Answer
$$x=y^2-8y+12, 1\leq y\leq 6$$
See the attached graph.
Work Step by Step
$$x=t^2+4t$$
$$\implies (2-y)^2+4(2-y)$$
$$x=y^2-8y+12$$
Hence,
$$x=y^2-8y+12, 1\leq y\leq 6$$
See the attached graph.
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