Here's the problem -
400-6x^2
Find the slope of the tangent to this curve at this point (1, 394)
HELP!!
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 $\displaystyle a$.
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 =/