Finding the slope of the tangent line to the parabola I have to find the slope of the tangent line to the parabola
$y=4x - x^2$ at the point (1,3),
using defintion one which is 
$$\frac{f(x)-f(a)}{x-a}.$$
To me this means $\frac{(4x-x^2)-3}{x-1}$ or alternatively a 1 instead of the 3. Neither of these give me the correct answer and I am not sure how to approach problems that do not involve point 1,1 since my book only gives examples using 1,1
I am also suppose to be able to use the definition $2$ which is $\frac{f(a+h)-f(a)}{h}$. 
This is even more confusing for me as I do not know what $h$ represents and it is not defined in my book anywhere.
 A: You are a bit confused. The slope of the tangent is not merely
$$\frac{f(x)-f(a)}{x-a},$$
it's the limit as $x\to a$ of that fraction. 
In this case, the point in question is $(1,3)$, which means that $x=1$, and that $f(1)=3$. So you are trying to find
$$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{(4x-x^2)-3}{x-1} = \lim_{x\to 1}\frac{-(x^2-4x+3)}{x-1}.$$
Now, unsurprisingly, the numerator and denominator both evaluate to $0$ at $x=1$. Why "unsurprisingly"? Because this always happens when you try to compute the slope of the tangent. That's why we use limits. This being a rational function, you can always factor out $x-a$ from the numerator (in this case, $x-1$); so it was not that you "got lucky and accidentally found" that you could factor, with rational functions like this, a polynomial divided by a polynomial, that's what you should always be looking for. Indeed,
$$-(x^2-4x+3) = -(x-3)(x-1),$$
so
$$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{-(x^2-4x+3)}{x-1} = \lim_{x\to 1}\frac{-(x-1)(x-3)}{x-1} = \lim_{x\to 1}-(x-3),$$
and this limit can be evaluated simply by plugging in $x=1$.
Definition 2 is even easier. We want to find
$$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h},$$
where $a=1$. Now, $f(1) = 3$, we knew that already. What is $f(1+h)$? Plug in!
$$f(1+h) = 4(1+h) - (1+h)^2 = 4+4h - (1+2h+h^2) = 4+4h-1-2h-h^2 = 3+2h-h^2.$$
So we have:
$$\lim_{h\to 0}\frac{f(1+h)-f(1)}{h} = \lim_{h\to 0}\frac{(3+2h-h^2)-3}{h}
= \lim_{h\to 0}\frac{2h-h^2}{h}.$$
Why is this easier? Because here it should be obvious how to factor the numerator so you can cancel it with the $h$ in the denominator and simplify: you have $2h-h^2 = h(2-h)$. Cancel, and then do the resulting easy limit.
