consider the following parametric equations:\n\n$x = \\frac{t}{-3}$ and $y = t^{2}-3$\n\nstep 2 of 2…

Explanation:

Step1: Eliminate the parameter (t)

From (x = \frac{t}{-3}), we can solve for (t) as (t=-3x). Substitute (t = - 3x) into the equation (y=t^{2}-3). Then (y=(-3x)^{2}-3), which simplifies to (y = 9x^{2}-3).

Step2: Find the domain

For the function (y = 9x^{2}-3), since (x) can take any real - valued input (there are no restrictions such as division by zero or taking the square root of a negative number for the polynomial (y = 9x^{2}-3)), the domain is all real numbers. In interval notation, the domain is ((-\infty,\infty)).

Step3: Find the range

The function (y = 9x^{2}-3) is a quadratic function of the form (y = ax^{2}+bx + c) with (a = 9), (b = 0), and (c=-3). The vertex of a quadratic function (y = ax^{2}+bx + c) is at (x=-\frac{b}{2a}). Here, (x = 0). When (x = 0), (y=-3). Since (a=9>0), the parabola opens upwards. So the minimum value of (y) occurs at (x = 0). The range of the function (y = 9x^{2}-3) is ([-3,\infty)) (because (y) can be any value greater than or equal to (-3)).

Answer:

Domain: ((-\infty,\infty)) Range: ([-3,\infty))