Exponential growth and decay

Question

Cobalt-57 is used in a medical testing and has a half-like fed of about 272 days, which means after 272 days, half of the cobalt-57 has decayed to less radioactive and more stable form. The function for the half-life of an isotope is A(t)=A0(1/2)^t/h. A0 is the initial amount of the isotope, h is the half-life, t is the input or time elapsed, and A(t) is the output or amount of isotope remaining as a function of time.

1.What type of function does this seem to represent? Explain your reasoning.

2.lets say you have 500 grams of cobalt-57. Substitute the half-life and initial amount valued into the function. Enter the new function.

3.if the initial amount of cobalt-57 were to double, what type of transformation would that be from the function in question 2? Explain your reasoning and enter the new function.

4.if the half-life, h, were to double, what type of transformation would that be from the function is question 2? Explain your reasoning and enter the new function.

5.rewrite the half-life function from question 2 using the rules of exponents to make the exponent negative and the base of the function greater than 1. Enter your new function.

Answer 1

A(t) = A₀(1/2)^t/h

1) Exponential decay because the base (1/2) is less than one

2) A(t) = 500(1/2)^t/272

3) Doubling means the base would be 2 instead of 1/2: A(t) = 500(2)^t/272. Since 2 = (1/2)^-1, you could also write it as A(t) = 500((1/2)^-1)^t/272 = 500(1/2)^-t/272. This would be a reflection of the original function across the y-axis

4) A = 500(1/2)^t/544. This would be a horizontal stretch by a factor of 2

5) A(t) = 500(1/2)^t/272 = 500(2^-1)^t/272 = 500(2)^-t/272