# witch of Maria Agnesi

• Oct 15th 2009, 03:27 PM
miss.strw
witch of Maria Agnesi
The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5).

http://i35.tinypic.com/mma7bc.gif

y =

It looks so weird (Speechless)
• Oct 15th 2009, 03:30 PM
skeeter
Quote:

Originally Posted by miss.strw
The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5).

http://i35.tinypic.com/mma7bc.gif

slope, $m = y'(-2)$

now use the point-slope form of a linear equation ...

$y - y_1 = m(x - x_1)$ where $(x_1,y_1)$ is the given point.
• Oct 15th 2009, 03:37 PM
miss.strw
Quote:

Originally Posted by skeeter
slope, $m = y'(-2)$

now use the point-slope form of a linear equation ...

$y - y_1 = m(x - x_1)$ where $(x_1,y_1)$ is the given point.

Do i need to plug in -2 into the equation to get m? or u're telling me that m = -2? thanks for helpin! (Wink) please reply soon!

i do not know how to get the y like i don't know how to do tat.
• Oct 15th 2009, 04:03 PM
skeeter
Quote:

Originally Posted by miss.strw
Do i need to plug in -2 into the equation to get m? or u're telling me that m = -2? thanks for helpin! (Wink) please reply soon!

i do not know how to get the y like i don't know how to do tat.

the slope is the derivative of the function evaluated at x = -2

what does this mean? ...
Quote:

like i don't know how to do tat.
you don't know how to find a derivative?
• Oct 15th 2009, 04:12 PM
miss.strw
Quote:

Originally Posted by skeeter
the slope is the derivative of the function evaluated at x = -2

what does this mean? ...

you don't know how to find a derivative?

Not for those one. i don't know is it like 1/1+x^2 = 2x?

so it's like

y-(1/5)=(2x)(x-(-2)) ?

thank you so much for helping!
• Oct 15th 2009, 04:17 PM
skeeter
Quote:

Originally Posted by miss.strw
Not for those one. i don't know is it like 1/1+x^2 = 2x? no

so it's like

y-(1/5)=(2x)(x-(-2)) ? no

$y = \frac{1}{1+x^2}$

use the quotient rule to find y' , or ...

$y = (1+x^2)^{-1}$

... take the derivative using the chain rule.
• Oct 15th 2009, 04:25 PM
miss.strw
Quote:

Originally Posted by skeeter
$y = \frac{1}{1+x^2}$

use the quotient rule to find y' , or ...

$y = (1+x^2)^{-1}$

... take the derivative using the chain rule.

g(-2)h(-2)-g(-2)h(-2)
---------------------------
[h(-2)]^2

this doesn't make sense to me :(

if i do the derivative

1
-------
(1+x^2)

is derivative is x^2 = 2x
• Oct 15th 2009, 04:28 PM
skeeter
you need to find the derivative first, then evaluate it at x = -2