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Suppose you graphed every single point of the form $(2t + 3, 3-3t)$ . 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.

Thanks! HALP:)

 Aug 22, 2018
edited by Guest  Aug 22, 2018
 #1
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Suppose you graphed every single point of the form $(2t + 3, 3-3t)$ . 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.

 

 

we have a graph defined by the locus of all points (2t+3, 3-3t)

 

so

x= 2t +3  (1a)        and         y=3-3t     (2a)

I am going to sove these simultaneously in order to eliminate t.

3x=6t+9   (1b)      and         2y=6-6t     (2b)

 

add equations    2a and 2b

 

3x+2y=15

 

This is the equation of a line.

All points on this line will satisfy the given condtion and no point that is not on this line will satisfy it.

 Aug 25, 2018

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