i tried it, but i got the answer of 6, not 2.
g(x) = x^2-4x , (3,-3)
use the limit process to find the slope of the graph of the function at the specified point.
i think my math is totally off
If you have a functiony = f(x) the slope at x = a is calculated by:
$\displaystyle y'=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$
With your example you have:
$\displaystyle g'(3)=\lim_{h\to 0}\dfrac{((3+h)^2-4(3+h))-(3^2-4\cdot 3)}{h}$
Expand the brackets, factor out h in the numerator, cancel h, and then calculate the value if h approaches zero:
$\displaystyle \lim_{h\to 0}\dfrac{((3+h)^2-4(3+h))-(3^2-4\cdot 3)}{h} =\lim_{h\to 0} \dfrac{9+6h+h^2-12-4h-9+12}{h}=$ $\displaystyle \lim_{h\to 0}\dfrac{2h+h^2}{h}=\lim_{h\to 0}(2+h) = 2$