Question

Which point(s) are on the graph of the following parametric equations?
x(t)=2t
y(t)=t^2
A.) (-4,4)
B.) (-2,-4)
C.) (0,0)
D.) (1,2)
E.) (6,9)
There should be 3 of these that are right

Answer 1

x(t)=2t
y(t)=t^2
A) (-4,4),
x(t)=2t=-4, t=-2
y(-2)=(-2)²=4 , True

B) (-2,-4)
x(t)=2t=-2, t=-1
y(-1)=(-1)²=1 False. we have y=-4 

C) (0,0)
x(t)=2t=0, t=0
y(0)=(0)²=0   True 

D)(1,2)
x(t)=2t=1, t=1/2
y(1/2)=(1/2)²=1/4 False. we have y=2

E) (6,9)
(-2,-4)
x(t)=2t=6, t=3
y(3)=(3)²=9 True.     

Correct answers is A, C, E.

Answer 2

The points that belong to the graph of the given parametric equations are:

A, C, and E.

Which points are on the graph of the parametric equations?

Here the parametric equations are:

x(t) = 2*t

y(t) = t^2

The points on the graph must be of the form:

(2t, t^2)

Now, let's see if we can find values of t such that we get the points shown in the options.

A) (-4, 4)

We must find a value of t such that:

2t = -4

t^2 = 4

The value of t = -2 meets this condition, so (-4, 4) belongs to the graph.

B) (-2,-4)

Notice that:

t^2 = -4

Has no real solutions (the square of a number is always positive), so this point does not belong to the graph.

C) (0, 0)

Here we can just take t = 0.

2*0 = 0

0^2 = 0

Then this point belongs to the graph.

D) (1, 2)

We have the equations:

2*t = 1

t^2= 2

The solution of the first one is:

t = 1/2

Replacing this in the other equation we get:

(1/2)^2 = 1/4 ≠ 2

So there is no value of t that meets the two conditions, then this point does not belong to the graph.

E) (6, 9)

We must have:

2*t = 6

t^2 = 9

If we take t = 3 we have:

2*3 = 6

3^2 = 9

Then this point belongs to the graph.

Concluding, the correct options are A, C, and E.

If you want to learn more about parametric equations, you can read:

https://brainly.com/question/12695467