**QUESTION**

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The height of the first 5 seconds that a drone, released $2 \mathrm{~m}$ above the ground, as it goes up in the air is modelled by the parabola: 2 metres, plus the time squared (in seconds)

e) Graph the relation and its inverse on the same grid (use different colours, label axes, title your graph, use graph paper, etc.) f) Determine what would be the rule of the relation that models the situation if the drone was released from the ground and: - Describe the transformation that occurs from the original model. - How does this transformation affect the domain and range of the relation? - How does this transformation affect the domain and range of the relation's inverse?

Question is as follows I've used this solution but the e and f parts are missing in it please upload the solution for this too

Here's the answer i've referred to:

Solution: Let the height be denoted by ' $h$ ', and time be denoted by ' $t$ '. a) Variables: $h \& t$ b) According to question, the rule which swits the model is as follows: \[ h=2+t^{2} \] c) As time cannot be negative, $t \geqslant 0$. \[ \text { Thus, Aomain }=\{t: t \geqslant 0\} \& t \leqslant 5\} \] \[ \begin{array}{l} \text { At } t=0, h=2+0^{2}=2 \\ \text { At } t=5, h=2+5^{2}=27 \\ \text { Range }=\{h: 2 \leqslant h \leqslant 27\} \\ \text { Domain }=\{t: 0 \leqslant t \leqslant 5\} \end{array} \] d) \[ \begin{array}{l} \hbar=2+t^{2} \\ t^{2}=h-2 \\ t= \pm \sqrt{h-2} \end{array} \] But, $t$ cannot be negative. \[ \therefore t=\sqrt{h-2} \Rightarrow \text { Inverse: } h=\sqrt{t-2} \] Domain $=\{t: 0 \leq t \leq 2 t \geqslant 2\}$ Range $=\{h: h \geqslant 0\}$