Suppose you graphed every single point of the for (2t+3,3-3t) . For example, when t=2, we have 2t+3=7 and 3-3t=-3 , so (7,-3) is on the graph. Explain why the graph is a line, and find an equation whose graph is this line.
In order for this graph to be a line, we need to verify two things: first, that all the points on the graph are on the proposed line and second, that all points on the proposed line are on the graph. Be sure to deal with both.
What happens after 11:00 EST?
The graph is a line because both of those functions to plot points are linear. So every point being plotted would generate a graph that is a line.
An equation where the graph is a line is anything on the first order, so y=x or y=2x or y =3x. Those are all linear functions.
(2t + 3, 3 - 3t) = ( x, y)
This implies that
2t + 3 = x subtract 3 from both sides
2t = x - 3 divide both sides by 2
t = (x - 3) / 2 (1)
And we also have that
y = 3 - 3t substituting (1) in for t, we get that
y =3 - 3 ( x - 3) / 2 simplify
y= 3 - (3/2)(x - 3)
y= 3 - (3/2) x + 9/2
y= 6/2 - (3/2)x + 9/2
y= -(3/2)x + 15/2
Thus...we have a line with a slope of -3/2 and a y intercept of 15/2
Note that when x = 7 y = -(3/2)(7) + 15/2 = -21/2 + 15/2 = -6/2 = -3
And we can verify the y intercept.....when x = 0 ...... 2t = -3 ⇒ t = -3/2
Plugging this into 3 - 3(-3/2) = 3 + 9/2 = 6/2 + 9/2 = 15/2
Here is a graph of the parameterized line and the linear equation : https://www.desmos.com/calculator/ruklcsfe62
Note that the graphs coincide