BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

15th Edition

ISBN: 9781608408405

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Cengage Learning

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Question

Chapter 2.4, Problem 33E

To determine

Interpret the meaning of the vertex in this situation.

Expert Solution

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 yof students absent from school due to the flu each day x.

BIG IDEAS MATH Algebra 2: Common Core 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.

To determine

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

Expert Solution

Answer to Problem 33E

The equation of the parabola is f(x)=−0.5(x−6)2+19.

f(10)=11.

Explanation of Solution

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

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 2.4, Problem 33E , additional homework tip 2

Calculation :

From the graph,

V(h,k)=(6,19).h=6 and k=19 so the equation of the parabola is, f(x)=a (x−h) 2 +k. f(x)=a (x−6) 2 +16.

Since parabola passes through (0,1).

a (0−6) 2 +19=1.36a+19=1.36a=−18.a=−18/39=−1/2=−0.5 f(x)=−0.5 (x−6) 2 +19. f(10)=−0.5 (10−6) 2 +19.f(10)=−0.5×16+19=−8+19=11.f(10)=11.

To determine

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

Expert Solution

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 yof students absent from school due to the flu each day x.

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 2.4, Problem 33E , additional homework tip 3

Calculation :

From the graph,

V(h,k)=(6,19).h=6 and k=19 so the equation of the parabola is, f(x)=a (x−h) 2 +k. f(x)=a (x−6) 2 +16.

Since parabola passes through (0,1).

a (0−6) 2 +19=1.36a+19=1.36a=−18.a=−18/39=−1/2=−0.5 f(x)=−0.5 (x−6) 2 +19. f(10)=−0.5 (10−6) 2 +19.f(10)=−0.5×16+19=−8+19=11.

m1=(f(6)−f(0))/(6−0)=(19−1)/6=3.m2=(f(11)−f(6))/(11−6)=(6.5−19)/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 2.4, Problem 32E

Chapter 2.4, Problem 34E

Chapter 2 Solutions

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - Prob. 51E Ch. 2.1 - Prob. 52E Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - 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