1. **State the problem:** We are given data points for $x$ and $y$ and asked to draw the graph of $y$ against $x$ for $0 \leq x \leq 15$ (part c), and to find a possible value of $k$ such that Alex breaks even within 10 to 15 months (part d). 2. **Recall the function from part (b):** From the data points $(1,3)$, $(5,5)$, and $(10,10)$, we find the function $y$ in terms of $x$. 3. **Find the function $y=f(x)$:** Calculate the rate of change (slope) between points: Between $(1,3)$ and $(5,5)$: $$m=\frac{5-3}{5-1}=\frac{2}{4}=0.5$$ Between $(5,5)$ and $(10,10)$: $$m=\frac{10-5}{10-5}=\frac{5}{5}=1$$ Since slopes differ, the data is not perfectly linear, but for simplicity, assume a linear function passing through $(0, y_0)$ and $(10,10)$. Using points $(0,y_0)$ and $(10,10)$: Assuming $y_0$ is the initial cost at $x=0$, from the problem setup, initial cost is $20,000$ (equipment and renovation), but the table shows $y$ values as 3,5,10 which likely represent thousands or a simplified scale. Given the problem context, the total cost function is: $$y = 20000 + 6000x$$ 4. **Part (c) Draw the graph:** Plot the linear function: $$y = 20000 + 6000x$$ for $x$ from 0 to 15. 5. **Part (d) Find $k$ for break-even:** Break-even means total income equals total cost. Total cost after $x$ months: $$y = 20000 + 6000x$$ Total income after $x$ months: $$kx$$ Set income equal to cost: $$kx = 20000 + 6000x$$ Rearranged: $$kx - 6000x = 20000$$ $$x(k - 6000) = 20000$$ $$k - 6000 = \frac{20000}{x}$$ $$k = 6000 + \frac{20000}{x}$$ Since Alex wants to break even between 10 and 15 months, choose $x=10$: $$k = 6000 + \frac{20000}{10} = 6000 + 2000 = 8000$$ Or $x=15$: $$k = 6000 + \frac{20000}{15} \approx 6000 + 1333.33 = 7333.33$$ So a possible value of $k$ is any number between approximately 7333.33 and 8000. **Final answers:** - (c) Graph of $y = 20000 + 6000x$ for $0 \leq x \leq 15$. - (d) Possible value of $k$ is between 7333.33 and 8000.