Question bank / Nonlinear functions

Math · Advanced MathMedium
xx f(x)f(x)
0 640
1 480
2 360
3 270

The table shows four values of xx and their corresponding values of f(x)f(x) for the function ff. Which equation could define ff?

A

f(x)=640−160xf(x) = 640 - 160x

B

f(x)=640(0.25)xf(x) = 640(0.25)^x

C

f(x)=640(0.75)xf(x) = 640(0.75)^x

D

f(x)=480(0.75)xf(x) = 480(0.75)^x

Show answer and explanation

Answer: C

Each value of f(x)f(x) is 0.750.75 times the one before it: 480640=360480=270360=0.75\frac{480}{640} = \frac{360}{480} = \frac{270}{360} = 0.75. So ff is exponential with base 0.750.75, and f(0)=640f(0) = 640 is the starting value. Thus f(x)=640(0.75)xf(x) = 640(0.75)^x.

Choice A fits the first two rows, but it gives f(2)=320f(2) = 320, not 360360.

Why the other choices are wrong

  • A. This fits the first two rows, but it gives f(2) = 320, not 360. The values drop by different amounts each time.
  • B. 0.25 is the fraction lost each step, not the fraction kept. Each value is 0.75 times the one before.
  • D. 480 is f(1). The starting value, f(0), is 640.
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