Question 1037654: x = 2t
y = t + 5, -2 ≤ t ≤ 3
Graph these two pairs of parametric equations
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Notice how x = 2t and how t is restricted to be between -2 and 3 (inclusive)
The smallest that t can be is t = -2. If t = -2, then x = 2*t = 2*(-2) = -4
So the smallest that x can be is x = -4. This is the left endpoint of the domain
The largest that t can be is t = 3. If t = 3, then x = 2*t = 2*(3) = 6
So the largest that x can be is x = 6. This is the right endpoint of the domain
The domain of this function is therefore 
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Step 1) Solve for t in the first equation
x = 2t
2t = x
2t/2 = x/2
t = x/2
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Step 2) Use substitution. Replace every copy of 't' in the second equation with x/2
y = t + 5
y = x/2 + 5 ... replace t with x/2
y = (1/2)*x + 5
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After doing the substitution, notice how the t term goes away. We're left with an equation in the form of x and y. So we can now graph it. This is simply a linear equation since it is in the form  where in this case
m = 1/2 is the slope
b = 5 is the y intercept
So start at (0,5) on the y axis. This is the first point. Move up 1, then over to the right 2 units to land on the point (2,6).
Plot the two points (0,5) and (2,6). Draw a straight line through them.
Because the domain is , we can only graph for x values between -4 and 6 (inclusive). So what you'll need to do is erase the portion of the graph to the left of x = -4. Also, erase the portion to the right of x = 6.
In short, this graph is basically a segment from (-4,3) to (6,8) as shown below
Side Note: the endpoints are closed circles to incidicate we are including the endpoints. Open endpoints would have been used if we wanted to exclude the endpoints. However we are not using open endpoints here.
Another side note: The two original equations combine to form one equation, therefore making one graph. It may be counter-intuitive at first. Do not fall into the trap and think that the two original equations will form two separate graphs.
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