Big Ideas Math Algebra 2, 2014
BI
Big Ideas Math Algebra 2, 2014 View details
4. Transformations of Exponential and Logarithmic Functions
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Exercise 54 Page 324

Practice makes perfect
a Let's analyze the given function.

P=100(0.99997)^t We know that t represents the time and P represents the weight of the element. Therefore, either of them cannot be negative. Moreover, because the initial amount of plutonium-239 is 100 grams and the given function is an exponenetial decay, the range cannot be greater than 100. With this, the domain and range can be stated as below. Domain:& 0 < t Range:& 0< P < 100

b To find the amount of plutonium-239 after 12 000 years, we will substitute t=12 000 in the function. Let's do it!
P=100(0.99997)^t
P=100(0.99997)^(12 000)
P=100(0.69767...)
P=69.76725...
P≈ 69.77
Therefore, the amount of plutonium-239 is about 69.77 grams.
c Let's first write the function for 550 grams plutonium-239.

P_(100)=100(0.99997)^t P_(550)=550(0.99997)^t The new amount is 5.5 time the old amount. To write the function for 550 grams plutonium-239, the output P_(100) needs to be multiplied by a constant greater than 1. P_(550)=5.5P_(100) Therefore, the transformation is a vertical stretch by a factor of 5.5.

d Because the vertical stretch by a factor of 5.5 changes the initial amount from 100 grams to 550 grams, the maximum value of the range also changes.

0< P < 100 ⇒ 0< P < 550 Therefore, the vertical stretch affects the range but it does not affect the domain.