Question

match each exponential function to its graph.
m(x) = 3(2)^x + 4

g(x) = 2(3)^x + 4

h(x) = 3(3)^x + 2

k(x) = 4(3)^x + 2

Explanation:

Step1: Analyze Vertical Asymptote (Horizontal Shift)

For an exponential function of the form \( f(x) = a(b)^x + c \), the horizontal asymptote is \( y = c \). Let's find \( c \) for each function:

• \( m(x) = 3(2)^x + 4 \): Asymptote \( y = 4 \)
• \( g(x) = 2(3)^x + 4 \): Asymptote \( y = 4 \)
• \( h(x) = 3(3)^x + 2 \): Asymptote \( y = 2 \)
• \( k(x) = 4(3)^x + 2 \): Asymptote \( y = 2 \)

So functions with \( c = 4 \) (m, g) correspond to graphs with asymptote \( y = 4 \); functions with \( c = 2 \) (h, k) correspond to graphs with asymptote \( y = 2 \).

Step2: Analyze Initial Value (x=0)

Calculate \( f(0) \) for each function:

• \( m(0) = 3(2)^0 + 4 = 3(1) + 4 = 7 \)
• \( g(0) = 2(3)^0 + 4 = 2(1) + 4 = 6 \)
• \( h(0) = 3(3)^0 + 2 = 3(1) + 2 = 5 \)
• \( k(0) = 4(3)^0 + 2 = 4(1) + 2 = 6 \)

Step3: Match to Graphs

• Asymptote \( y = 4 \):
• \( m(0) = 7 \), \( g(0) = 6 \). So graph with asymptote \( y = 4 \) and \( y(0)=7 \) is for \( m(x) \); \( y(0)=6 \) is for \( g(x) \).
• Asymptote \( y = 2 \):
• \( h(0) = 5 \), \( k(0) = 6 \). So graph with asymptote \( y = 2 \) and \( y(0)=5 \) is for \( h(x) \); \( y(0)=6 \) is for \( k(x) \).

Now, looking at the graphs:

• Top - Left: Asymptote \( y = 4 \), \( y(0) \approx 7 \) → \( m(x) = 3(2)^x + 4 \)
• Top - Right: Asymptote \( y = 4 \), \( y(0) \approx 6 \) → \( g(x) = 2(3)^x + 4 \)
• Bottom - Left: Asymptote \( y = 2 \), \( y(0) \approx 5 \) → \( h(x) = 3(3)^x + 2 \)
• Bottom - Right: Asymptote \( y = 2 \), \( y(0) \approx 6 \) → \( k(x) = 4(3)^x + 2 \)

Answer:

Top - Left Graph: \( m(x) = 3(2)^x + 4 \)
Top - Right Graph: \( g(x) = 2(3)^x + 4 \)
Bottom - Left Graph: \( h(x) = 3(3)^x + 2 \)
Bottom - Right Graph: \( k(x) = 4(3)^x + 2 \)