a gym offers three levels of membership based on the number of visits per month, as represented by the…

Explanation:

Step1: Define cost - function for each plan

Let $x$ be the number of visits per month. For the first plan (max 12 visits): The cost function $C_1(x)=15 + 5x$ for $0\leq x\leq12$. For the second plan (from 13 to 16 visits): The cost for the first 12 visits is $15+5\times12 = 75$. For $13\leq x\leq16$, the cost function $C_2(x)=75+4(x - 12)=75+4x-48=27 + 4x$. For the third plan (from 17 to 24 visits): The cost for the first 16 visits is $75+4\times(16 - 12)=75 + 16=91$. For $17\leq x\leq24$, the cost function $C_3(x)=91+3(x - 16)=91+3x-48=43 + 3x$.

Step2: Analyze the slope and break - points

The first function $C_1(x)$ has a slope of 5 and starts at the y - intercept of 15. At $x = 12$, $C_1(12)=15+5\times12=75$. The second function $C_2(x)$ has a slope of 4 and starts at the point $(13,C_2(13))=(13,27 + 4\times13)=(13,79)$. The third function $C_3(x)$ has a slope of 3 and starts at the point $(17,C_3(17))=(17,43+3\times17)=(17,94)$. The graph will have a steeper slope for the first 12 visits, then a less - steep slope from 13 to 16 visits, and an even less - steep slope from 17 to 24 visits.

We need to visually inspect the given graphs to find the one that has these characteristics of changing slopes at $x = 12$ and $x = 16$. Without the actual options for the graphs shown completely, we can't give a specific graph name, but the correct graph will have a line with slope 5 for $0\leq x\leq12$, then a line with slope 4 for $13\leq x\leq16$ and a line with slope 3 for $17\leq x\leq24$.

If we assume we are choosing from typical piece - wise linear graph options: We look for a graph that has a y - intercept of 15, then changes its slope at $x = 12$ and then again at $x = 16$.

Answer:

The graph with a y - intercept of 15, slope of 5 for $0\leq x\leq12$, slope of 4 for $13\leq x\leq16$ and slope of 3 for $17\leq x\leq24$.