Draw the secants of a curve (line?)

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The discussion revolves around drawing secants from the point (2, 60) on the line defined by the equation d = 30t to various points on the graph at t = 3, 2.5, 2.1, and 1.01. Participants express confusion about the concept of secants in relation to a linear function, questioning how a line can have secants since they typically apply to curves. It is suggested that the task involves drawing secants to the calculated points (3, d(3)), (2.5, d(2.5)), etc. The consensus indicates that all secant lines will overlap and have the same slope, highlighting the unique characteristics of linear functions. The discussion emphasizes the understanding of secants in the context of linear equations versus curves.
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Homework Statement


An automobile's distance is given by d = 30t (t in seconds, d in metres):

"Draw the secants from the point (2, 60) to each of the points on the graph ending at t = 3, 2.5, 2.1, 1.01"

"Determine the equations of each of the secants above."

Homework Equations


The Attempt at a Solution



To my understanding, d = 30t is a line.
According to the text, "A line that intersects a curve in two places is called a secant".
How can a line have any secants?
 
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A curve in its most liberal interpretation can also include a straight line.
 
zeion said:
"Draw the secants from the point (2, 60) to each of the points on the graph ending at t = 3, 2.5, 2.1, 1.01"

How can a point end? :confused:
 
Uh I'm not sure, but I'm guessing it's asking me to draw the secant from (2, 60) to these points (3, d(3)), (2.5, d(2.5)), etc.

Though I still don't see the point since they're all going to overlap anyway lol.
 
zeion said:
Uh I'm not sure, but I'm guessing it's asking me to draw the secant from (2, 60) to these points (3, d(3)), (2.5, d(2.5)), etc.

Though I still don't see the point since they're all going to overlap anyway lol.

Perhaps the point is to show that for a straight line, all secant lines have the same slope.
 
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