Answer:
[tex]\displaystyle{y=2\sqrt{x+4}}[/tex]
Step-by-step explanation:
We are given parametric equations of:
[tex]\displaystyle{x=t^2-4}[/tex] and [tex]\displaystyle{y=2t}[/tex]
Isolate t-variable in [tex]\displaystyle{x=t^2-4}[/tex] first by adding both sides by 4:
[tex]\displaystyle{x+4=t^2-4+4}\\\\\displaystyle{x+4=t^2}[/tex]
Square root both sides then write plus-minus:
[tex]\displaystyle{\sqrt{x+4}=\sqrt{t^2}}\\\\\displaystyle{t=\pm \sqrt{x+4}}[/tex]
Since the choice determines y-term as a function and only positive t-value exists, substitute only [tex]\displaystyle{t=\sqrt{x+4}}[/tex] in [tex]\displaystyle{y=2t}[/tex].
Therefore, the answer is [tex]\displaystyle{y=2\sqrt{x+4}}[/tex]
