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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.

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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.2011-09-11
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    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 $x = a$.2011-09-11
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    @Jordan: Please try to write your entire question in one go, instead of half the question first, and the second half fourteen minutes later.2011-09-11

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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.

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    This is really confusing for me, whatever comes after f is considered to be x? As in if I have f(a+b+c) it would be a(x)+b(x)+c(x)?2011-09-11
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    @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 $f(a+b) = 4(a+b)-(a+b)^2$.2011-09-11
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    @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.2011-09-11
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    Thanks that makes a lot more sense, hopefully I can squeeze out a C on this test!2011-09-11
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    One thing I am still confused one, what is a. What is the definition of a, how do I find it. Is it just y?2011-09-11
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    @Jordan: In the formula $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a},$$ $a$ is the value of $a$ at which you are trying to find the tangent.2011-09-11
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    But what is a if I have y=x^2 and point 6,4 so it is y and the value is 4?2011-09-11
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    @Jordan: If you have $y=x^2$, then the point $(6,4)$ is **not** on the graph, so you cannot be finding the tangent through $(6,4)$.2011-09-11
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    So then the point (2,4)2011-09-11
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    @Jordan: The point on the graph is $(a,f(a))$. If you are looking for the slope of the tangent through $(2,4)$, then $a$ is $2$.2011-09-11
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    So a is the slop of the tangent?2011-09-11
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    @Jordan: No. $a$ is the $x$-value of the point **at which** you are trying to find the (slope of the) tangent. The slope of the tangent is the value of the limit $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$2011-09-11
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    @Jordan: I am not trying to be rude, but are you sure you are ready to be taking a calculus course? One cannot do well in calculus without basic algebra skills, any more than one can write a novel without knowing how to talk or write. Your basic algebra skills just don't seem to be yet up to the challenge of calculus.2011-09-11
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    Well I don't have time to waste on going back. Yeah maybe I should just drop out and go work in a factory or whatever but I don't have the time or money to go back 6 math classes.2011-09-11
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    I did not say "drop out". But with poor algebra skills, you will be wasting the time and money on taking a calculus class, because you would not be ready for it and will not be able to learn the material. This does not reflect on your intelligence or ability, only on your preparation.2011-09-11
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    I am 24, I have taken college algebra twice. If that isn't good enough than I mind as well just drop out.2011-09-11
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    This is not a dichotomy, and there are plenty of options between "take calculus **now**" and "drop out". Again, I am *trying* to be helpful, but if such suggestions are taken, as they seem to be, as exhortations to "drop out and go work in a factory", then (i) you are not understanding what I'm trying to say; and (ii) I won't make them anymore.2011-09-11
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    I just don't think taking college algebra for a third time would be helpful if the first two weren't. Especially if I want to go on in math.2011-09-11