+0

# Help need this done before 11:00 EST. (40 min) AHHHHH

-2
151
4

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.

Aug 10, 2019

#1
+2

What happens after 11:00 EST?

Aug 10, 2019
#3
0

That’s when their homework is due ;)

Guest Aug 10, 2019
#2
+1

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.

Aug 10, 2019
#4
+1

(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   Aug 10, 2019
edited by CPhill  Aug 10, 2019