Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015
Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015
1st Edition
ISBN: 9781680331165
Author: HOUGHTON MIFFLIN HARCOURT
Publisher: Houghton Mifflin Harcourt
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Question
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Chapter 2.4, Problem 33E

a.

To determine

Interpret the meaning of the vertex in this situation.

a.

Expert Solution
Check Mark

Answer to Problem 33E

The vertex is (6,19) shows that the maximum number of students absent from school due to flu is 19 and this happens in day 6.

Explanation of Solution

Given information: The graph shows the number y of students absent from school due to the flu each day x.

  Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015, Chapter 2.4, Problem 33E , additional homework tip  1

Calculation :

From the graph,

The vertex is (6,19) shows that the maximum number of students absent from school due to flu is 19 and this happens in day 6.

b.

To determine

Write an equation for the parabola to predict the number of students absent on day 10.

b.

Expert Solution
Check Mark

Answer to Problem 33E

The equation of the parabola is f(x)=0.5(x6)2+19.

  f(10)=11.

Explanation of Solution

Given information: The graph shows the number y of students absent from school due to the flu each day x.

  Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015, Chapter 2.4, Problem 33E , additional homework tip  2

Calculation :

From the graph,

  V(h,k)=(6,19).h=6andk=19sotheequationoftheparabolais, f(x)=a (xh) 2 +k. f(x)=a (x6) 2 +16.

Since parabola passes through (0,1).

   a (06) 2 +19=1.36a+19=1.36a=18.a=18/39=1/2=0.5 f(x)=0.5 (x6) 2 +19. f(10)=0.5 (106) 2 +19.f(10)=0.5×16+19=8+19=11.f(10)=11.

c.

To determine

Compare the average rates of change in the students with the flu from 0 to 6 days and 6 to 11 days.

c.

Expert Solution
Check Mark

Answer to Problem 33E

The rate of students getting flu per day was 3 between day 0 and day 6, while from day 6 to day 11 decreased by 2.5 per day.

Explanation of Solution

Given information: The graph shows the number y of students absent from school due to the flu each day x.

  Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015, Chapter 2.4, Problem 33E , additional homework tip  3

Calculation :

From the graph,

  V(h,k)=(6,19).h=6andk=19sotheequationoftheparabolais, f(x)=a (xh) 2 +k. f(x)=a (x6) 2 +16.

Since parabola passes through (0,1).

   a (06) 2 +19=1.36a+19=1.36a=18.a=18/39=1/2=0.5 f(x)=0.5 (x6) 2 +19. f(10)=0.5 (106) 2 +19.f(10)=0.5×16+19=8+19=11.

  m1=(f(6)f(0))/(60)=(191)/6=3.m2=(f(11)f(6))/(116)=(6.519)/5=2.5.

So,

The rate of students getting flu per day was 3 between day 0 and day 6, while from day 6 to day 11 decreased by 2.5 per day.

Chapter 2 Solutions

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

Ch. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.1 - Prob. 30ECh. 2.1 - Prob. 31ECh. 2.1 - Prob. 32ECh. 2.1 - Prob. 33ECh. 2.1 - Prob. 34ECh. 2.1 - Prob. 35ECh. 2.1 - Prob. 36ECh. 2.1 - Prob. 37ECh. 2.1 - Prob. 38ECh. 2.1 - Prob. 39ECh. 2.1 - Prob. 40ECh. 2.1 - Prob. 41ECh. 2.1 - Prob. 42ECh. 2.1 - Prob. 43ECh. 2.1 - Prob. 44ECh. 2.1 - Prob. 45ECh. 2.1 - Prob. 46ECh. 2.1 - Prob. 47ECh. 2.1 - Prob. 48ECh. 2.1 - Prob. 49ECh. 2.1 - Prob. 50ECh. 2.1 - Prob. 51ECh. 2.1 - Prob. 52ECh. 2.2 - Prob. 1ECh. 2.2 - Prob. 2ECh. 2.2 - Prob. 3ECh. 2.2 - Prob. 4ECh. 2.2 - Prob. 5ECh. 2.2 - Prob. 6ECh. 2.2 - Prob. 7ECh. 2.2 - Prob. 8ECh. 2.2 - Prob. 9ECh. 2.2 - Prob. 10ECh. 2.2 - Prob. 11ECh. 2.2 - Prob. 12ECh. 2.2 - Prob. 13ECh. 2.2 - Prob. 14ECh. 2.2 - Prob. 15ECh. 2.2 - Prob. 16ECh. 2.2 - Prob. 17ECh. 2.2 - Prob. 18ECh. 2.2 - Prob. 19ECh. 2.2 - Prob. 20ECh. 2.2 - Prob. 21ECh. 2.2 - Prob. 22ECh. 2.2 - Prob. 23ECh. 2.2 - Prob. 24ECh. 2.2 - Prob. 25ECh. 2.2 - Prob. 26ECh. 2.2 - Prob. 27ECh. 2.2 - Prob. 28ECh. 2.2 - Prob. 29ECh. 2.2 - Prob. 30ECh. 2.2 - Prob. 31ECh. 2.2 - Prob. 32ECh. 2.2 - Prob. 33ECh. 2.2 - Prob. 34ECh. 2.2 - Prob. 35ECh. 2.2 - Prob. 36ECh. 2.2 - Prob. 37ECh. 2.2 - Prob. 38ECh. 2.2 - Prob. 39ECh. 2.2 - Prob. 40ECh. 2.2 - Prob. 41ECh. 2.2 - Prob. 42ECh. 2.2 - Prob. 43ECh. 2.2 - Prob. 44ECh. 2.2 - Prob. 45ECh. 2.2 - Prob. 46ECh. 2.2 - Prob. 47ECh. 2.2 - Prob. 48ECh. 2.2 - Prob. 49ECh. 2.2 - Prob. 50ECh. 2.2 - Prob. 51ECh. 2.2 - Prob. 52ECh. 2.2 - Prob. 53ECh. 2.2 - Prob. 54ECh. 2.2 - Prob. 55ECh. 2.2 - Prob. 56ECh. 2.2 - Prob. 57ECh. 2.2 - Prob. 58ECh. 2.2 - Prob. 59ECh. 2.2 - Prob. 60ECh. 2.2 - Prob. 61ECh. 2.2 - Prob. 62ECh. 2.2 - Prob. 63ECh. 2.2 - Prob. 64ECh. 2.2 - Prob. 65ECh. 2.2 - Prob. 66ECh. 2.2 - Prob. 67ECh. 2.2 - Prob. 68ECh. 2.2 - Prob. 69ECh. 2.2 - Prob. 70ECh. 2.2 - Prob. 71ECh. 2.2 - Prob. 72ECh. 2.2 - Prob. 73ECh. 2.2 - Prob. 74ECh. 2.2 - Prob. 75ECh. 2.2 - Prob. 76ECh. 2.2 - Prob. 77ECh. 2.2 - Prob. 78ECh. 2.2 - Prob. 79ECh. 2.2 - Prob. 80ECh. 2.2 - Prob. 81ECh. 2.2 - Prob. 82ECh. 2.2 - Prob. 83ECh. 2.2 - Prob. 84ECh. 2.2 - Prob. 85ECh. 2.2 - Prob. 86ECh. 2.2 - Prob. 87ECh. 2.2 - Prob. 88ECh. 2.2 - Prob. 1QCh. 2.2 - Prob. 2QCh. 2.2 - Prob. 3QCh. 2.2 - Prob. 4QCh. 2.2 - Prob. 5QCh. 2.2 - Prob. 6QCh. 2.2 - Prob. 7QCh. 2.2 - Prob. 8QCh. 2.2 - Prob. 9QCh. 2.2 - Prob. 10QCh. 2.2 - Prob. 11QCh. 2.2 - Prob. 12QCh. 2.2 - Prob. 13QCh. 2.2 - Prob. 14QCh. 2.3 - Prob. 1ECh. 2.3 - Prob. 2ECh. 2.3 - Prob. 3ECh. 2.3 - Prob. 4ECh. 2.3 - Prob. 5ECh. 2.3 - Prob. 6ECh. 2.3 - Prob. 7ECh. 2.3 - Prob. 8ECh. 2.3 - Prob. 9ECh. 2.3 - Prob. 10ECh. 2.3 - Prob. 11ECh. 2.3 - Prob. 12ECh. 2.3 - Prob. 13ECh. 2.3 - Prob. 14ECh. 2.3 - Prob. 15ECh. 2.3 - Prob. 16ECh. 2.3 - Prob. 17ECh. 2.3 - Prob. 18ECh. 2.3 - Prob. 19ECh. 2.3 - Prob. 20ECh. 2.3 - Prob. 21ECh. 2.3 - Prob. 22ECh. 2.3 - Prob. 23ECh. 2.3 - Prob. 24ECh. 2.3 - Prob. 25ECh. 2.3 - Prob. 26ECh. 2.3 - Prob. 27ECh. 2.3 - Prob. 28ECh. 2.3 - Prob. 29ECh. 2.3 - Prob. 30ECh. 2.3 - Prob. 31ECh. 2.3 - Prob. 32ECh. 2.3 - Prob. 33ECh. 2.3 - Prob. 34ECh. 2.3 - Prob. 35ECh. 2.3 - Prob. 36ECh. 2.3 - Prob. 37ECh. 2.3 - Prob. 38ECh. 2.3 - Prob. 39ECh. 2.3 - Prob. 40ECh. 2.3 - Prob. 41ECh. 2.3 - Prob. 42ECh. 2.3 - Prob. 43ECh. 2.3 - Prob. 44ECh. 2.3 - Prob. 45ECh. 2.3 - Prob. 46ECh. 2.3 - Prob. 47ECh. 2.3 - Prob. 48ECh. 2.3 - Prob. 49ECh. 2.3 - Prob. 50ECh. 2.3 - Prob. 51ECh. 2.3 - Prob. 52ECh. 2.3 - Prob. 53ECh. 2.3 - Prob. 54ECh. 2.3 - Prob. 55ECh. 2.3 - Prob. 56ECh. 2.3 - Prob. 57ECh. 2.3 - Prob. 58ECh. 2.3 - Prob. 59ECh. 2.3 - Prob. 60ECh. 2.4 - Prob. 1ECh. 2.4 - Prob. 2ECh. 2.4 - Prob. 3ECh. 2.4 - Prob. 4ECh. 2.4 - Prob. 5ECh. 2.4 - Prob. 6ECh. 2.4 - Prob. 7ECh. 2.4 - Prob. 8ECh. 2.4 - Prob. 9ECh. 2.4 - Prob. 10ECh. 2.4 - Prob. 11ECh. 2.4 - Prob. 12ECh. 2.4 - Prob. 13ECh. 2.4 - Prob. 14ECh. 2.4 - Prob. 15ECh. 2.4 - Prob. 16ECh. 2.4 - Prob. 17ECh. 2.4 - Prob. 18ECh. 2.4 - Prob. 19ECh. 2.4 - Prob. 20ECh. 2.4 - Prob. 21ECh. 2.4 - Prob. 22ECh. 2.4 - Prob. 23ECh. 2.4 - Prob. 24ECh. 2.4 - Prob. 25ECh. 2.4 - Prob. 26ECh. 2.4 - Prob. 27ECh. 2.4 - Prob. 28ECh. 2.4 - Prob. 29ECh. 2.4 - Prob. 30ECh. 2.4 - Prob. 31ECh. 2.4 - Prob. 32ECh. 2.4 - Prob. 33ECh. 2.4 - Prob. 34ECh. 2.4 - Prob. 35ECh. 2.4 - Prob. 36ECh. 2.4 - Prob. 37ECh. 2.4 - Prob. 38ECh. 2.4 - Prob. 39ECh. 2.4 - Prob. 40ECh. 2.4 - Prob. 41ECh. 2 - Prob. 1CRCh. 2 - Prob. 2CRCh. 2 - Prob. 3CRCh. 2 - Prob. 4CRCh. 2 - Prob. 5CRCh. 2 - Prob. 6CRCh. 2 - Prob. 7CRCh. 2 - Prob. 8CRCh. 2 - Prob. 9CRCh. 2 - Prob. 10CRCh. 2 - Prob. 11CRCh. 2 - Prob. 12CRCh. 2 - Prob. 13CRCh. 2 - Prob. 14CRCh. 2 - Prob. 15CRCh. 2 - Prob. 16CRCh. 2 - Prob. 1CTCh. 2 - Prob. 2CTCh. 2 - Prob. 3CTCh. 2 - Prob. 4CTCh. 2 - Prob. 5CTCh. 2 - Prob. 6CTCh. 2 - Prob. 7CTCh. 2 - Prob. 8CTCh. 2 - Prob. 9CTCh. 2 - Prob. 10CTCh. 2 - Prob. 11CTCh. 2 - Prob. 1CACh. 2 - Prob. 2CACh. 2 - Prob. 3CACh. 2 - Prob. 4CACh. 2 - Prob. 5CACh. 2 - Prob. 6CACh. 2 - Prob. 7CACh. 2 - Prob. 8CACh. 2 - Prob. 9CA
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