Question

What is the rectangular form of the parametric equations?x(t)=t+3, y(t)=2t^2−4, where t is on the interval [−4,0].Enter your answer, in standard form, by filling in the boxes.

Answer 1

$\begin{gathered} x=t+3 \\ y=2t^2-4 \end{gathered}$

1. Solve t in the first equation:

$\begin{gathered} x=t+3 \\ t=x-3 \end{gathered}$

2. Substitute the t in the second equation by (x-3):

$\begin{gathered} y=2t^2-4 \\ y=2(x-3)^2-3 \end{gathered}$

3. Write the equation you get in previous step in standard form:

$\begin{gathered} (a-b)^2=a^2-2ab+b^2 \\ \\ y=2(x^2-2*x*3+3^2)-3 \\ y=2(x^2-6x+9)-3 \\ y=2x^2-12x+19-3 \\ \\ y=2x^2-12x+16 \end{gathered}$

To find the inverval of x use the equation x=t+3 and the interval of t

$\begin{gathered} t\lbrack-4,0\rbrack \\ x=t+3 \\ x=-4+3=-1 \\ x=0+3=3 \\ \\ x\lbrack-1,3\rbrack \end{gathered}$

Answer:

$\begin{gathered} y=2x^2-12x+16 \\ \lbrack-1,3\rbrack \end{gathered}$