Question

score on last try: 0.8 of 2 pts. see details for more. next question get a similar question you can retry this questi match each equation with a graph above \\( y = 2(1.27)^x \\) a. red (r) \\( y = 3(1.27)^x \\) b. green (g) \\( y = 2(1.48)^x \\) c. blue (b) \\( y = 2(0.71)^x \\) d. black (k) \\( y = 2(0.83)^x \\) e. orange (o)

Explanation:

Step1: Analyze exponential growth/decay

For exponential functions \( y = a(b)^x \), if \( b>1 \), it's growth; if \( 0 < b < 1 \), it's decay. Also, larger \( a \) (initial value) or larger \( b \) (growth factor) means steeper growth (or less steep decay).

Step2: Match \( y = 2(1.27)^x \)

• \( a = 2 \), \( b = 1.27>1 \) (growth). Compare with \( y = 3(1.27)^x \) (larger \( a \)) and \( y = 2(1.48)^x \) (larger \( b \)). The blue (B) graph is a growth curve with \( a = 2 \) (since \( y = 3(1.27)^x \) would be steeper than \( y = 2(1.27)^x \), and \( y = 2(1.48)^x \) steeper than both). So \( y = 2(1.27)^x \) matches blue (B)? Wait, no—wait, the green (G) or black? Wait, re - evaluate:

Wait, \( y = 2(1.27)^x \): \( a = 2 \), growth. The blue graph (B) has \( a \) around 2? Wait, maybe I messed up. Let's re - sort:

Growth functions: \( y = 2(1.27)^x \), \( y = 3(1.27)^x \), \( y = 2(1.48)^x \) (all \( b>1 \)). Decay: \( y = 2(0.71)^x \), \( y = 2(0.83)^x \) (\( 0 < b < 1 \)).

For growth, larger \( a \) (3 vs 2) or larger \( b \) (1.48 vs 1.27) makes steeper. So \( y = 3(1.27)^x \) (a = 3) is steeper than \( y = 2(1.27)^x \), and \( y = 2(1.48)^x \) (b = 1.48) is steeper than \( y = 2(1.27)^x \).

Decay: \( y = 2(0.71)^x \) (b = 0.71) decays faster than \( y = 2(0.83)^x \) (b = 0.83). The red (R) and black (K) are decay. Red (R) decays faster (steeper decay) so \( y = 2(0.71)^x \) (faster decay) matches red (R), \( y = 2(0.83)^x \) matches black (K)? Wait, no, the original problem's options:

Wait, the first equation \( y = 2(1.27)^x \): let's check the graphs. The blue (B) is a growth curve with \( a = 2 \), \( y = 3(1.27)^x \) (a = 3) would be green (G) (since it's steeper than blue), \( y = 2(1.48)^x \) (b = 1.48) is orange (O) (steeper than \( y = 2(1.27)^x \) and \( y = 3(1.27)^x \)? Wait, no, maybe I had it wrong.

Wait, the correct matches (assuming standard exponential graph matching):

• \( y = 2(1.27)^x \): blue (B) – c. blue (B)
• \( y = 3(1.27)^x \): green (G) – b. green (G) (since \( a = 3>2 \), steeper than \( y = 2(1.27)^x \))
• \( y = 2(1.48)^x \): orange (O) – e. orange (O) (since \( b = 1.48>1.27 \), steeper than \( y = 2(1.27)^x \))
• \( y = 2(0.71)^x \): red (R) – a. red (R) (faster decay, steeper)
• \( y = 2(0.83)^x \): black (K) – d. black (K) (slower decay, less steep)

Answer:

• \( y = 2(1.27)^x \): c. blue (B)
• \( y = 3(1.27)^x \): b. green (G)
• \( y = 2(1.48)^x \): e. orange (O)
• \( y = 2(0.71)^x \): a. red (R)
• \( y = 2(0.83)^x \): d. black (K)