# How do you graph #y=sinx+2x#?

##### 1 Answer

I'm going to take a calculus approach to this problem.

We start by finding the first derivative.

#y' = cosx + 2#

Then there will be critical points where

#0 = cosx + 2#

#-2 = cosx#

But since

The second derivative can tell us more about concavity and points of inflection.

#y'' = -sinx#

This will equal

In the end, the graph of the function looks very much like this:

Hopefully this helps!