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algebra: concepts and connections - plc vertical stretches and shrinks …

Question

algebra: concepts and connections - plc vertical stretches and shrinks of exponential functions which is a shrink of an exponential growth function? $f(x)=3\left(\frac{1}{3}\
ight)^x$ $f(x)=\frac{1}{3}(3)^x$ $f(x)=3(3)^x$ $f(x)=\frac{1}{3}\left(\frac{1}{3}\
ight)^x$

Explanation:

Step1: Recall exponential form

The standard exponential function is $f(x)=ab^x$, where:

  • $a$ = vertical scale factor
  • $b>1$ = growth function, $0
  • Vertical shrink: $0<|a|<1$

Step2: Identify growth base

First, filter functions with $b>1$ (growth):

  • $f(x)=3(3)^x$: $b=3>1$ (growth)
  • $f(x)=\frac{1}{3}(3)^x$: $b=3>1$ (growth)
  • The other two have $b=\frac{1}{3}<1$ (decay, eliminate)

Step3: Check vertical shrink

For vertical shrink, need $0

  • $f(x)=3(3)^x$: $a=3>1$ (vertical stretch)
  • $f(x)=\frac{1}{3}(3)^x$: $0<\frac{1}{3}<1$ (vertical shrink)

Answer:

$\boldsymbol{f(x)=\frac{1}{3}(3)^x}$