For the set of parametric equations
x = 2t+1 and y = t^2-5, name the initial point on the graph on the interval -5<t<5.
a. (-9, -30)
b. ( -5, 5)
c. (11, 20)
d. (-9, 20) (I chose this one)
I would have chosen that one too
To find the initial point on the graph of the parametric equations x = 2t+1 and y = t^2-5, we need to substitute the given interval values into the equations and then select the point that corresponds to the initial value of t.
Let's start by substituting t = -5 into the equations:
x = 2(-5)+1 = -10+1 = -9
y = (-5)^2-5 = 25-5 = 20
So the point corresponding to t = -5 is (-9, 20).
Now, let's substitute t = 5 into the equations:
x = 2(5)+1 = 10+1 = 11
y = (5)^2-5 = 25-5 = 20
So the point corresponding to t = 5 is (11, 20).
Since we are looking for the initial point, we need to select the point that corresponds to the smallest value of t. In this case, t = -5, and the corresponding point is (-9, 20).
Therefore, the correct answer is d. (-9, 20).