What is the rectangular form of the parametric equations? x(t) = t - 3, y(t) = 2t^2 − 1, where t is on the interval [−2, 2].

Enter your answer, in standard form, by filling in the blanks.

y = , where x is on the interval [ __,__ ].

The rectangular form is;

y = 2x² + 12x + 17

The intervals are;

y = 7 and x = (-5,1)

Step-by-step explanation:

We are given that x(t) = t - 3 and that y(t) = 2t² − 1

Let's make t the subject of the formula in the first equation.

So, t = x + 3

Now, let's put this for t in the second equation to get;

y = 2(x + 3)² − 1

Expanding, we have;

y = 2(x² + 6x +9) - 1

y = 2x² + 12x + 18 - 1

y = 2x² + 12x + 17

So, the rectangular form is;

y = 2x² + 12x + 17

Since t is on the interval of (-2,2)

Thus,when t = - 2,since x(t) = t - 3

Then, x(-2) = - 2 - 3 = - 5

Also,x(2) = 2 - 3 = 1

Similarly,since y(t) = 2t² − 1, then;

y(-2) = 2(-2)² - 1 = 7

Also,y(2) = 2(2)² - 1 = 7

So the intervals of x and yare;

y = 7 and x = (-5,1)

the answer is (-4,-4)

hope this ! : d