oh well :)
This past week we reviewed for our exams and presented our power point presentations. I felt like i learned a lot about different jobs and how much math does play a part in everyday life. But onto the real stuff.
Some things i'm finally catching on to:
- Tangent lines
- Derivatives in general (finallyyyyyyyyy)
- Limits (sometimes)
Of course i'm still not getting some things also:
- Optimization- i understand the steps and all so don't comment me telling me them. I just can't find the two formulas at the beginning. If anyone has any ideas or can help me with this I NEEED IT
- Looking at graphs determining derivatives and such
- Vertical asymptotes
- And of course the easy stuff that i always seem to mess up on that we will not speak of
So now that that was said i'll give an example of something i understand :)
g(x)=x^2 - 4x + 9
Write the equation of the tangent line to the graph of g at x=3.
First you can simply just find the y value so you can later put your information in point slope form. Since you are given the x value at x=3, you just plug it into the original formula to find the y.
g(3)=3^2 - 4(3) + 9
g(3)=6
So your point is (3, 6)
Now you need a slope, so guess what we do now.
Take a derivative, which is what we'll be doing everyday for what feels like the rest of our lives so do ittttt!
g'(x)=2x-4
Now plug in the x value into the derivative.
g'(3)=2(3) - 4
g'(3)=2
Now that you have a point and a slope, you put it into point-slope form because it is asking for an equation of the tangent line.
Your final answer will be y - 6=2(x - 3)
Easy enough, right?
Oh and another thing i learned this week is how to use the table function and that you should treat it like your bible basically (Thanks John!)
If anyone can help me with anything please let me know. I'm always struggling with something. But thanks to all of these study guides i think i'm finally getting a hang of some of these things.
Now for yet another week of Calculus
To find verticle asymptote, you first have to factor out factor out the top an the bottom to make sure nothing cancels out. Whatever you are left with at the bottom, you set equal to zero and your answer(s) is a vertical asymptote.
ReplyDeleteEXAMPLE: x^2+2x+1/ x^2-1
the top factors to (x+1) (x+1)
the bottom factors to (x-1) (x+1)
Therefor the (x+1)'s cancel leaving you with
(x+1)/(x-1)
When you set the bottom equal to zero: you get x=1 so there is a vertical asymptote at x=1.
ok so when you look at a graph to find what the graph of the next derivative will look like. first, you determine what the function of the graph given will be. like, if you are given a diagnol line, figure out if it's increasing or decreasing. looking from the left to the right.
ReplyDeletewell, when you take the derivative of a diagnol line, you get a horizontal line. the horizontal line will be either positive or negative depending on whether or not it is increasing or decreasing.
:)