Posted by **Vanessa** on Thursday, July 10, 2008 at 12:08pm.

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.

- Calculus -
**Reiny**, Thursday, July 10, 2008 at 10:56pm
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 (4.41-4)/(2.1-2)

= 4.1

Now you do the rest in the same way.

Do you notice what is happening?

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