How to find equation of tangent line?

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Homework Help Overview

The discussion revolves around finding the equation of the tangent line to the function f(x) = 3x^3 + 2 at the point x = 1. Participants explore the necessary steps and concepts related to derivatives and tangent lines.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of finding the derivative and its significance in determining the slope of the tangent line. There are questions about the correct interpretation of the tangent line's equation and the role of the y-intercept. Some participants express confusion about the requirements for the tangent line and the definitions involved.

Discussion Status

The discussion includes attempts to clarify the process of finding the tangent line, with some participants providing guidance on using derivatives. There is an ongoing exploration of the correct values and methods, with various interpretations being considered. Some participants express uncertainty about their understanding and the correctness of their answers.

Contextual Notes

One participant notes a typo in the answer choices, which may have contributed to confusion. There is also mention of a lack of formal calculus instruction for one participant, affecting their understanding of derivatives and tangent lines.

salma17
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Find the equation of the line tangent to f(x)= 3x^3 + 2 at x = 1.
a) y = 9x-4
b) y = 9x+5
c) y = 3x 2
d) y = 3x+1
e) Not enough information given.

Im confused on this one, but I am thinking about E, because it doesn't specify if the equation should be parallel or perpendicular to the function they give.

Any thoughts?
 
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salma17 said:
Find the equation of the line tangent to f(x)= 3x^3 + 2 at x = 1.
a) y = 9x-4
b) y = 9x+5
c) y = 3x 2
d) y = 3x+1
e) Not enough information given.

Im confused on this one, but I am thinking about E, because it doesn't specify if the equation should be parallel or perpendicular to the function they give.

Any thoughts?

Welcome to the PF.

Do you know how to find the tangent to a line using calculus? Have you learned that yet? If so, show us the equation for the tangent line. BTW, is there a typo in your writing of answer c)?
 
salma17 said:
Find the equation of the line tangent to f(x)= 3x^3 + 2 at x = 1.
a) y = 9x-4
b) y = 9x+5
c) y = 3x 2
d) y = 3x+1
e) Not enough information given.

Im confused on this one, but I am thinking about E, because it doesn't specify if the equation should be parallel or perpendicular to the function they give.
An equation is not parallel or perpendicular to anything. The graph of an equation might be parallel to a line or perpendicular to it.


salma17 said:
Any thoughts?
 
What IS the tangent line to your graph? Do you have the tools to find it? You should.
 
Find the derivative at that point and then plug in your value to y=mx+b.
 
Yea, that was a typo. Choice C is supposed to say 3x+2.

I solved for the derivative and got 0.
So now I have to plug in 0 for m??
 
Last edited:
salma17 said:
Yea, that was a typo. Choice C is supposed to say 3x+2.

I solved for the derivative and got 0.
So now I have to plug in 0 for m??

The derivative is not zero. How are you taking the derivative?

-Dave K
 
Never mind, forget the zero. I put 3x^3 + 2 in my calculator and got 9x^2 for the derivative.
Is this right?
 
salma17 said:
Never mind, forget the zero. I put 3x^3 + 2 in my calculator and got 9x^2 for the derivative.
Is this right?

Yes, do you know how to take this on your own? It's a very easy derivative (power rule).

-Dave K
 
  • #10
Nope, I actually don't know how to do it without a calculator.

So now, my next step would be to plug in the x value (1)?
 
  • #11
salma17 said:
Nope, I actually don't know how to do it without a calculator.

Shouldn't you have learned that? http://www.math.ucdavis.edu/~hass/Calculus/HTAC/excerpts/node19.html
So now, my next step would be to plug in the x value (1)?

Yes, it would be. Then you would have the slope of the needed line. What else would you need to find the equation of a line? (There's a hint here with the word "would" that things are about to take a turn).

-Dave K
 
  • #12
Ok so I plugged in 1 in 9x^2, giving me 9.
so then that's the slope?
and I think we use y=mx+b, correct?
so y=9x+b?..is b the y-int?
 
  • #13
I am confused. Are you not taking a Calculus course? If you are, one of the first things you should have learned is that the derivative, at a given point, is the slope of the tangent line. And you should have learned what "y intercept" means well before starting a Calculus class.
 
  • #14
Nope, I am not taking a calc course. i know what a y-intercept is though. to find the y-int, I just plug in x to solve for y. so here, I plugged in x, which is 1, in 9x^2, and ended up with 9. so then that's my y-int. so i have y=9x+9. which actually doesn't make sense to me. I am not sure where i went wrong.
 
  • #15
Alright i got it. slope is 9, the point is (1,5). I plug it in y=mx+b, getting y=9x+b. Plugging in my x and y values, I end up with y=9x-4.
finally.
 
  • #16
salma17 said:
Nope, I am not taking a calc course.

Well that would be it then. You're missing some basic definitions and stuff. Self-teaching can work, but where are you learning from? (I'd suggest khanacademy for some good explanations)

i know what a y-intercept is though. to find the y-int, I just plug in x to solve for y. so here, I plugged in x, which is 1, in 9x^2, and ended up with 9. so then that's my y-int.

Do you know what a derivative is supposed to give you?

It's the *slope* of a tangent line at a particular point. So when you take your derivative and plug in x - all you have is a slope - not a y intercept.

so i have y=9x+9. which actually doesn't make sense to me. I am not sure where i went wrong.

Notice the definition of a derivative I gave you above.. The derivative gives you the *slope* of a tangent line at a *point*. The slope is one bit of information.

-Dave K

[deleted a bit of nonsense i wrote here earlier]
 
Last edited:
  • #17
salma17 said:
Alright i got it. slope is 9, the point is (1,5). I plug it in y=mx+b, getting y=9x+b. Plugging in my x and y values, I end up with y=9x-4.
finally.

Oh, nevermind. re-read the initial post. It was not worded like I'm used to. Sorry :redface:
 
  • #18
Im very confused right now. Did I not get the correct answer? I thought I finally understood how to do it. :/
 
  • #19
salma17 said:
Im very confused right now. Did I not get the correct answer? I thought I finally understood how to do it. :/

Yes you did.
 

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