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(i)
First, translate $f(x)$ 3 units in the positive x-direction.
Then scale $f(x)$ by a factor of $\frac{1}{4}$ units parallel to the y-axis.

(ii)

(iii)

Explanation:

$f(x) = \frac{1}{4}(x-3)^2 , 1 \textless x \le 3$

$1 + f(\frac{x}{2}) = 1 + \frac{1}{4}(\frac{x}{2}-3)^2, 1 \textless \frac{x}{2} \le 3$

$1 + f(\frac{x}{2}) = 1 + \frac{1}{4}(\frac{x}{2}-3)^2, 2 \textless x \le 6$