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What does the graph of the parametric equations x(t)=2−t and y(t)=(t+3)^2, where t is on the interval [−4,0], look like? Drag and drop the answers to the boxes to correctly complete the statements.

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Martebi

The graph state that the initial point is (6.9), and the terminal point is (2,1). The vertex of the parabola is  (3,0). Arrows along the parabola to indicate motion right to left.

What is the parametric equations about?

Let take x = 4 - t    =t  =  4 - x       as equation  (1)

And y = (t - 1)^2       as equation (2)

Then input equation  (1)  into equation   (2)

y  = ( 4 - x - 1)^2

y = ( 3 - x)^2

y = ( x - 3)^2

So therefore, y = a ( x -3)^2  +  0

It can be written as y  = a(x - h)^2  + k      

Where, a = 1    

The vertex  known to be (h, k)  will be = (3,0).

The starting point is  if t = -2

Therefore, x  = (4 - -2)   = 6      

  y =  (-2 - 1)^2  = (-3)^2    = 9  

So x, y =  (6,9).

Note that if t  = 2

Then x = (4 - 2) =  2      

  y  =  (2 - 1)^2   = 1^2  

=  1    

Therefore, x, y =(2, 1).

Therefore, The graph of the parametric equations  state that the initial point is (6.9), and the terminal point is (2,1). The vertex of the parabola is  (3,0). Arrows along the parabola to indicate motion right to left.

See full question below

What does the graph of the parametric equations x(t)=4−t and y(t)=(t−1)^2, where t is on the interval [−2,2], look like?

The parametric equations graph as a portion of a parabola. The initial point is _____, and the terminal point is _____. The vertex of the parabola is ____. Arrows are drawn along the parabola to indicate motion ____.

Learn more about parametric equations from

https://brainly.com/question/51019

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