/*! 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 14 Two parabolic arches are being b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 14

Two parabolic arches are being built. The first arch can be modelled by the function \(y=-x^{2}+9,\) with a range of \(0 \leq y \leq 9\) The second arch must span twice the distance and be translated 6 units to the left and 3 units down. a) Sketch the graph of both arches. b) Determine the equation of the second arch.

Short Answer

Expert verified
The equation of the second arch is \(y = -\frac{1}{4}(x+6)^2 + 6\).

Step by step solution

01

- Understand the First Arch

The equation of the first arch is given by: \(y=-x^{2}+9\). This is a downward-opening parabola with its vertex at (0, 9) and it spans the y-values between 0 and 9, inclusive.
02

- Sketch the First Arch

To sketch the first arch, plot the vertex (0, 9) and recognize that the parabola is symmetric about the y-axis. It intersects the y-axis at y = 9 and the x-axis at \(x=卤3\).
03

- Transformations for the Second Arch

The second arch must span twice the distance and be translated 6 units to the left and 3 units down. Spanning twice the distance means we need to horizontally stretch the parabola by a factor of 2.
04

- Horizontal Stretch

The horizontal stretching factor modifies the equation \(y = -x^2 + 9\) to \(y = -\frac{1}{4}x^2 + 9\). This is because stretching by a factor of 2 means we replace \(x\) with \(\frac{x}{2}\) in the original equation.
05

- Translation Left and Down

Next, translate the stretched parabola 6 units to the left. This modifies the equation to: \(y = -\frac{1}{4}(x+6)^2 + 9\). Finally, translate it 3 units down, resulting in the equation: \(y = -\frac{1}{4}(x+6)^2 + 6\).
06

- Sketch the Second Arch

To sketch the second arch, start by plotting its new vertex at \((-6, 6)\) and use the transformed equation to plot points.
07

- Final Graph

Combine both sketches by placing both parabolas on the same coordinate plane to visualize their positions and shapes relative to each other.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

graphing parabolas
Graphing parabolas is a foundational skill in algebra. A parabola is a symmetric curve that can open upwards or downwards. It is defined by a quadratic function of the form \( y = ax^2 + bx + c \). The sign of the coefficient \(a\) determines the direction of the opening鈥攑ositive for upward and negative for downward. To graph parabolas effectively, it is helpful to identify key features such as the vertex, axis of symmetry, and intercepts. For the equation \(y = -x^2 + 9\), the vertex is at \( (0, 9) \), and the parabola opens downward, indicating that the highest point is at the vertex.
vertex form
The vertex form of a parabola's equation allows easy identification of its vertex and transformations. It is written as \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex. In the original problem, the first arch's equation in vertex form is \( y = -x^2 + 9 \), easily showing the vertex at \( (0, 9) \). When equations are presented in vertex form, it simplifies graphing since the vertex and orientation of the parabola are immediately clear.
transformations of functions
Transformations alter the position or shape of the graph of a function. These can be translations, reflections, stretches, or compressions. For example, translating a function left or right shifts the graph horizontally, while translating up or down affects it vertically. In the given exercise, the second arch's graph involves a horizontal stretch, a leftward shift, and a downward shift. Each transformation can be identified and applied step-by-step to arrive at the final equation.
horizontal stretch
A horizontal stretch changes the width of the parabola. Stretching by a factor \( k \) involves replacing \( x \) with \( \frac{x}{k} \) in the equation. For instance, stretching the original parabola \( y = -x^2 + 9 \) by a factor of 2 modifies it to \( y = -\frac{1}{4}x^2 + 9 \). The factor \( \frac{1}{4} \) accounts for the doubled distance between points on the x-axis.
translations in geometry
Translations in geometry move every point of a figure the same distance in a specified direction. For the second arch, the horizontal stretch is followed by translations to the left by 6 units and down by 3 units. This modifies the intermediate equation \( y = -\frac{1}{4}x^2 + 9 \) to \( y = -\frac{1}{4}(x+6)^2 + 6 \). Translating 6 units to the left shifts \( x \) to \( x+6 \), and translating 3 units downward decreases the constant term by 3.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For each function, state the values of \(h\) and \(k,\) the parameters that represent the horizontal and vertical translations applied to \(y=f(x)\). a) \(y-5=f(x)\) b) \(y=f(x)-4\) \(y=f(x+1)\) d) \(y+3=f(x-7)\) e) \(y=f(x+2)+4\)

For each function, state two ways to restrict the domain so that the inverse is a function a) \(f(x)=x^{2}+4\) b) \(f(x)=2-x^{2}\) b) \(f(x)=2-x^{2}\) c) \(f(x)=(x-3)^{2}\) d)\(f(x)=(x+2)^{2}-4\)

The key point (-12,18) is on the graph of \(y=f(x) .\) What is its image point under each transformation of the graph of \(f(x) ?\) a) \(y+6=f(x-4)\) b) \(y=4 f(3 x)\) c) \(y=-2 f(x-6)+4\) d) \(y=-2 f\left(-\frac{2}{3} x-6\right)+4\) e) \(y+3=-\frac{1}{3} f(2(x+6))\)

An object falling in a vacuum is affected only by the gravitational force. An equation that can model a free-falling object on Earth is \(d=-4.9 t^{2},\) where \(d\) is the distance travelled, in metres, and \(t\) is the time, in seconds. An object free falling on the moon can be modelled by the equation \(d=-1.6 t^{2}\) a) Sketch the graph of each function. b) Compare each function equation to the base function \(d=t^{2}\)

Gil is asked to translate the graph of \(y=|x|\) according to the equation \(y=|2 x-6|+2\) He decides to do the horizontal translation of 3 units to the right first, then the stretch about the \(y\) -axis by a factor of \(\frac{1}{2},\) and lastly the translation of 2 units up. This gives him Graph 1. To check his work, he decides to apply the horizontal stretch about the \(y\) -axis by a factor of \(\frac{1}{2}\) first, and then the horizontal translation of 6 units to the right and the vertical translation of 2 units up. This results in Graph 2. a) Explain why the two graphs are in different locations. b) How could Gil have rewritten the equation so that the order in which he did the transformations for Graph 2 resulted in the same position as Graph \(1 ?\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.