1. ## Derivative question

Hello everyone, seeing as I'm not a natural English speaker, I'll try my best to express my problem in English terms, fingers crossed .

The image of the function:
$f(x)=x^3-4x^2+x-1$
is a line 'l'.

a) write an equation of the straight line that is tangent to 'l' in the abcissa point 2.

b)determine the points of line 'l' in which the tangent is parallel to the straight line of the following equation:
$1+4x+1=0$

I hope this sounds ok'ish to you guys, I'm not very familiar with English maths terms.

Anyway, a) was easy enough, I reached the conclusion that the answer is -3.

b) is where I don't know where to turn, I'm still learning this basic maths stuff (I've had no high school maths education before reaching college), so there are still a few topics that I need to explore.

Would anyone be kind of enough to help me out here step by step?

If it helps, I have the final answer for b): $(1, -3)$ and $(\frac53, -\frac{57}{27})$

edit:
Apologies if this is in the wrong section.

2. Originally Posted by Cudaz
Hello everyone, seeing as I'm not a natural English speaker, I'll try my best to express my problem in English terms, fingers crossed .

The image of the function:
$f(x)=x^3-4x^2+x-1$
is a line 'l'.
"curve l" would be better-the word "line" implies straight.

a) write an equation of the straight line that is tangent to 'l' in the abcissa point 2.

b)determine the points of line 'l' in which the tangent is parallel to the straight line of the following equation:
$1+4x+1=0$

I hope this sounds ok'ish to you guys, I'm not very familiar with English maths terms.

Anyway, a) was easy enough, I reached the conclusion that the answer is -3.
The problem asks for an "equation of a the straight line". "-3" is not an equation! The equation of the tangent line will be y= f'(2)(x- 2)+ f(2).

b) is where I don't know where to turn, I'm still learning this basic maths stuff (I've had no high school maths education before reaching college), so there are still a few topics that I need to explore.
Should one of those "1"s be a "y"? What you have written would be a vertical line- and this graph never has vertical tangent line. If it was y+ 4x+ 1= 0 (or 1+ 4x+ y= 0), then the straight line has slope -4. Then tangent to the curve will be parallel to that if f'(x)= -4.

Would anyone be kind of enough to help me out here step by step?

If it helps, I have the final answer for b): $(1, -3)$ and $(\frac53, -\frac{57}{27})$

edit:
Apologies if this is in the wrong section.
By the way, your English is excellent- far better than my [Put the language of your choice here]!

3. Hey HallsofIvy, you are totally right, the right expression is y+ 4x+ 1= 0

What I got from a) (following the derivative rules) was -3x-1, this answer was confirmed on the textbook answers, is this wrong then?

Could you maybe explain to me b) in more detail please?