Tangent Line Problem: Find Point P & Compute Slope m_L

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Homework Help Overview

The problem involves finding a point P on the curve defined by the equation C: y=8x^5+5x+1, where a line L through the origin is tangent to the curve at that point. Participants are tasked with determining the coordinates of P and computing the slope of line L.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for the derivative of the function to find the slope and mention using point-slope form. There are questions about the role of sketches and clarifications regarding the distinction between tangent and secant lines. Some participants express uncertainty about how to proceed after sketching the function.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have offered guidance on writing expressions for the slopes and equating them, while others are still questioning the initial steps and concepts involved.

Contextual Notes

Participants note constraints such as the difficulty of using LaTeX on mobile devices and the potential confusion between tangent and secant lines. There is also mention of needing to clarify the setup of the problem before proceeding.

CrossFit415
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I'm on mobile so I can't use latex.

Let C: y=8x^5+5x+1 and suppose L is a line through the origin tangent to C at a point P=(a,f(a)) on C.

-Find the coordinates of P

-Compute the slope m sub L of L

Where should I begin? I'm guessing I would need the derivative of the equation f(x)? Then I would use the point slope form to figure out point P?
 
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CrossFit415 said:
I'm on mobile so I can't use latex.

Let C: y=8x^5+5x+1 and suppose L is a line through the origin tangent to C at a point P=(a,f(a)) on C.

-Find the coordinates of P

-Compute the slope m sub L of L

Where should I begin?
Maybe a sketch of the function?
CrossFit415 said:
I'm guessing I would need the derivative of the equation f(x)?
One would think that might somehow enter into things.
CrossFit415 said:
Then I would use the point slope form to figure out point P?
 
After I sketch the eqiation what would I do after? So does that mean L is a secant line through the equation?
 
L is a line that is tangent to the graph of the function. It's not a secant line (a line that hits a curve at two points).

BTW, "a secant line through the equation" doesn't make much sense, unless you actually draw a line through an equation, as opposed to drawing a line through the graph of an equation.
 
Mark44 said:
L is a line that is tangent to the graph of the function. It's not a secant line (a line that hits a curve at two points).

BTW, "a secant line through the equation" doesn't make much sense, unless you actually draw a line through an equation, as opposed to drawing a line through the graph of an equation.

Haha I'll be more precise. I misread I thought it was going through the graph of the equation thinking its a secant line.

So I got the derivative which is f'(x) = (40x^4) + 5.

I wouldn't be able to apply the slope formula since I'm figuring out the coordinates of P. What methods are there?
 
CrossFit415 said:
suppose L is a line through the origin tangent to C at a point P=(a,f(a)) on C.

So you have a line from (0, 0) to (a, f(a)) on the graph of your curve. Write an expression that represents the slope of the line.

Write another expression that represents the slope of the tangent at any point on your curve.

Equate the two expressions.
 

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