Question

I'm wondering whether I should take Pre-Calculus or just take Calculus without taking Pre-Calc. Is Pre-Calc really necessary/important for me to take Calculus?

Answer 1

pre calculus is just trigonometry rules

Answer 2

and some annoying algebra :3

Answer 3

and yes, you need to know trigonometry rules if your going to be taking their derivatives and integrating them

Answer 4

annoying algebra = factoring ridiculously high-degree polynomials :D

Answer 5

I can do factoring :p

Answer 6

Oh?
Then factor this...
\[\Large x^5 + 2x^4+4x^3+8x^2+16x+32\]

Answer 7

LOL
nvm..
though do it if you're feeling bored...

Answer 8

lol i'm wondering whether i want 2 or not :3 Point is, let's say I do review trigonometry rules and over algebra 2 which I'm almost finishing in like less than 2 weeks, would i be fine in calculus ? o.O

Answer 9

Uhh... hard to say...lol
it's called precalc for a reason... hang on...

Answer 10

Trying to recall my calculus 1

Answer 11

I don't think I did anything quite strenuous as far as the lessons in precalc are concerned...
how about you, @amistre64 ?
Do you remember your own calc 1 experiences? ^_^

Answer 12

Ugh lol got to take pre-calc anyways -_- on flvs it's a prerequisite :p answer solved lol :) but yeah just 2 know I'm curious about calc I :)

Answer 13

Now, to spam your thread... I wonder if you ever dreamed this'd be how to do it...
\[\Large x^5 + 2x^4+4x^3+8x^2+16x+32\]\[\Large \color{red}{(x-2)}(x^5 + 2x^4+4x^3+8x^2+16x+32)\color{red}{(x-2)^{-1}}\]\[\Large \color{red}{(x^6-64)}{(x-2)^{-1}}\]\[\Large \color{red}{(x^3+8)(x^3-8)}{(x-2)^{-1}}\]\[\Large (x^3+8)\color{red}{(x-2)(x^2+2x+4)}{(x-2)^{-1}}\]\[\Large \color{blue}{(x+2)(x^2-2x+4)(x^2+2x+4)}{}\]

Answer 14

BTW, you wanna know what you'll start with in calc 1?

Answer 15

yeah sry 4 late reply :p was afk

Answer 16

First thing you'll do (they'll probably assume you've taken precalc) is introduce the concept of the limit, albeit intuitively...

Answer 17

For instance, a tame function, like a linear one,
\[\large f(x) = 3x + 2\]

Answer 18

The concept of the limit of a function as x approaches a certain number... let's let x approach 0.

Answer 19

k :p i'll take a look into that :)

Answer 20

1 is (somewhat) close to zero, so the value of f(x) when x = 1 is...5.
-1 is also that close to zero, so the value of the function when x = -1 is -1.
Let's go closer...
0.5 is closer to zero, the value of the function at that point is 3.5
-0.5 is also that close to zero, the value of the function at that point is 0.5

I'll not go further, suffice it to say, as x goes closer and closer to zero, the value of the function gets closer and closer to 2.

:P

Answer 21

It will be denoted...
\[\Large \lim_{x\rightarrow 0}[3x+2]=2\]

Answer 22

lol ok sounds fun :p so will the value ever reach 2? :p or will it almost but never?

Answer 23

In this case, it does.
You might say that the limit of the function as x goes to zero is just the function itself, evaluated at zero...after all,
\[\large f(0) = 2\]
So why bother with limits? ^_^

Answer 24

The answer is that a limit may exist, even though the function itself is undefined at a point.
Study... and try to evaluate this limit...
\[\Large \lim_{x\rightarrow 2}\frac{x^2-4}{x-2}\]

Answer 25

k i see :p & lol what math r u learning atm? :D

Answer 26

If you must know, it's abstract algebra (really fun), an introduction to modern geometries, combinatorics, introduction to complex analysis :3

Answer 27

Am I just supposed 2 factor (x^2-4)/x-2? That'd make it & cool ok ^.^