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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 ____.

 

 May 15, 2020
 #1
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x = 4 - t   ⇒   t  =  4 - x        (1)

y = (t - 1)^2       (2)

 

Put  (1)  into (2)

 

y  = ( 4 - x - 1)^2

y = ( 3 - x)^2

y = ( x - 3)^2

 

y = a ( x -3)^2  +  0

 

In the  form   y  = a(x - h)^2  + k       

 

a = 1     and the  vertex  is  (h,k)  = (3,0)

 

The initial point is  when t = -2

 

x  = (4 - -2)   = 6         y =  (-2 - 1)^2  = (-3)^2    = 9  ⇒   (6,9)

 

When t  = 2

 

x = (4 - 2) =  2          y  =  (2 - 1)^2   = 1^2   =  1    ⇒  (2, 1)

 

Since  (6.9)  is to  the right of (2,1)   the motion is  right to left

 

 

cool cool cool

 May 15, 2020

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