1. **State the problem:** We are given a linear relationship between the total cost $y$ and the number of months $x$ for operating a café. The cost function is $y = 20000 + 6000x$. 2. **Express $y$ as a function of $x$:** The total cost $y$ includes a fixed startup cost of 20000 and a monthly operating cost of 6000 per month. Thus, the function is: $$y = 20000 + 6000x$$ 3. **Graph the function for $0 \leq x \leq 15$:** The graph is a straight line starting at $y=20000$ when $x=0$ and increasing by 6000 for each additional month. This linear function can be plotted with $x$ on the horizontal axis and $y$ on the vertical axis. 4. **Find a possible value of $k$ for break-even between 10 and 15 months:** Break-even means total income equals total cost: $$kx = 20000 + 6000x$$ Solve for $k$: $$kx = 20000 + 6000x$$ $$k = \frac{20000 + 6000x}{x}$$ Simplify by canceling $x$ in the denominator and numerator where possible: $$k = \frac{20000}{x} + 6000$$ Evaluate $k$ at $x=10$: $$k = \frac{20000}{10} + 6000 = 2000 + 6000 = 8000$$ Evaluate $k$ at $x=15$: $$k = \frac{20000}{15} + 6000 \approx 1333.33 + 6000 = 7333.33$$ Since $k$ must be constant and break-even occurs between 10 and 15 months, a possible value of $k$ is any number between 7333.33 and 8000. **Final answers:** - (b) $y = 20000 + 6000x$ - (c) Graph is a line starting at $(0,20000)$ with slope 6000 for $0 \leq x \leq 15$ - (d) Possible $k$ values for break-even between 10 and 15 months: $7333.33 \leq k \leq 8000$