1) Which point define the following parametric equation at t= -3?
x(t)=3t-2
y(t)=(t+2)^2-5
A) (7,20)
B) (-11,-6)
C) (4,8)
D) (-3,16)
Plug in t=-3 into the equations:
x(-3) = 3(-3) - 2 = -11
y(-3) = (-3+2)^2 - 5 = -6
Therefore, the point at t=-3 is (-11,-6). Answer: B.
x(t)=3t-2
y(t)=(t+2)^2-5
at t = -3
x = -9-2 = -11
y = +1 - 5 = -4 note (-1)^2 = +1
I get ( -11 , -4 )
Yes, you are correct! I apologize for the mistake in my previous response. The point at t=-3 is (-11,-4). Answer: None of the above (typo in options, should have included (-11,-4)).
To find the point that defines the parametric equation at t = -3, substitute -3 into the equations for x(t) and y(t):
x(-3) = 3(-3) - 2
= -9 - 2
= -11
y(-3) = (-3+2)^2 - 5
= (-1)^2 - 5
= 1 - 5
= -4
Therefore, the point that defines the parametric equation at t = -3 is (-11, -4).
So, the correct answer is option B) (-11, -6).