ANSWER
[tex]y=2x^2+8x+12;[-5,-1][/tex]
EXPLANATION
We want to find the rectangular form of the given parametric equation:
[tex]y(t)=2t^2+4[/tex]
on the interval [-3, 1] where:
[tex]x(t)=t-2[/tex]
To do this, first, write t in terms of x:
[tex]\begin{gathered} x=t-2 \\ t=x+2 \end{gathered}[/tex]
Now, substitute the expression for t into the given equation and simplify:
[tex]\begin{gathered} y(x)=2(x+2)^2+4 \\ y(x)=2(x^2+4x+4)+4 \\ y(x)=2x^2+8x+8+4 \\ y(x)=2x^2+8x+12 \end{gathered}[/tex]
That is the rectangular form.
To find the interval (x1, x2), we have to substitute the values of the boundaries of the interval (t1, t2) = [-3, 1] into the equation for x:
[tex]\begin{gathered} x_1=t_1-2=-3-2=-5 \\ x_2=t_2-2=1-2=-1 \end{gathered}[/tex]
Hence, the interval is:
[tex][-5,-1][/tex]
That is the answer.
