Question

Given the parametric equations, x = 2t + 5 and y = 3t^2 - 1, find the point on the graph at time, t = 4.
Please explain how you got your answer.

Answer 1

Answer: (13, 47)

The expression t = 4 means that 4 units of time have come off the clock (possibly 4 seconds). Plug this value into each equation given for x and y

x = 2*t + 5 = 2*4 + 5 = 8 + 5 = 13

y = 3*t^2 - 1 = 3*4^2 - 1 = 3*16 - 1 = 48 - 1 = 47

In short, if t = 4 then it leads to x = 13 and y = 47 at the same time. These x and y values pair up to get the final answer (13,47)

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Here's an update:

If t = 2, then,

• x = 2t+5 = 2*2+5 = 4+5 = 9
• y = 3t^2-1 = 3*2^2-1 = 3*4-1 = 12-1 = 11

In short, if t = 2, then x = 9 and y = 11

Answer 2

Parametric equations,  $x=2t+5, y=3t^{2} -1$ at t = 4

The point on the graph at time, t = 4 is (13,47).

Given:

$x=2t+5\\y=3t^{2} -1\\t=4$

The expression t = 4 means that 4 units of time have come off the clock (possibly 4 seconds). Plug this value into each equation given for x and y$x=2t+5\\x=2(4)+5=13\\y=3t^{2} -1\\y=3(4)^{2} -1=47$

In short, if t = 4 then it leads to x = 13 and y = 47 at the same time. These x and y values pair up to get the final answer (13,47).

Therefore, the point on the graph at time, t = 4 is (13,47).

For more information:

https://brainly.com/question/22341280