Derivatives: Finding Slopes and Tangent Lines

[Dustobusto]

Homework Statement

Let f(x) = 2x² -3x -5. Show that the slope of the secant line through (2, f(2)) and (2+h, f(2+h)) is 2h + 5. Then use this formula to compute the slope of :

(a) The secant line through (2, f(2)), and (3, f(3))
(b) The tangent line at x = 2 (by taking a limit)

Homework Equations

too many to count

The Attempt at a Solution

Ok, so the first part I can do. I do f(x) - f(a) over x - a. In this case, for the numerator you plug 2+h into all of the x's in the f(x) formula given, and 2 into a, and then the denominator is x - a or in this case, (2+h) - 2 which = h. Factor it out, simplify, you get 2h + 5.

My question is, once I have this, is part (a) asking me to solve using 2h + 5? I'm not sure how I would go about this.

And part b, if its also asking me to solve using 2h + 5, then my GUESS would be:

the limit of (2h + 5) as x approaches 2 = 9 ,because you would just plug 2 into h.

Edit: Okay so..

P = (a, f(a) and
Q = (x, f(x)

so then for (2, f(2)), and (3, f(3)) a would be 2 and x would be 3. In the book it says that h = x - a.

So if we plug that into 2h + 5 we get 2(3-2) + 5 and that = 7. Likewise, if we just plug 3 and 2 into the formula f(x)-f(a) over x - a, we get 7/1. So that answer must be right. I think I'm on to something here...

 

[rock.freak667]

f'(a) is defined as the limit as h→0 of [f(a+h)-f(a)]/h


So far you have calculated [f(2+h)-f(2)]/h = 2h+5

so as h→0, what does 2h+5 go to?

 

[Dustobusto]

I see.

So f'(a) = lim of 2h+5 as h → 0 would be 5.

i.e. the derivative is 5. right?

 

[Dick]

I see.

So f'(a) = lim of 2h+5 as h → 0 would be 5.

i.e. the derivative is 5. right?

Yes, the derivative is 5.