Question

21 mark for review for the exponential function f, the value of f(1) is k, where k is a constant. which of the following equivalent forms of the function f shows the value of k as the coefficient or the base? a f(x) = 50(1.6)^{x+1} b f(x) = 80(1.6)^{x} c f(x) = 128(1.6)^{x - 1} d f(x) = 204.8(1.6)^{x - 2}

Explanation:

Step1: Recall the form of exponential function

The general form of an exponential function is \( f(x) = a(b)^{x - h}+k \), but for our case, we need to find \( f(1) = k \), so substitute \( x = 1 \) into each function and see which one has \( k \) as coefficient or base.

Step2: Analyze Option A

For \( f(x)=50(1.6)^{x + 1} \), substitute \( x = 1 \):
\( f(1)=50(1.6)^{1 + 1}=50(1.6)^{2}=50\times2.56 = 128 \)
The coefficient here is 50, not related to \( k = 128 \) as coefficient or base in the form.

Step3: Analyze Option B

For \( f(x)=80(1.6)^{x} \), substitute \( x = 1 \):
\( f(1)=80(1.6)^{1}=80\times1.6 = 128 \)
The coefficient is 80, not equal to \( k = 128 \) as coefficient or base in the form.

Step4: Analyze Option C

For \( f(x)=128(1.6)^{x-1} \), substitute \( x = 1 \):
\( f(1)=128(1.6)^{1 - 1}=128(1.6)^{0}=128\times1 = 128 \)
Here, when \( x = 1 \), the exponent becomes 0, and the coefficient is 128, which is \( k \). So this form shows \( k = 128 \) as the coefficient.

Step5: Analyze Option D

For \( f(x)=204.8(1.6)^{x - 2} \), substitute \( x = 1 \):
\( f(1)=204.8(1.6)^{1-2}=204.8(1.6)^{-1}=\frac{204.8}{1.6}=128 \)
The coefficient is 204.8, not equal to \( k = 128 \) as coefficient or base in the form.

Answer:

C. \( f(x) = 128(1.6)^{x-1} \)