Calculus
posted by Vanessa .
For the function ƒ(x) = x2 find the slope of secants from the point (2, 4) to each of the following points.
(3, __ )
(2.5 , __ )
(2.1 , __ )
(2.01, __ )
(2.001, __ )
Using the pattern from question 1, what do you think the slope of a tangent at the point (2, 4) would be? Explain you answer.

I will do the third point for you
x = 2.1 so f(2.1) = (2.1)^2 = 4.41
so slope of line between (2,4) and (2.1,4.41) is (4.414)/(2.12)
= 4.1
Now you do the rest in the same way.
Do you notice what is happening?
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