Tangent line equation. (a, f(a)), f(x) = x^2, with a = 2.

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Tangent line equation. (a, f(a)), f(x) = x^2, with a = 2.

Mathematics
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there are 2 ways to solve this problem
ok!
1) By using the definition and I forgot abt the other way lol

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hahaha , ok, lets use the definition
first : (a, f(a) ) is going to be : (2, 4)<-- you'll use these points later to find the equation we'll have to find the slope first
>_< calculus I lol, alright, remind me, m = limit of that function
f(x) - f(a) = f'(a) . (x - a)
LOL! that's it! find the derivative!
you have f(a) = 4, and a = 2
are you sure that you have to find the derivative? I remember something with limits
\[\ m= lim_{h \rightarrow 0}f(a) - f(a+1)/h\] that's the equation needed if we want to find the slope using the definition , right?
yeap!
alright! all we have to do is plug in the values we have
hmmm if f(x) = x^2, then f(a) = a^2 , right?
that's correct.
then,\[= \lim_{h \rightarrow 0} (a^2 -a^2-1)/h\] = -1/h...hmm there's something wrong. h must be canceled
LOL! no no no wait!
\[\lim_{h \rightarrow 0} f(a) - f(a+h)/h \] that's the right equation LOL! sorry ^^"
np
so you'll get:\[=\lim_{h \rightarrow 0} (a^2 -a^2 -h)/h\]\[=\lim_{h \rightarrow 0} -h/h\] m = -1 ^_^ there we have found m and we have the points (2,4) <-- you know how I got them right?
a = 2, f(a) = (2)^2 = 4 (2,4)
ahhhh
i got it. thanks!
soooooooooooo, we'll use the equation which is : y -yo = m (x-xo) y - 4 = = -1(x-2) ^_^ LOL np :)
sorry abt the mess though >_<
np :)
^_^ good luck

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