1. **Problem statement.** Analyze the function $f(x)=\sqrt{8.58^2 - \frac{(x-11.1)^2}{1.24^2}}$ and describe domain, range, intercepts, and maximum; identify its relation to an ellipse. 2. **Formula and rules.** For real-valued $f(x)$ defined by a square root we require the radicand to be nonnegative, i.e. $\text{radicand} \ge 0$. 3. **Write equation with $y=f(x)$.** Let $y=f(x)$ so $y=\sqrt{8.58^2 - \frac{(x-11.1)^2}{1.24^2}}$. 4. **Square to remove root.** Square both sides to get $$y^2 = 8.58^2 - \frac{(x-11.1)^2}{1.24^2}$$ 5. **Rearrange into ellipse form.** Move terms: $$y^2 + \frac{(x-11.1)^2}{1.24^2} = 8.58^2$$ 6. **Divide by $8.58^2$ to get standard form.** $$\frac{(x-11.1)^2}{(1.24\cdot 8.58)^2} + \frac{y^2}{8.58^2} = 1$$ 7. **Identify semi-axes.** From the previous line the semi-axis in $y$ is $b=8.58$ and the semi-axis in $x$ is $a=1.24\cdot 8.58 = 10.6392$. 8. **Domain (where radicand >=0).** Solve $$\frac{(x-11.1)^2}{1.24^2} \le 8.58^2$$ which gives $$|x-11.1| \le 1.24\cdot 8.58 = 10.6392$$ and hence $$x\in [11.1-10.6392,\ 11.1+10.6392] = [0.4608,\ 21.7392].$$ 9. **Range.** Because the square root returns nonnegative values and maximum occurs at center we have $$0 \le y \le 8.58$$. 10. **x-intercepts.** Set $y=0$ to get $x = 11.1 \pm 10.6392$ so the intercepts are $(0.4608,\ 0)$ and $(21.7392,\ 0)$. 11. **y-intercept.** Evaluate $x=0$ gives radicand $8.58^2 - \frac{11.1^2}{1.24^2}$ which is negative so there is no real y-intercept. 12. **Critical point / maximum.** Differentiate $y$ using $y=\sqrt{R(x)}$ to get $$y' = \frac{- (x-11.1)}{1.24^2\, y}.$$ Setting $y'=0$ gives $x=11.1$ which yields $y=8.58$, the global maximum on the domain. 13. **Geometric interpretation.** The curve is the top half of an ellipse centered at $(11.1,0)$ with semi-axes $a=10.6392$ in $x$ and $b=8.58$ in $y$, forming the "watermelon" top. 14. **Final answer (summary).** Domain: $[0.4608,\ 21.7392]$. Range: $[0,\ 8.58]$. x-intercepts: $(0.4608,\ 0)$ and $(21.7392,\ 0)$. Maximum: $(11.1,\ 8.58)$.