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

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- Feb 1st 2009, 01:01 AM>_<SHY_GUY>_<Slope of a graph
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 - Feb 1st 2009, 01:18 AMred_dog
The slope is

$\displaystyle g'(3)=\lim_{x\to 3}\frac{g(x)-g(3)}{x-3}=\lim_{x\to 3}\frac{x^2-4x+3}{x-3}=$

$\displaystyle =\lim_{x\to 3}\frac{(x-1)(x-3)}{x-3}=\lim_{x\to 3}(x-1)=2$ - Feb 1st 2009, 01:24 AMearboth
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:

$\displaystyle \lim_{h\to 0}\dfrac{((3+h)^2-4(3+h))-(3^2-4\cdot 3)}{h} =\lim_{h\to 0} \dfrac{9+6h+h^2-12-4h-9+12}{h}=$ $\displaystyle \lim_{h\to 0}\dfrac{2h+h^2}{h}=\lim_{h\to 0}(2+h) = 2$