# How to find equation of tangent line?

1. Nov 28, 2012

### salma17

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 im thinking about E, because it doesnt specify if the equation should be parallel or perpendicular to the function they give.

Any thoughts?

2. Nov 28, 2012

### Staff: Mentor

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)?

3. Nov 28, 2012

### Staff: Mentor

An equation is not parallel or perpendicular to anything. The graph of an equation might be parallel to a line or perpendicular to it.

4. Nov 28, 2012

### Zondrina

What IS the tangent line to your graph? Do you have the tools to find it? You should.

5. Nov 28, 2012

### iRaid

Find the derivative at that point and then plug in your value to y=mx+b.

6. Nov 29, 2012

### salma17

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: Nov 29, 2012
7. Nov 29, 2012

### dkotschessaa

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

-Dave K

8. Nov 29, 2012

### salma17

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

9. Nov 29, 2012

### dkotschessaa

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

-Dave K

10. Nov 29, 2012

### salma17

Nope, I actually dont know how to do it without a calculator.

So now, my next step would be to plug in the x value (1)?

11. Nov 29, 2012

### dkotschessaa

Shouldn't you have learned that? power rule
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. Nov 29, 2012

### salma17

Ok so I plugged in 1 in 9x^2, giving me 9.
so then thats the slope?
and I think we use y=mx+b, correct?
so y=9x+b?..is b the y-int?

13. Nov 29, 2012

### HallsofIvy

Staff Emeritus
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. Nov 29, 2012

### salma17

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 thats my y-int. so i have y=9x+9. which actually doesnt make sense to me. im not sure where i went wrong.

15. Nov 29, 2012

### salma17

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. Nov 29, 2012

### dkotschessaa

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)

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.

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: Nov 29, 2012
17. Nov 29, 2012

### dkotschessaa

Oh, nevermind. re-read the initial post. It was not worded like I'm used to. Sorry

18. Nov 29, 2012

### salma17

Im very confused right now. Did I not get the correct answer? I thought I finally understood how to do it. :/

19. Nov 29, 2012

Yes you did.