Solution
Given the following parametric equation
[tex]x(t)=t-3,y(t)=2t^2-1[/tex]
To find the rectangular form of the given parametric equations, make x the subject as shown below
[tex]\begin{gathered} x=t-3 \\ t=x+3 \end{gathered}[/tex]
Substitute for t into y(t),
[tex]\begin{gathered} y=2(x+3)^2-1 \\ y=2(x(x+3)+3(x+3))-1 \\ y=2(x^2+3x+3x+9)-1 \\ y=2(x^2+6x+9)-1 \\ y=2x^2+12x+18-1 \\ y=2x^2+12x+17 \end{gathered}[/tex]
For the interval of x, where t = -2
[tex]x=-2-3=-5[/tex]
Where t = 2
[tex]x=2-3=-1[/tex]
Hence, the answer is
[tex]y=2x^2+12x+17\text{ where x ix on the interval }(-5,-1)[/tex]
