \) and \( k = -5 \). Hence, the vertex is found at the point \( (0, -5) \).
Understanding the location of the vertex is essential, as it serves as a reference point for sketching the entire parabola. It also aids greatly in understanding its behavior and orientation, which we'll discuss next.
Horizontal Parabola
Unlike the traditional vertical parabolas, horizontal parabolas open sideways. This means that instead of opening upwards or downwards, the curve extends to the left or the right. The distinctive form for horizontal parabolas is \( x = a(y - k)^2 + h \).
This style of parabola is less common, but understanding its characteristics is crucial. In this form, if \( a \), the coefficient of \((y-k)^2\), is positive, the parabola opens to the right. If \( a \) is negative, such as in our equation \( x = -2(y+5)^2 \), it opens to the left.
This is why identifying the form of the equation is such a vital step. It tells us immediately what direction the parabola will open without having to graph it first. Recognizing and understanding these details make graphing parabolas a quicker and more logical process.
Orientation of Parabolas
The orientation of a parabola indicates its direction and overall shape on a coordinate plane. This orientation is determined directly by the coefficient \( a \) in the parabola's equation.
For horizontal parabolas, the coefficient of \((y-k)^2\) defines the direction:
  • If \( a \) is positive, the parabola opens to the right.
  • If \( a \) is negative, it opens to the left.
In our specific example of \( x = -2(y+5)^2 \), the parabola opens to the left because \( -2 \) is negative.
The orientation not only affects the graph's direction but also its symmetry and spread. Understanding whether a parabola opens left or right helps in predicting and verifying its graphical representation, making it more intuitive to sketch accurately.

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Most popular questions from this chapter

Recall that in business, a demand function expresses the quantity of a commodity demanded as a function of the commodity's unit price. \(A\) supply function expresses the quantity of a commodity supplied as a function of the commodity's unit price. When the quantity produced and supplied is equal to the quantity demanded, then we have what is called market equilibrium. (Graph can't copy) The demand function for a certain style of picture frame is given by the function $$ p=-2 x^{2}+90 $$ and the corresponding supply function is given by $$ p=9 x+34 $$ where \(p\) is in dollars and \(x\) is in thousands of units. Find the equilibrium quantity and the corresponding price by solving the system consisting of the two given equations.

Cindy Brown, an architect, is drawing plans on grid paper for a circular pool with a fountain in the middle. The paper is marked off in centimeters, and each centimeter represents 1 foot. On the paper, the diameter of the "pool" is 20 centimeters, and "fountain" is the point \((0,0)\) a. Sketch the architect's drawing. Be sure to label the axes. b. Write an equation that describes the circular pool. c. Cindy plans to place a circle of lights around the fountain such that each light is 5 feet from the fountain. Write an equation for the circle of lights and sketch the circle on your drawing.

Rationalize each denominator and simplify, if possible. See Section 10.5 $$ \frac{10}{\sqrt{5}} $$

If you are given a list of equations of circles and parabolas and none are in standard form, explain how you would determine which is an equation of a circle and which is an equation of a parabola. Explain also how you would distinguish the upward or downward parabolas from the left-opening or right- opening parabolas.

The graph of each equation is a circle. Find the center and the radius, and then sketch. See Examples \(5,6,\) and \(8 .\) $$ (x-3)^{2}+y^{2}=9 $$

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