# Slope of a graph

• 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 AM
red_dog
The slope is

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

$=\lim_{x\to 3}\frac{(x-1)(x-3)}{x-3}=\lim_{x\to 3}(x-1)=2$
• Feb 1st 2009, 01:24 AM
earboth
Quote:

Originally Posted by >_<SHY_GUY>_<
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:

$y'=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$

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