Calculus: Finding the slope and equation

Hi,
I need some help with the following:

Find the slope and the equation of the tangent line drawn to the curve y=12x / (x-3) at the point (-1,3)

I know the equation is limit of f(x+h) - f(x) /h as h approaches 0. and I know that x is -1. I keep messing up setting up the equation, I'm not quite sure what x+h would be.

Help please, I've spent way too much time on this one problem!!

Thank you

f(x+h) - f(x) /h
= (12(-1 + h) / (-1 + h - 3)) - (12(-1) / (-1 - 3)) / h
= (-12 + 12h / (-4 + h)) - (-12 / -4) / h
= (-12 + 12h / h(-4 + h)) - (3 / h)
= (-12 / h(-4 + h)) + (12h / h(-4 + h)) - (3/h)
= (-12 / h(-4 + h)) + (12 / (-4 + h)) - (3/h)
= (-12 / h(-4 + (0)) + (12 / (-4 + (0)) - (3/h)
= (-12 / -4h) + (12 / -4) - (3/h)
= (3/h) - (3/h) + (-3)
= - 3

So the slope at point (-1,3) would be -3.
Now we can simple create a line equation for a line with a slope of -3 passing through the point (-1,3)
Just use the form y = ax - b, where "a" is the slope and "b" the x intercept

y = (-3)x - b

Now we just need to solve for "b" by plugging the values for our point (-1,3)

(3) = (-3)(-1) - b
b = 3 - 3
b = 0

So,
y = -3x - 0
y = -3x


In summary:
The slope of the line at the point (-1,3) is -3 and the tangent line is y = -3x

Voila!

it is {f(x+h) -f(x)}/h. It is not f(x+h) -f(x)/h, thats just sloppy.
By convention you always divide/multiply before add/subtract: f(x+h) -f(x)/h =f(x+h) - {f(x)/h}.
y=f(x) = 12x/(x-3) so 12(x+h)/(x+h-3) - 12x/(x-3) =limit*h
expanding and x-multiplying {(12x +12h)(x-3) -12x(x+h-3)}/{(x+3)*(x+h-3)}
{12x²+12xh-36x -36h -12x²-12xh+36x}/{(x-3)*(x+h-3)}
-36h /{(x-3)*(x+h-3)}
-36h /{x²+xh -3x-3x-3h+9} = limit * h
so limit = -36/{x²+xh -6x+3h+9}
limit = -36/{x²+xh +3h-6x+9}
as h →0, xh and 3h →0 so
limit = -36/{x-3}²