# 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 $((4x-x^2)-3)/(x-1)$ 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 (f(a+h)-f(a))/h

This is even more confusing for me as I do not know what h or h represents and it is not defined in my book anywhere.

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I think I got a little further with the problem. I got really lucky and accidently found that I can factor it into (x-1)(-x+3) I just don't see how -x+3 helps me. Unless it is implied that the limit is x->1 so then that means it would be -1+3 which is a correct answer but now I am just guessing. – user138246 Sep 11 '11 at 19:27
The equation you list should have $\lim_{x \rightarrow a}$ in front of it. This is the definition of the derivative of $f$ at the point $a$ and gives you the slope of the line tangent to the graph of $f$ at the point $x = a$. – Austin Mohr Sep 11 '11 at 19:33
@Jordan: Please try to write your entire question in one go, instead of half the question first, and the second half fourteen minutes later. – Arturo Magidin Sep 11 '11 at 19:39

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.
@Jordan: A formula that says $f(x) = 4x-x^2$ is a set of "instructions" on how to evaluate the function $f$. It says "whatever you are given as an input, multiply it by $4$; then take the result of that, and subtract the result of squaring what you were given as an input". If you are given $1$ as an input, $f(1)$, then you evaluate $4(1)-1^2$. If you are given $\pi$ as an input, $f(\pi)$, then you evaluate $4(\pi)-pi^2$. If you are given $t$ as an input, then you evaluate as $f(t) = 4(t) - t^2$. If you are given $a+b$, then you evaluate $f(a+b) = 4(a+b)-(a+b)^2$. – Arturo Magidin Sep 11 '11 at 19:44
@Jordan: but, no. If you have $f(a+b+c)$, then you plug in "$a+b+c+$" for $x$; $a(x)+b(x)+c(x)$ makes no sense. You don't have functions $a$, $b$, and $c$. If you are told to evaluate $f(a+b+c)$, then you take $a+b+c$, multiply it by $4$, then take $a+b+c$, square it; and then take the difference. That is, if $f(x)=4x-4x^2$, then$$f(a+b+c) = 4(a+b+c) - (a+b+c)^2.$$ You replace the $x$ in the formula with whatever the input is. – Arturo Magidin Sep 11 '11 at 19:46