All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)

$y = \frac{(x-2)^3}{2} - 6$

When $x = 0, y = -10$

When $y = 0, \frac{(x-2)^3}{2} = 6$

$\Rightarrow x-2 = \sqrt[3]{12}$

$\Rightarrow x = 2 + \sqrt[3]{12}$

Coordinates are $(0,-10) \text{~and~} (2 + \sqrt[3]{12}, 0)$

(ii)

Some students wasted time to find the expression $f^{-1}$, which shows they have poor knowledge. Students should label the axial intercepts coordinates, $(0,-10) \text{~and~} (2 + \sqrt[3]{12}, 0)$.