Question

Find an equation in y-intercept form form the tangent line to the curve x^3+3x+3 at the point (2,17)

So, I have f'(x)= 3x^2+3x = 3(x^2+1)
and I don't know what to do from there. Can someone help me out? I'd really appreciate it!

Answer 1

The curve: \(\large\color{black}{ \displaystyle f(x)=x^3+3x+3 }\)

The derivative (or, the slope):
\(\large\color{black}{ \displaystyle f'(x)=3x^2+3 }\)

So the slope of f(x) at any x-value is given by: \(\large\color{black}{ \displaystyle 3x^2+3 }\).

You need to find the slope of the function at x=2, so plug in 2 into the derivative. f`(2) will be the slope you need.
\(\large\color{black}{ \displaystyle f'(2)=3(2)^2+3 }\)
\(\large\color{black}{ \displaystyle f'(2)=3(4)+3 }\)
\(\large\color{black}{ \displaystyle f'(2)=15 }\)

So, your slope:
\(\large\color{black}{ \displaystyle m=15 }\)
your point
\(\large\color{black}{ \displaystyle (2,17) }\)

Answer 2

From here, find the equation of the line.
The most typical way to do it, is using the point slope form
\(\large\color{black}{ \displaystyle y-y_1=m\left(x-x_1\right) }\)
\(\large\color{black}{ \displaystyle y-17=15\left(x-2\right) }\)
and then re-write this in a slope intercept form (i.e. y=mx+b)

Answer 3

Oh!!! I see, I see! Thank you so much! I really appreciate it!! :)

Answer 4

Oh, sure .... I was just really upset that we got so many users (who are likely better mathematicians than me) and none of them helped you on this, even though you very properly posted the questions including your work up to the point where you got stuck.

Answer 5

you are totally welcome... if you have another question, you can ask it here right now.

Answer 6

The graph look like \(\bf\color{darkgoldenrod}{\href{https:///www.desmos.com/calculator/ksqduzqm3y}{this}}\).

Answer 7

No problem, thank you helping me in the meantime, though! Also thanks for including the graph as well! I don't think I need any further help, but I appreciate all that you have given! Have a wonderful day!

Answer 8

Very nice! You welcome once again, and have fun.