consider the following parametric equations:\n\nx = \\sqrt{t - 2} and y = 3t + 1\n\nstep 2 of 2: determine…

Explanation:

Step1: Eliminate the parameter (t)

From (x = \sqrt{t - 2}), we can square both sides to get (x^{2}=t - 2), then (t=x^{2}+ 2). Substitute (t=x^{2}+2) into (y = 3t+1). So (y=3(x^{2}+2)+1=3x^{2}+6 + 1=3x^{2}+7). Since (x=\sqrt{t - 2}), and the square - root function (x=\sqrt{t - 2}\geq0) (the domain of (x) in terms of the parametric equation: (t-2\geq0) gives (t\geq2), and (x=\sqrt{t - 2}\geq0)).

Step2: Find the domain

For the function (y = 3x^{2}+7) obtained after eliminating the parameter, considering the non - negativity of (x) from the parametric form (because (x=\sqrt{t - 2})), the domain of (x) is (x\geq0). In interval notation, the domain is ([0,\infty)).

Step3: Find the range

The function (y = 3x^{2}+7) is a quadratic function. The coefficient of (x^{2}) is (a = 3>0), so the parabola opens upwards. The vertex form of a quadratic function (y=ax^{2}+bx + c) is (y=a(x - h)^{2}+k) (for (y = 3x^{2}+7), (h = 0) and (k = 7)). When (x = 0), (y=3\times0^{2}+7=7). As (x) increases (since (x\geq0)), (y) also increases. So the range of (y) is ([7,\infty)).

Answer:

Domain: ([0,\infty)), Range: ([7,\infty))