when solving equations that begin in fraction form and then are changed through multiplying by a LCD into quadratic form you must always state your domain restrictions to ensure that you aren't dividing by zero.

10.3 solving equations in quadratic form

2(4m-3)^2+7(4m-3)+5=0
2u^2+7u+5=0 let 4m-3=u
(2u+5)(u+1)=0 factor zero factor prop.
2u+5=0 or u+1=0


u=-5/2 or u=-1

4m-3=-5/2 or 4m-3=-1 substitute 4m-3 for u

4m=1/2 or 4m=2 solve for m

m=1/8 or m=1/2

More on 10.3!

k= SQRT of 6k-8

-take the square of both sides.

k^2=6k-8

k^2-6k+8=0

(k-4)(k-2)=0

k=4 or k=2

-to check, plug into ORIGINAL equation.

x+ SQRT of x=6

-isolate the square root term.

-"work our magic" with squaring terms.

SQRT of x=6-x

x=(6-x)^2

x=(6-x)(6-x)

x=36-12x+x^2

-x -x

x^2-13x+36=0

(x-9)(x-4)=0

x=9 or x=4

when checking, 9 doesn't work so 4 is the only solution.

10.3

1/x+1/x-1=7/12

1. Clear fractions using LCD.

LCD=12(x)(x-1)

12(x)(x-1) over x + 12(x)(x-1) = 7(12(x)(x-1)

12(x-1)+12x=7x(x-1)

12x-12+12x=7x^2-7x

24x-12=7x^2-7x

2. Put into quadratic form: ax^2+bx+c=0

7x^2-31x+12=0

(7x-3)(x-4)=0

7x-3=0 or x-4=0

x=3/7 or x=4

3. You could...

factor

complete the square

quadratic form

square root property

*you must show the domain restrictions!

x cannot equal o or 1

Ok so on problem 27 in the hw for tonight the answer is -2+_ sr i2/3
when I solved the problem I got -2+_ sr i19 / 3
as you can see the only thing I had simplified differently was the 38, can some on please help me

The Discriminant

"b^2-4ac" > 0 2 distinct real solutions
"b^2-4ac" = 0 1 real solution (double root)
"b^2-4ac" < 0  0 real solutions

Imaginary Numbers!

-The square root of a negative number is imaginary, (i)

-The square root of -1 is i.

How you write i as an answer, Ex. 12i, not i12.

Ex. 9q^2-6q+5=0

has no real solutions but has two imaginary solutions.

10.3 Equations Quadratic in form

7x^2-31x+12 Standard form

(7x-3)(x-4)=0 Factor

7x-3=0 or x-4=0 zero-factor property

x=3/7 or x=4 solve each equation


Check by substituting in original equation

Ike's Panera Bread Fun!

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HAYO!

http://www.panerabread.com/about/company/history.php

Discriminant

(b^2-4ac) tells the number of REAL and/or UNIQUE solutions.

b^2-4ac > 0, we have 2 distinct, real solutions

b^2-4ac = 0, there is 1 real solution (double root)

b^2-4ac < 0, there are 0 real solutions

Ex. Find the discriminant of each equation, then, determine the number of real solutions.

a. 2x^2+4x+1=0

ax^2+bx+c

a=2,b=4,c=1

(4)^2-4(2)(1)

16-8=8

2 real solutions.

b. 2x^2+4x+2=0

ax^2+bx+c

a=2,b=4,c=2

(4)^2-4(2)(2)

16-16=0

There is 1 real solution

c. 2x^2+4x+3=0

ax^2+bx=c

a=2,b=4,c=3

(4)^2-4(2)(3)

16-24=-8

There are 0 real solutions

Flickr Diary -Need Points?

Flickr Diary


This is your chance to tell a algebraic graphic story.  You should use pictures that are bold and speak to the viewer.  Add a caption below your picture. More instructions below!


Tag all photos a2a11. Additional tags are listed in parenthesis. If no, parenthesis then tag using word next to the number.  Also tag with your name.

Also, note how each picture visual associates concepts learned for you personally.

Due by the final exam period.

1. Rectangular Coordinate Grid (PiGrid)
2. A Parabola (PiParab)
3. Polynomial (PiPoly)
4. Exponents (PiExp)
5. Real life representation -- graph of Function (PiFunction)
6. Real life symmetry (PiSymmetry) -spell correctly!
7. PiReflect
8. PiMonomial
9. Common Factor (PiCommon)
10. piquadratic
11. PiRatio
12. PiProportion
13. PiFunction
14. PiFunctionMachine
15. PiLine
16. PiAllometric
17. PiRadicalFunction
18. PiAbsvalue
19. PiCubicfuncgraph
20. Functions in Athletics (PiSport)
21. Piinequality
22. Piprojectile

Quadratic Formula

-b+/- SQRTb^2-4ac

2a

-There are two solutions because it can either be negative or positive

-Solutions same as roots

-zeros and or x-intercepts

Ex. x^2+5x-14=0

a=1, b=5, c=-14

x=-b-5+/-SQRT5^2-4(1)(-14)

2a

-5 +/- SQRT 25-(-56)

2

= -5 +/- SQRT 81

2

=5+/- 9

2

x=2 and -7 = {2,-7}

Rationalize the Denominator

-a solution is not simplified if there is an irrational number in the denominator of a fraction.

Ex. The square root of 121/7, the same thing is the square root of 121 over the square root of 7.

=11/the square root of 7

Next, multiply this fraction by the square root of 7 over the square root of 7, this equals 1.

=11 SQRT 7/7

Now, the denominator has a rational number, 7.

10.1 example equations

X^2=-17
√17, √-17

(k+5)^2=-100
√-100
-5+√10
-5,-10

2r^2-4r+1=0
2r^2+4r=-1x4
-2x2=4

3z^2-6z-2-0
3z^2=2+9
z=3+√11/3

HW problem #35

Some people had trouble with this problem.
Here it is step by step
#35

3w^2-w-24=0
3w^2-w=24
3(w^2-1/3w+(-1/6)^2)=24 +3(-1/6)^2
3(w^2-1/3w+1/36)=24+3/36
3(w-1/6)(w-1/6)=24+3/36
-1/6+-1/6=1/36
-1/6+-1/6=-2/6=-1/3
root(w-1/6)^2= root 289/36
w-1/6= = +- root 289/ root 36
w-1/6= +- 17/6
w=1/6 + 17/6 =18/6=3 OR -16/6= -8/3
{3, -8/3}

Completing the Square

Completing the Square

-forces any quadratic expression to factor

-use “completing the square” to solve quadratic equations

To Complete the Square:

Ex. x^2+6x+16=0

Note: a=1, b=6, c=-16

Step 1. Move the constant term ( c ) to one side and every other term to the other side.

x^2+6x-16=0 → x^2+6x=16

Step 2: Factor out the leading coefficient (a)

*will be more relevant when a does not equal 1.

1(x^2+6x)=16

Step 3: Add (1/2 x b)^2 to side with (a) and (b) term. Then add a(1/2 x b)^2 to the side with constant term.

Simplify:

1(x^2+6x+(6/2)^2)= 16+1(6/2)^2

x^2+6x+9=16+9

x^2+6x+9=25

Step 4: Factor side with (a) and (b) terms

(x+3)(x+3)=25 → (x+3)^2=25

*At this point you can either solve for x

Something to watch out for

If the solution is less than or equal to zero then using the square root property always produces two square roots one positive and one negative.

10,1 Completing the Square

Simple steps to Completing the Square
To solve ax^2+bx+c=0 (a greater than or equal to zero)
Steps
Step 1: Be sure the squared term has a coefficient 1. If the coefficient of the squared term is some other nonzero number a, divide each side of the equation by a.
Step 2: Write the equation in the correct form so that the terms with variables are on one side of the equals sigh and the constant is on the other side.
Step 3: Square half the coefficient of the first-degree term.
Step 4: Add the square to each side.
Step 5: Factor the perfect square trinomial. One side should now be a perfect square trinomial. Factor it as the square of a binomial. Simplify the other side.
Step 6: Solve the equation. Apply the square root property to complete the solution.

10.1!

10.1 The Square Root Property and Completing the Square

-the square root property allows you to eliminate the squared property of a number.

a means, “what two of the exact same number can be multiplied to get a?”

Ex. x^2=36

x^2= 36

x= 6 and x=-6

Because 6x6=36 and -6x-6=36

Standard Form of Quadratic Equation

ax^2+bx+c=0

A quadratic equation is a second degree equation because the term with the largest sum of exponents is ax^2 and its exponent is 2.

Ex. 4(x^2)+(-48x)+8=0

2(x62)+36x+6=0

Zero Factor Property

That is, if ab=0, then a=0 or b=0.

(3x+7)(x-4)=0

(a) x (b)=0

Either quantity a has to equal zero or quantity b has to equal zero in order for this statement to be true.



Properties of Square Roots, only used if a is greater than or equal to zero or b is greater than or equal to zero.


Ex. Simplify the following radical.
20 *Find the largest perfect square that goes into 20 evenly.

20=
4x5= 4x5= 25

Solve a quadratic equation in the form (ax^2+b)^2=c

Ex. (x-5)^2=36
(x-5)2= 36
((x-5)2)= 36
(x-5) =6 or (x-5)=-6
x=11 or x=-1
0 the square root could be positive or negative.

Verticle and Horizontal line test

Verticle line test: -Functions are only functions if they pass the verticle line test. The verticle line test is when you run a verticle line over a graph and see if in any places on the graph, the line hits more than one point. Horizontal Line test: You use the horizontal line test to see if a function is one-to-one. Similar to the vertical line test, you run a horizontal line across the graph and if the line gets to a point where it hits two points, then the funciton isn't one-to-one.

Transformations of graphs

Test Review Transformations of graphs: f(x) = - a (x +/- h) +/- k - The negative sign is the reflection a - stretches or flattens +/- h - Shifts left or right (opposite of sign) +/- k - Shifts up or down (same as sign)

Inverse Equation

I was confused in class today about the whole switching of variables when you use the inverse equation. If you think about it like the original variables are in the top and bottom equations and the switched are in the two middle equations it gets much easier.

Extra Help

Remember when a number is on the inside of parentheses it is moved to the same number of units to the opposite side.  If it's a positive number it shifts to the left and if it's a negative number it shifts to the right.  

If  the number is on the outside of parentheses, you move it up if it's a positive number and down if it's a negative number.

easy way to remember composition of functions

when you think about f(g(x)) it can sometimes be confusing which part to work on first, especially if the variable is left as x, no number substituted. When I looked back over all my notes I saw the f of g of x and it made so much more sense to think about it that way, like working from right to left

Remember

The domain of the original function is the range of its inverse function.  The range of the original function is the domain of its inverse function.

vertical and horizontal line tests

The vertical line test proves if the graph is a function.  The horizontal line test proves whether the function is one to one.

Functions

Here is a link that should help with the test if you are having a little trouble

http://www.youtube.com/user/khanacademy#p/c/7AF1C14AF1B05894/39/VhokQhjl5t0

composition functions

here is an explanation of composition functions

http://www.youtube.com/watch?v=S4AEZElTPDo&feature=related

1 to 1

If the graph doesnt pass the horizontal line test then it is not 1 to 1

Inverse Functions

f(x)= Ix-4I-3

x f(x)
-2 3
-1 2
0 1
1 0
2 1

Inverse: [(3,-2),(2,-1),(1,0),(0,1),(1,2)]
the domain comes the range and the range becomes the domain in inverse functions.

f(x)=Ix-4I-3 GRAPH

-the inverse of a functions graph reflects upon the y=x

g(x)=2x+5
1. Replace function name with y, y=2x+5
2. Switch x and y x=2y+5
3.Solve for y=
y=x-5/2
4. Rename ywith inverse function name g(x)=x-5/2=1/2x-5/2

-If the original function is a function and the inverse of that is a function it is called an original function one to one.

Inverse Functions

If f is a function from a set A to a set B, an inverse function is from B to A. If input x produces an output y then an inverse function would be input y and output x
Domain-all possible set of input usually (x) values
Range-all possible set of outpt usually y values
Purplemath discusses what weere goin over in class today which is how to find a function's domain and range
http://www.purplemath.com/modules/fcns2.htm

Vertical line test

You also know if a set of coordinates is a function if it passes the vertical line test.  Draw a vertical (straight up and down) line through a line and it only intersects at one point, then it's a function.

On the daily quiz to day many of us forgot how to factor out problems like...
(x+3)^3. All you do to solve these problems is set up the problem so it looks like (x+3) (x+3) (x+3), next you factor the first (x+3) in to the second (x+3) giving you (x+3) (x^2+6x+9). Than distribute the last (x+3) throughout.
x^3+6x^2+9x+3x^2+18x+27
combine like terms and your answer is...
x^3+9x^2+27x+27

More on Composition Functions

Composition Functions

-when you apply a function rule to the result of another function rule you compose a function.

-you MUST know the notation to compose.

Notation: f(g)(x)= f(g(x))
pronounce- “f of g of x”
f composed of g composed of x.

-all functions are composed of x, x is the input.

f(g(x)) is the same as (fog)(x)

f(x)= x^2+4 g(x)= 2x+3

now instead of putting x as the input, put g(x).
f(g(x))=(2x+3) then apply all of the x values from the x problem
f(g(x))= (2x+3)^2+4
simplify
=(2x+3)(2x+3)+4
=4x^2+6x+9+4
f(g(x))= 4x^2+12x+13

Ex1: Find (gof)(2)
=g(f(2))
1. Always work inside out by plugging the input values in.
=g(f(2))= g(8)

Test Review!

Here is some review for the test on friday! So far we have done:

-Re-Introduction to Functions 

If you want more practice on this, you should go to the websites that we had some homework problems on that is posted on the blog and look at the slideshow again. The homework problems are really helpful because they also show the answers!

-12.1 Graphs of Functions and Composition Functions

To review this section it would be helpful to study all of the graph parent functions and review what other people have posted on the blog to help remember all of the different graphs. To review composition functions it would be helpful to do problems 7-10 in the chapter review (p.775). 

-11.1 Inverse Functions

We will probably go over this today in class, but it would be helpful to look at the notes we took on this section for homework last night. 

Parabola

y=ax^2+bx+c

y=a(x-h)^2+k

It opens up if a >0
down if a <0
The vertex is (h,k)

It has a x^2 term not y squared

functions

A function is one to one when each x-value corresponds to only one y-value and each y-value corresponds to just one x-value.

One-to-One Functions

One-to-one functions come up in section 11.1. A one-to-one function is a function where each x-value corresponds to only one y-value, and each y-value corresponds to only one x-value.

An example of this is:

a. (1,2), (3,8), (9,0)

This IS a one-to-one function because each x and y-value only appears once throughout the ordered pairs.

b. (4,7), (8,6), (4,3)

This IS NOT a one-to-one function because the 4 shows up throughout the ordered pairs twice as the x-value.

Helpful website for conic sections

http://www.intmath.com/plane-analytic-geometry/conic-sections-summary.php
General Form of the Circle

An equation which can be written in the following form (with constants D, E, F) represents a circle:

x2 + y2 + Dx + Ey + F = 0
Formal Definition A circle is the locus of points that are equidistant from a fixed point (the center).
Conic Section



If we slice one of the cones with a plane at right angles to the axis of the cone, the shape formed is a circle.

While doing the HW problems today I forgot that terms like (x+5)^2 are written as (x+5)(x+5) - I was writing the problem as x^2+25. This is a small but critical error that could make all the difference on the up coming test.
That's all...Weeeeerd Up!

Composition Functions

When you apply a function rule to the result of another function rule you compose a function.

-you MUST know the notation to compose.

Notation: f(g)(x)= f(g(x))
pronounce- “f of g of x”
f composed of g composed of x.

-all functions are composed of x, x is the input.

f(g(x)) is the same as (fog)(x)

f(x)= x^2+4 g(x)= 2x+3

now instead of putting x as the input, put g(x).
f(g(x))=(2x+3) then apply all of the x values from the x problem
f(g(x))= (2x+3)^2+4
simplify
=(2x+3)(2x+3)+4
=4x^2+6x+9+4
f(g(x))= 4x^2+12x+13

Ex1: Find (gof)(2)
=g(f(2))
1. Always work inside out by plugging the input values in.
=g(f(2))= g(8)

12.1 functions

f(x)=x^2- the squaring function

f(x)=|x|- absolute value function

f(x)=1/x the reciprocal function

f(x)=√ x square root function

12.1 exercises HW

for problems 5-18 you can put the function in your caculator and see what the graph is supposed to look like. It really helps