This is a calculus problem in disguise. I'm guessing you're in Precalculus or an introductory calculus class. I'll try to make the definition as painless as possible.
The last time you probably saw the phrase rate of change in mathematics was back in Algebra 1 (or an equivalent) where you were discussing the concept of "slope" with respect to linear equations. In the same vein, this problem is asking for you to find an "average rate of change" by doing the exact same thing as what you did back when you were in Algebra 1.
If you recall, the function for rate of change was:
m = (y2 - y1) / (x2 - x1)
Where you had two points, (x1, y1) and (x2, y2). In this case they're asking for the same exact thing, except our points have now become (x, f(x)) and (x+h, f(x+h))
So with those substitutions in mind, we get the following slope equation:
m = (f(x+h) - f(x)) / (x+h - x)
I'm going to simplify this now on the bottom of this fraction before I continue the problem further:
m = (f(x+h) - f(x)) / (h)
We now do a substitution of x+h into f(x) and x into f(x)
f(x+h) = 2(x+h)^2 + 1 and f(x) = 2(x)^2 + 1
m = (2(x+h)^2 + 1 - (2(x)^2 + 1)) / h
I'm going to drop the 1s in this case, they cancel out
m = (2(x+h)^2 - 2(x^2)) / h
m = (2 * (x^2 + 2xh + h^2) - 2(x^2)) / h
m = (2 * (x^2 + 2xh + h^2 - x^2)) / h
m = (2 * (2xh + h^2)) / h (by the way, here's where you stop if it's not guaranteed that h is not 0)
m = 4x + h
That's your answer. As to why this delves into calculus, here's the long and short of it:
Finding the instantaneous rate of change of a curve is done by performing the algebra as seen above to whatever function you're given, then reducing h to 0 using a limit. Obviously mathematics doesn't like you dividing by zero, but taking the limit as you approach a value of 0 is approaching infitesmal quantities of "h", which allows us to drop it. In the above example, the overall derivative would be dy/dx = 4x.
This is not the last time you will see the limit definition of a derivative, they use this quite often until they eventually feel confident in your algebra abilities to give you the shortcuts.
