# [SOLVED] Slopes of Tangents and Secants

• Feb 12th 2009, 03:53 PM
Dickson
[SOLVED] Slopes of Tangents and Secants
Find the slope of the tangent line of the parabola 2x - x^2 at point (2,0):

So I have to use the forumla

m = lim f(a + h) - f(a) / h
h->0

So we have to find f(a) which = 0.

Then put in that to f(a + h) which = 2(0 + h) - (0+h)^2

So then you get for the equation:

(2h - 0) / h

Which would then equal 2, but the answer is -2...can anyone help?
• Feb 12th 2009, 04:00 PM
Dickson
Sorry it'd be (2h - h^2) on top wouldn't it? But that still doesn't work
• Feb 12th 2009, 04:08 PM
skeeter
Quote:

Originally Posted by Dickson
Find the slope of the tangent line of the parabola 2x - x^2 at point (2,0):

So I have to use the forumla

m = lim f(a + h) - f(a) / h
h->0

So we have to find f(a) which = 0.

Then put in that to f(a + h) which = 2(0 + h) - (0+h)^2

So then you get for the equation:

(2h - 0) / h

Which would then equal 2, but the answer is -2...can anyone help?

$\displaystyle f(x) = 2x - x^2$

$\displaystyle f(2+h) = 2(2+h) - (2+h)^2$

$\displaystyle f(2) = 0$

$\displaystyle \lim_{x \to 2} \frac{f(2+h) - f(2)}{h}$

$\displaystyle \lim_{x \to 2} \frac{2(2+h) - (2+h)^2 - 0}{h}$

$\displaystyle \lim_{x \to 2} \frac{4+2h -(4 + 4h + h^2)}{h}$

$\displaystyle \lim_{x \to 2} \frac{4+2h -4 - 4h - h^2}{h}$

$\displaystyle \lim_{x \to 2} \frac{-2h - h^2}{h}$

$\displaystyle \lim_{x \to 2} \frac{h(-2 - h)}{h}$

$\displaystyle \lim_{x \to 2} (-2 - h) = -2$
• Feb 12th 2009, 04:14 PM
Dickson
I thought you were supposed to use the value you got from f(a) and put that in for the A value in f(a + h)
• Feb 12th 2009, 04:25 PM
skeeter
Quote:

Originally Posted by Dickson
I thought you were supposed to use the value you got from f(a) and put that in for the A value in f(a + h)

no, it's the same "a" ... a = 2.
• Feb 12th 2009, 04:34 PM
Dickson
Quote:

Originally Posted by skeeter
no, it's the same "a" ... a = 2.

Okay thanks a lot