Terrio's Reflection

Aleks blog...so, finding the slope of a line when the equation is given to you:

Notes - Take the equation given, turn into slope intercept for, pull out the slope from equation, that's the answer.

Example: find the slop of the line 4 x +5 y = 3

You start out by converting the equation to slope intercept form, which is y = m x + b. ( m stands for the slope )

4 x + 5 y = 3
5 y = 3 - 4 x
y = ( 3 - 4 x ) / 5
y = ( 3 / 5 ) - ( 4 x / 5 )
y = - ( 4 /5 ) x + ( 3 / 5 )

Now that it is converted into slope intercept form you take ( m ) out of the equation, - ( 4 /5 ), and that's your slope.

So the answer is - ( 4 /5 )

Reflection

In calculus this past week we did stuff with substitution, so you need to know how to integrate and find bounds for some substitution problems. Lately we have been doing a lot of stuff with integrals, so we definitely need to know how to integrate. Here is an example of integration.

EXAMPLE:

1. Find the integral of: x^3 + 3x
When finding the integral, you have to add one to the exponent and divide that number.
x^4/4 + 3(x^2/2)
1/4x^4 + 3/2x^2 + c

2. Find the integral of: 6x
6[x^2/2]
3x^2 + c
Also when solving for integrals, do not forget to add the +c.

When finding the bounds, you set the equations in the problem equal too each other and then solve for x.

Ex :

x^2 + 2x = 3x

x^2 - x = 0

x ( x - 1 ) = 0

the bounds are x = 1, 0

blog march 24

Well i just finished 28 aleks and it was very sick nasty, this stuff is so simple but the simple mistakes are starting to kill me. Its driving me insane, but at least i am done and will be getting my 28 points, but ill review some things i got wrong
Factoring a quadratic with a leading coefficient greater than one...
Ex :
5y^2 + 13y - 6
Use the FOIL method backwards...
The coefficient of is . Using whole numbers, only 1 and 5 can be multiplied to get .
So you have to find two integers and such that 5y^2 + 13y - 6 = ( 1y + p) ( 5y + q )
Using the FOIL method to expand the right-hand side of this equation, you get that ( pq ) = -6 and q + 5p = 13
You'll find that 3 and -2 work...3 in place of p and -2 in place of q.
Factoring a quadratic with a leading coefficient greater than one...
Ex :
5y^2 + 13y - 6
Use the FOIL method backwards...
The coefficient of is . Using whole numbers, only 1 and 5 can be multiplied to get .
So you have to find two integers and such that 5y^2 + 13y - 6 = ( 1y + p) ( 5y + q )

Using the FOIL method to expand the right-hand side of this equation, you get that ( pq ) = -6 and q + 5p = 13

You'll find that 3 and -2 work...3 in place of p and -2 in place of q.

The answer is

Terrio's Reflection

Don't really understand what we're doing this week too well sooooo, review blog? I'll review chain rule, veritcal asymptotes, and derivatives:

-When solving for a vertical asymptote of a graph, you factor out the top and bottom and set the bottom equal to zero.

Examples:
1) f(x) = x^2/x^2-4
= x(x)/(x+2)(x-2)
(x+2)(x-2) = 0
x = -2, 2

2) g(x) = 1/2(x+1)
= 2(x+1) = 0
2x+2 = 0
x = -1

-When solving chain rule, you take the derivative from outside in: derivative of outside, recopy inside, multiply by derivative of inside.

Examples:
1) y = cos3x^2
y' = -sin3x^2 X (6x)
= -6xsin(3x^2)

2) g(x) = 3(4-9x)^4
= 12(4-9x)^3 X (-9)
= -108(4-9x)^3

-When taking regular derivatives you can use a few different strategies, sometimes you may have to use the product rule or quotient rule.

Examples:
1) f(x) = 3x^5 + 4x +8
f'(x) = 3(5)x^(5-1) + 4 +0
= 15x^4 + 4

2) f(x) = 3xy + 4x^2
f'(x) = [3x(1) + y(3)] + 8x
= 3x + 3y + 8x
= 11x + 3y

3/20 Reflection

Last week we learned something new, Integration by Substitution. This is, in a way, the product/quotient rule of integration because it allows you to integrate terms that are being multiplied and divided.

To use substitution, one function must be the derivative of another.
*If you are only off by a number, tho, it can be fixed.

Here is an example of substitution:

x ( x^2 + 1 )^3

-First, find your "u" and your "du"

u = x^2 + 1 du = 2x

The problem has to become 1/2 S 2x ( x^2 + 1 )^3 because:
du = 2x (du is the derivative of u) that means that on the front side of the integration symbol, you must put the reciprical of 2. This is to balance the eqn out because you had to add a 2 to the problem.

-Now replace the terms with u and du:
1/2 S u^3 du

-Now integrate:
1/2 (1/4u^4) + c

-Now plug in for u and simplify and you get:
1/8 (x^2 + 1)^4 + c

BLOG MARCH 20

This week in cal we did some integrating thing, which im having a little difficulty with but i am slowly starting to get it. I didnt bring my notes home on it, so ill talk about some aleks stuff.
Factoring a quadratic with a leading coefficient greater than one...
Ex :
5y^2 + 13y - 6
Use the FOIL method backwards...
The coefficient of is . Using whole numbers, only 1 and 5 can be multiplied to get .
So you have to find two integers and such that 5y^2 + 13y - 6 = ( 1y + p) ( 5y + q )
Using the FOIL method to expand the right-hand side of this equation, you get that ( pq ) = -6 and q + 5p = 13
You'll find that 3 and -2 work...3 in place of p and -2 in place of q.
Tomorrow we have a quiz on the integration stuff, i hope i do good.

Reflection

So I forgot to bring my notes home but I will attempt to this blog and get it close to being right. This week in calculus we did more Aleks, and on there we are reviewing simple things which we have learned before like dealing with negative exponents and even factoring…basically easy. But we began to learn a new section in class which has to deal with substitution. When substituting integrals, you have to first find the u and du. The du is the derivative of u, but there is an exception where if they are one number off then you can still use those numbers; if it is off by more than one number then you have to do the reciprocal of a common number. During the substitution process, you have to use the process of integration also. By the end of the problem, something out of the problem will disappear and it will end with +c.