witch of Maria Agnesi

**miss.strw** · October 15th 2009, 03:27 PM

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).

y =

How do I do this? Please help me out

It looks so weird

**skeeter** · October 15th 2009, 03:30 PM

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).

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.

**miss.strw** · October 15th 2009, 03:37 PM

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!

please reply soon!

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

**skeeter** · October 15th 2009, 04:03 PM

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!

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? ...

like i don't know how to do tat.

you don't know how to find a derivative?

**miss.strw** · October 15th 2009, 04:12 PM

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!

**skeeter** · October 15th 2009, 04:17 PM

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.

**miss.strw** · October 15th 2009, 04:25 PM

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

**skeeter** · October 15th 2009, 04:28 PM

you need to find the derivative first, then evaluate it at x = -2

Thread: witch of Maria Agnesi