Here's the problem -

400-6x^2

Find the slope of the tangent to this curve at this point (1, 394)

HELP!!

Printable View

- Oct 30th 2008, 01:14 PMlava1980help with a slope of the tangent to a curve
Here's the problem -

400-6x^2

Find the slope of the tangent to this curve at this point (1, 394)

HELP!! - Oct 30th 2008, 01:33 PMskeeter
how are you supposed to find the slope? ... straight power rule derivative, or derivative from the limit of a difference quotient?

- Oct 30th 2008, 01:35 PMVaritron
The question is how to find the equation of the line tangent to a specified curve, $\displaystyle f(x)$, at a specified point, $\displaystyle (a, f(a))$. The answer is quick and easy:

$\displaystyle

y \ = \ f(a) \ + \ f'(a)(x-a),

$

where $\displaystyle f'(a)$ is the derivative of $\displaystyle f(x)$ evaluated at $\displaystyle a$. - Oct 30th 2008, 01:36 PMlava1980
so the derivative of 400-6x^2 is -12x

then where do I go? - Oct 30th 2008, 01:43 PMVaritron
- Oct 30th 2008, 01:48 PMlava1980
- Oct 30th 2008, 01:51 PMVaritron
- Oct 30th 2008, 01:54 PMlava1980
- Oct 30th 2008, 02:05 PMVaritron
- Oct 30th 2008, 02:11 PMlava1980
- Oct 30th 2008, 02:44 PMKhaali91
i think what he is trying to say is...

remember the derivative is going to be the slope...

if the slope is -12 then m=-12 and u have (x,y)-> (1, 394)

so solve for b in y=mx+b

394=-12(1)+b

then u have the equation for the line tangent to the point

correct me if im wrong, i just learned this last week =/ - Oct 30th 2008, 02:48 PMVaritron
- Oct 30th 2008, 02:57 PMlava1980
- Oct 30th 2008, 03:01 PMKhaali91