Slope of a graph

**>_<SHY_GUY>_<** · February 1st 2009, 01:01 AM

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

**red_dog** · February 1st 2009, 01:18 AM

The slope is

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

$=\lim_{x\to 3}\frac{(x-1)(x-3)}{x-3}=\lim_{x\to 3}(x-1)=2$

**earboth** · February 1st 2009, 01:24 AM

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:

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

With your example you have:

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

Expand the brackets, factor out h in the numerator, cancel h, and then calculate the value if h approaches zero:

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

Thread: Slope of a graph

LinkBack

Thread Tools

Display

Slope of a graph

Similar Math Help Forum Discussions

For what x value does the tangent of the graph of f(x) = xe^2x have a slope = 0?

slope at a point on a graph

Help with slope of sin graph?

Finding the slope of this graph

Slope/Graph--Algebra ?'s

Search Tags