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Which of the following curves exhibits modified exponential decay, has a y of 110 when t is 0, and approaches the value y=30 as t goes to infinity?

a.) y = 110e^{7t} - 80

b.) y = 110 - 80e^{-5t}

c.) y = 30 + 80e^{-2t}

d.) y = 110 - 80e^{2t}

e.) y = 30 - 80e^{-4t}

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Answer:

C

Step-by-step explanation:

Let's check the first condition [y=110, t=0] for all the choices and eliminate:

A)

[tex]y = 110e^{7t} - 80\\y = 110e^{0} - 80\\y=30[/tex]

NO

B)

[tex]y = 110 - 80e^{-5t}\\y = 110 - 80e^{0}\\y=30[/tex]

NO

C)

[tex]y = 30 + 80e^{-2t}\\y = 30 + 80e^{0}\\y=110[/tex]

YES

D)

[tex]y = 110 - 80e^{2t}\\y = 110 - 80e^{0}\\y=30[/tex]

NO

E)

[tex]y = 30 - 80e^{-4t}\\y = 30 - 80e^{0}\\y=-50[/tex]

NO

Note: e^0 = 1

Only C is satisfied. But let's make sure the 2nd condition applies as well.

[tex]y = 30 + 80e^{-2t}\\y = 30 + 80e^{-\infty}\\y=30+\frac{80}{e^{\infty}}\\y>>30[/tex]

Hence, y approaches 30 as t goes to infinity. 2nd condition is satisfied as well.

Answer is C

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