Secant/tangent line..help me! I don't even know how to start

[Guest]

(1 pt) The point P(16, 6) lies on the curve y=x^1/2 + 2. If Q is the point (x1, x1^1/2 +2), find the slope of the secant line PQ for the following values of x.

If x1=16.1, the slope of is:
and if x1=16.01, the slope of PQ is:
and if x1=15.9, the slope of PQ is:
and if x1=15.99, the slope of PQ is:

Based on the above results, guess the slope of the tangent line to the curve at P(16, 6).



PLEASE! Any help would be greatly appreciated

 

[Gene]

Start with the formula
(f(b)-f(a))/(b-a)
a=16,b=16.1

 

[Guest]

Thank you for replying...

so I would use the function... y = x^1/2 + 2???

and the formula gives me the slope correct?

 

[Gene]

Correct. The slope of the secant.
---------------
Gene

 

[Guest]

I tried those answers and they are incorrect.... any ideas?

 

[Gene]

My first idea is show your work. Hint: they should all be near .125

 

[stapel]

I tried those answers and they are incorrect.

Tried which answers? (You were provided with methods, not solutions.) On what basis are you saying that "they" are incorrect?

Please reply with specifics, including all the steps in your work.

Thank you.

Eliz.

 

[Guest]

Yes, I computed all of the answers and they all come out to be around .125.

This question is from an internet homework assignment... and it tells me if my answers are correct or not, and it is telling me that they are incorrect... all of them.

Here is my work:

f(b) - f(a) / b-a

[(16.1^1/2) + 2 - (6)] / 16.1 - 16 = .125

[(16.01^1/2) + 2 -(6)] / 16.01 - 16 = .120

etc. etc.

 

[Gene]

The method is right. My guess is that you are rounding too short. Try .12498 for the first. If It still won't take it go all the way with .1249804749 but I would think 5 places would be enough. Didn't they tell you how many places they want?

 

[Guest]

Yes, you were correct... i had my calculator set to too little of decimal places! Thank you for all your help! At least now I definitely understand the concept!