Draw the secants of a curve (line?)

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Homework Help Overview

The problem involves determining secant lines for the linear function d = 30t, specifically drawing secants from the point (2, 60) to various points on the graph at t = 3, 2.5, 2.1, and 1.01. The context centers around understanding the nature of secants in relation to a straight line.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the definition of a secant and question how a line can have secants. There is also confusion regarding the phrasing of the problem, particularly the concept of a point "ending."

Discussion Status

Some participants are attempting to interpret the problem and clarify the meaning of secants in the context of a linear function. There is an exploration of the implications of drawing secants from a point on a straight line, with some suggesting that all secant lines would have the same slope.

Contextual Notes

Participants note the potential overlap of secant lines due to the linear nature of the function, raising questions about the purpose of the exercise.

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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?
 
Last edited:
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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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