Equation for the tangent line

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Homework Help Overview

The discussion revolves around finding the equation of the tangent line to the curve defined by the function f(x) = 2x³ - 5x - 3 at the point where x = 2. Participants are exploring the concepts of derivatives and their application in determining the slope of the tangent line.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the role of the derivative in finding the slope of the tangent line and the necessity of having a point on the curve. There are questions about how to proceed with the calculations and what specific values to use.

Discussion Status

The conversation is ongoing, with some participants providing guidance on the relationship between derivatives and tangent lines. There is a mix of understanding and confusion, as participants seek clarity on the steps involved in finding the tangent line equation.

Contextual Notes

There is a sense of urgency due to an impending assignment deadline, which may be influencing the participants' requests for help and the level of detail in their questions.

Sophia1787
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Homework Statement



Find the equation for the tangent line to the curve f(x) = 2x3 - 5x - 3 at
x = 2



Homework Equations


How do i start this? I'm really confused and have an assignment with a bunch of these questions due tomorrow by midnight... if someone could help me that would be greatly appreciated..



The Attempt at a Solution

 
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The derivative f'(x) will give you the slope of the tangent. x and f(x) will give you a point on the curve. If you have a point and a slope, you have the equation for the tangent line, right?
 
One of the very first things you should learn in Calculus is that the derivative at a point is the slope of the tangent line. And even before Calculus you should have learned how to write the equation of a line given a point and the slope.
 
so basically 6x^2-5 ?
 
do i then plug in the 2? and solve?
 
Sophia1787 said:
so basically 6x^2-5 ?
"Basically" what about 6x^2- 5?

Sophia1787 said:
do i then plug in the 2? and solve?
Plug 2 into what? Solve what? Please ask complete questions!

I said before that the derivative at a given value of x is the slope of the tangent line at that x. What is the derivative of your function? What is its value at x= 2?

Now, how do you find the equation of a line through a given point with given slope? (And what "given point" does this line go through?
 

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