Question

which equation could represent the function? a. $f(x) = 0.2(1 - 0.5)^x$ b. $f(x) = 0.5(1 - 0.6)^x$ c. $f(x) = 0.3(1 - 0.4)^x$ d. $f(x) = 0.4(1 - 0.5)^x$

Explanation:

Step1: Analyze the y-intercept

The y-intercept occurs at \( x = 0 \). For an exponential function \( f(x)=a(1 - r)^x \), when \( x = 0 \), \( f(0)=a(1 - r)^0=a \). From the graph, at \( x = 0 \), the y - value is 0.4? Wait, no, looking at the graph, the first point at \( x = 0 \) (the blue dot on the y - axis) seems to be 0.4? Wait, no, let's check the options. Wait, the options are in the form \( f(x)=a(1 - r)^x \). Let's check \( x = 0 \): \( f(0)=a \). From the graph, when \( x = 0 \), the y - coordinate is 0.4 (the blue dot on the y - axis). So \( a = 0.4 \). Now check the base: \( 1 - r \). Let's check \( x = 1 \): the point at \( x = 1 \) has \( y\approx0.2 \). Let's test option D: \( f(1)=0.4(1 - 0.5)^1=0.4\times0.5 = 0.2 \), which matches. Let's check other options: Option A: \( f(1)=0.2\times0.5 = 0.1
eq0.2 \). Option B: \( f(1)=0.5\times0.4 = 0.2\)? Wait, \( 1 - 0.6 = 0.4 \), so \( f(1)=0.5\times0.4 = 0.2 \). Wait, but the y - intercept for B is 0.5, but the graph's y - intercept is 0.4. Option C: \( f(0)=0.3
eq0.4 \). Option D: \( f(0)=0.4 \), which matches the y - intercept (the blue dot at \( x = 0 \) is 0.4), and \( f(1)=0.4\times0.5 = 0.2 \), which matches the point at \( x = 1 \) (y = 0.2).

Step2: Verify with \( x = 1 \)

For option D: \( f(x)=0.4(1 - 0.5)^x=0.4(0.5)^x \). When \( x = 1 \), \( f(1)=0.4\times0.5 = 0.2 \), which matches the y - value at \( x = 1 \) (the blue dot at \( x = 1 \) has \( y = 0.2 \)). When \( x = 2 \), \( f(2)=0.4\times(0.5)^2=0.4\times0.25 = 0.1 \), which matches the point at \( x = 2 \) (y = 0.1). When \( x = 3 \), \( f(3)=0.4\times(0.5)^3=0.4\times0.125 = 0.05 \), which matches the point at \( x = 3 \) (y = 0.05). When \( x = 4 \), \( f(4)=0.4\times(0.5)^4=0.4\times0.0625 = 0.025 \), which matches the point at \( x = 4 \) (y = 0.025).

Answer:

D. \( f(x)=0.4(1 - 0.5)^x \)