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

Chapter 3, Problem 14CT

To determine

To calculate: The maximum height and distance of the horseshoe.

Expert Solution & Answer

Answer to Problem 14CT

The required distance is 35.615 feet.

Explanation of Solution

Given information:

The model of the tosses: y=−0.01x2+0.3x+2

Calculation:

The given model of tosses is y=−0.01x2+0.3x+2 .

Rewrite the above equation as:

y=−0.01x2+0.3x+2y=−0.01(x2−30x−200)

Simplify the expression x2−30x−200 by using perfect square.

y=−0.01(x2−30x−200)y=−0.01(x2+30x+225−425)y=−0.01[(x+15)2−425]y=−0.01(x+15)2+4.25

The above equation is vertex form.

So, the distance to the vertex is 15 and the maximum height is 4.25.

To find the maximum distance, solve the expression x2−30x−200=0 .

Use the quadratic formula x=−b±b2−4ac2a .

Substitute 1 for a , −30 for b and −200 for c in the above formula and simplify for x .

x=−(−30)±(−30)2−4(1)(−200)2(1)=30±900+8002=30±17002=30±41.232

Solve the above expression for x .

x=30+41.232=71.232=35.165

And, x=30−41.232=−11.232=−5.615

The distance cannot be negative.

Hence, the required distance is 35.615 feet.

Chapter 3, Problem 13CT

Chapter 3, Problem 1CA

Chapter 3 Solutions

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

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