Derivative without calculations

[kaebun]

graph the function in a square viewing window and without doing any calculations estimate the derivitive of the function at the given point by interperting it as the tangent line slope, if it exists at the point.

f(x)= 1/2(x)^2 +2x-5
at x=2

~~~~
I graphed this function but I don't understand what I'm supposed to do. I'm not even sure what a derivative is, we haven't learned about them yet, but there are some equations in my book but they don't realy mean anything to me....
Thank you for any help in advance

 

[pka]

f(x)= 1/2(x)^2 +2x-5 at x=2
I'm not even sure what a derivative is, we haven't learned about them yet, but there

Note that the point (2,1) is on the graph.
The derivative there is the slope of the tangent line to the graph there.
From the graph can you ‘picture’ a tangent touching at (2,1)?
Can you estimate the slope of such a line?
That number, the slope, is the derivative at x=2.

 

[stapel]

In this context, think of "derivative" as "slope of the line at that point". (The "tangent line" is the straight line through that point that has the same slope as the curvy line you're working with.)

Eliz.

 

[kaebun]

Thank you; that makes sense.