Find Equation for Tangent Line to y=e^x at Origin

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Homework Help Overview

The discussion revolves around finding the equation of a tangent line to the graph of y=e^x that passes through the origin. The subject area involves calculus, specifically the concepts of derivatives and tangent lines.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss how to compute the slope of the tangent line at any point on the graph and consider the implications of the tangent line passing through the origin. There is an exploration of the derivative of the function and its role in determining the tangent line's equation.

Discussion Status

Some participants have offered insights into the derivative and the general form of the tangent line equation. There is an ongoing exploration of what specific point on the graph would allow the tangent line to intersect the origin, with no explicit consensus reached yet.

Contextual Notes

Participants are grappling with the requirement that the tangent line must pass through the origin, which raises questions about the appropriate point of tangency on the curve y=e^x.

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Homework Statement


Find an equation for a line that is tangent to the graph of y=ex and goes through the origin.

Homework Equations


The Attempt at a Solution


y'=ex

That's about all I can think of. I don't know how to make the tangent line go through the origin. Can someone lead me in the right direction?
 
Last edited:
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iRaid said:

Homework Statement


Find an equation for a line that is tangent to the grpah of y=ex and goes through the origin.


Homework Equations





The Attempt at a Solution


y'=ex

That's about all I can think of. I don't know how to make the tangent line go through the origin. Can someone lead me in the right direction?

At any point (x,y) on the graph, how would you compute the slope of the tangent line?

RGV
 
Ray Vickson said:
At any point (x,y) on the graph, how would you compute the slope of the tangent line?

RGV

Find the derivative at that point.
 
And you have already said that the derivative is again [itex]e^x[/itex].

So the tangent line to [itex]y= e^x[/itex] at [itex]x= x_0[/itex] would be [itex]y= e^{x_0}(x- x_0)+ e^{x_0}[/itex]. Now, what must [itex]x_0[/b] be so that goes through (0, 0)?[/itex]
 
0? Not really sure..
 

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