## anonymous 5 years ago Tangent line equation. (a, f(a)), f(x) = x^2, with a = 2.

1. anonymous

there are 2 ways to solve this problem

2. anonymous

ok!

3. anonymous

1) By using the definition and I forgot abt the other way lol

4. anonymous

hahaha , ok, lets use the definition

5. anonymous

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

6. anonymous

>_< calculus I lol, alright, remind me, m = limit of that function

7. anonymous

f(x) - f(a) = f'(a) . (x - a)

8. anonymous

LOL! that's it! find the derivative!

9. anonymous

you have f(a) = 4, and a = 2

10. anonymous

are you sure that you have to find the derivative? I remember something with limits

11. anonymous

$\ m= lim_{h \rightarrow 0}f(a) - f(a+1)/h$ that's the equation needed if we want to find the slope using the definition , right?

12. anonymous

yeap!

13. anonymous

alright! all we have to do is plug in the values we have

14. anonymous

hmmm if f(x) = x^2, then f(a) = a^2 , right?

15. anonymous

that's correct.

16. anonymous

then,$= \lim_{h \rightarrow 0} (a^2 -a^2-1)/h$ = -1/h...hmm there's something wrong. h must be canceled

17. anonymous

LOL! no no no wait!

18. anonymous

$\lim_{h \rightarrow 0} f(a) - f(a+h)/h$ that's the right equation LOL! sorry ^^"

19. anonymous

np

20. anonymous

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?

21. anonymous

a = 2, f(a) = (2)^2 = 4 (2,4)

22. anonymous

ahhhh

23. anonymous

i got it. thanks!

24. anonymous

soooooooooooo, we'll use the equation which is : y -yo = m (x-xo) y - 4 = = -1(x-2) ^_^ LOL np :)

25. anonymous

sorry abt the mess though >_<

26. anonymous

np :)

27. anonymous

^_^ good luck