Polynomial+Functions

This is a discussion forum on topics from Test 2 and 3 on Polynomials Functions. It is designed for extra credit and also to give additional help.

=Begin each topic in this font. Let the games begin!=

=1. Finding x and y intercepts and critical points (Destiny)= Finding y-intercepts: In order to find y-intercepts, use the equation and plug "0" into x. If the equation is in the form of y=mx+b, then b is the y-intercept.
 * y-intercept**: point at which the function crosses the y axis

y=2x+8 y-int.= 8 __//OR//__ y=x^2-4x-12 y=(0)^2-4(0)-12 y=-12

Finding x-intercepts: To find the x-intercepts, or zeroes of an equation, factor (if possible), then find x of the factor. Another way is to plug "0" into y.
 * x-intercept**: point at which the function passes the x axis

y=x^2-5x-6 y=(x-6)(x+1) x=6 x=-1 Finding critical points: find the x-intercepts of the derivative, then substitute them into the original function to find the y values.
 * critical point**: point on the graph of a function at which the derivative is equal to zero of undefined

y=x^3-5x^2+7x y'=3x^2-10x+7 y'=3x^2-10x+7 =(3x^2-3x)(-7x+7) =3x(x-1)-7(x-1) =(3x-7)(x-1) x=7/3 x=1

y=(1)^3-5(1)^2+7(1) =1-5+7 =3 (1,3)

y=(7/3)^3-5(7/3)^2+7(7/3) =(343/27)-(245/9)+949/3) =(343/27)-(735/27)=(441/27) =49/27 (7/3,49/27)

**Click pictures referring to the topics for videos!**

=2. Long Division (Austin)= Long Division: In this instance we are using long division to get a quadratic equation with three terms. We usually start off with a cubic equation of four terms, which we’ll need to use long division to get our quadratic equation. After the long division you will be left with your answer from dividing and the factor you divided by. This factor is one of the three to the cubic function while to find the other two you have to factor the quadratic found through long division.

Equation: x^3-5x^2-2x+24

1. First we need to find a number that when substituted in for x makes the equation equal 0, this will give you the factor that you divide by.

X^3-5x^2-2x+24 x= -2 –> x + 2 = 0 (We will be dividing by x + 2) -2^3-5(-2)^2-2(-2)+24 -8-20+4+24  -28+28  0

2. Now we can divide the original cubic equation by x+2

X+2/ x^3-5x^2-2x+24 X goes into x3, x2 times so:

__X^2__ _ X+2/ x^3-5x^2-2x+24 Now that we have x2 we multiply x+2(x2) and get x3+2x2 - (x^3+2x^2 ) = -7x^2-2x

x goes into -7x^2, -7x times.

__x^2-7x__ _ x+2/ x^3-5x^2-2x+24 x+2(-7x)= -7x2-14x - (x^3+2x^2 ) = -7x^2-2x - (7x^2-14x) =12x+24 x goes into 12x 12 times.

__x^2-7x+12__ _ x+2/ x^3-5x^2-2x+24 x+2(12)= 12x+24 - (x^3+2x^2 ) = -7x^2-2x - (7x^2-14x) =12x+24 (Note: You shouldn’t be left with a remainder at the - (12x+24) end of the division)

x^2-7x+12 x^2-3x-4x+12 x(x-3)-4(x-3) (x-3) (x-4)

So our factors are -4,-3, and 2

http://www.youtube.com/watch?v=l6_ghhd7kwQ <-- Video link []

=3. Finding equation given 4 points (Spencer)=

Creating an equation using 4 points The first step in creating an equation using the points it contacts on a graph is to set the points into the equation y=ax3+bX2+cx+d. The points are inserted into the equation such that the point of x is set into all x variables and the point of y is set into y. Ex: point: (3,5) 5=a(3)^3+b(3)^2+c(3) +d Then the value of x is multiplied by its corresponding power. It is then multiplied by the coefficient in front of it forming an equation that looks like this, 5=27a+9b+3c+d. After this is done for all 4 given points the equations will be ready to be broken down in order to find the values of a, b, c, and d.

EX: Points: (0,-8) (-3,-140) (1,0) (3,-2)

-8= a(0)^3 +b(0)^2 +c(0)+d -8=d -140=a(-3)^3+b(-3)^2+c(-3)+d -140= -27a +9b +-3c – 8 0=a(1)^3+b(1)^2+c(1)+d 0= 1a +1b +1c – 8 -2=a(3)^3+b(3)^2+c(3)+d -2= 27a +9b + 3c – 8

The next step is to take the equations that have newly been created and use addition to cancel any variables you can. This allows you to isolate the remaining variables and find their value.

-27a +9b -3c =-132 __27a +9b +3c = 6 +__ 18b= -126 b= -7

If the variables do not have the same coefficient they can be multiplied or divided to suit the question but the entire equation must be altered in the same way y value included.

-9(0= 1a +1b +1c -8) -72= -9a -9b -9c -2= 27a + 9b +3c -8 __6= 27a +9b +3c +__ -66= 18a - 6c

-9(0=1a +1b +1c -8) -72= -9a -9b -9c -140= -27a +9b -3c -8 __-132= -27a +9b -3c__ -204= -36a -12c

2(-66= 18a -6c) -132= 36a -12c -204= -36a -12 __-204= -36a -12c__ -336= -24c c= 14 With 3 of the 4 variables now known the last can be found by plugging those 3 back into one of the original equations. 0= a(1)^3 -7(1)^2 +14(1) -8 a= 1

The values for the 4 variables are then set back into the equation y= ax3 +bx2 +cx +d to define the cubic equation. In this case that equation is y=x^3 -7x^2 +14x -8

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= 4. Solving Quadratic and Cubic Inequalities Taylor) = **To solve a polynomial inequality:** Locate the largest degree of the equation to know the shape of the graph. (ex: degree of 2 makes a parabola) 1. Factor the equation to find the x-intercepts. Make sure that all the variables are on one side. 2. Place the x-intercepts on a number line. 3. Find the placement of the graph between each interval by plugging a number in to the original equation that lies between the intervals. 4. State in interval notation/Graph if needed.
 * ** is used when the inequality is **>** or **<**.
 * []** is used when the inequality is **≤** or **≥**


 * Example:** x₂ - 2x - 8 ≥ 0

Factor: ( x+2 )( x-4 ) = 0

x intercepts: x= 4 and -2

Put the x intercepts on a number-line: Plug one number from each of the sections into the polynomial to determine the sign of the resulting answer: x = –3 x = 0 x = 5

( -3+2 )( -3-4 ) = 7 ( -3+2 )( -3-4 ) = -8 ( -3+2 )( -3-4 ) = 5

Use the chart to determine which sections fit the inequality. In this case, both positive sides answer the inequality so:

**Answer:**
(-∞,-2] U [4,∞) There's always a circular bracket by infinity because it doesn't include infinity because you'll never reach infinity- Austin

**Extra Help Video:**
media type="youtube" key="KFkWZHZTaDI" width="560" height="315" = = = = =** 5. Cubic and Quartic factoring (Richard) **=

** To solve a cubic polynomial function: **
Step 1: Re-arrange the expression so that it is in the form of ax^3 + bx^2 + cx + d Step 2: Find all possible factors of d. (ex. d=10, the factors are 1, 2, and 5 because they are the only things that 10 is divisible by.) Step 3: Find one of the factors that allows the expression to equal zero Step 4: Factor your root out of the rest of the equation (ex. borrow a root from the following term) Step 5: Continue factoring (ex. add all like factors together) Step 6: You will have only factored roots left, and those are your solutions Step 7: Plug in your solutions to the original equation to make sure you got it right!

Example:

-4x^2 + x^3 + 10 - 7x

(Re-arrange) x^3 - 4x^2 - 7x + 10

(Find all factors of d) 10 = 1, 2 and 5

(Find a factor that allows the equation to equal zero) 1^3 - 4(1)^2 - 7(1) + 10 = 1 - 4 - 7 + 10 = 0 (next do a little re-arranging) if x = 1, then thats the same as x - 1 = 0) (Factor your root out of the rest of the equation)  "(x - 1)" is our root. Let's see if we can factor it out of the rest of the equation. Let's take it one polynomial at a time.
 * Can we factor (x - 1) out of the x^ 3 ? No we can't. Be we can borrow a -x 2 from the second variable; then we can factor it: x^ 2 (x - 1) = x^ 3 - x^ 2.
 * Can we factor (x - 1) out of what remains from our second variable? No, again we can't. We need to borrow another little bit from the third variable. We need to borrow a 3x from -7x. This gives us -3x(x - 1) = -3x^ 2 + 3x.
 * Since we took a 3x from -7x, our third variable is now -10x and our constant is 10. Can we factor this? We can! -10(x - 1) = -10x + 10.
 * What we did was rearrange the variables so that we could factor out a (x - 1) out of the entire equation. Our rearranged equation looks like this: x^ 3 - x^ 2 - 3x^ 2 + 3x - 10x + 10 = 0, but it's still the same thing as x^ 3 - 4x^ 2 - 7x + 10 = 0.

(Continue to substitute by the factors)  Look at the numbers that we factored out using the (x - 1) in Step 5: x^ 2 (x - 1) - 3x(x - 1) - 10(x - 1) = 0. We can rearrange this to be a lot easier to factor one more time: (x - 1)(x^ 2  - 3x - 10) = 0. We're only trying to factor (x^ 2  - 3x - 10) here. This factors down into (x + 2)(x - 5).

(Plug your two solutions back into the original)
 * (x - 1)(x + 2)(x - 5) = 0 This gives us solutions of 1, -2, and 5.


 * Plug -2 back into the equation: (-2)^ 3 - 4(-2)^ 2 - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0.


 * Plug 5 back into the equation: (5)^ 3 - 4(5)^ 2 - 7(5) + 10 = 125 - 100 - 35 + 10 = 0.

=Quartic factoring=

Quartic equations with real roots: ex. 3x^4 + 6x^3 - 123x^2 -126x + 1080 = 0 First we simplify by dividing all terms by "a" Next we define the variable "f" (f = c - (3b^2/8) Now define "g" (g = d + (b^3/8) - (b*c/2) Define "h" (h = e - (3*b^4/256) + (b^2 * c/16) - (b*d^4) Next we plug "f" "g" and "h" into the following cubic function: y^3 + (f/2) * y^2 + ((f^2-4*h)/16) * y - g^2/64 = 0 Next we solve our cubic equation (y1, y2, y3) now let "p" and "q" be the square roots of any 2 non-zero roots The 4 roots for any quartic equation;

[[image:http://upload.wikimedia.org/math/5/2/1/521da41f04c14b9b3756838d089b9f9d.png caption="x^4+ax^3+bx^2+cx+d=0 ,"]]
==== where, , and. ====

The roots in terms of these four coefficients are given by:
==== ====

where
==== ====

[[image:http://upload.wikimedia.org/math/0/8/7/087bd9d0b5527e860a268279289e1ae9.png caption="Delta_1 = 2b^3 - 9abc + 27a^2 d + 27c^2 - 72bd"]]


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=** 6. Piecewise Functions (Brandan) **=

**Definition of Piecewise Function:**
Piecewise function is a function //f//(//x//) defined piecewise, that is //f//(//x//) is given by different expressions on various intervals.

**Solving Piecewise Function:**
To solve piecewise functions, solve each piece individually as you would any other function. If the solution for that piece doesn't satisfy the condition given for that piece, then it's not a solution for the entire function. Like so:


 * f(x) = x - 3, if x < 0**


 * 2x, if x __>__ 0**

Since this function has two parts. We will start off with f(x) = x - 3, if x < 0. Since x - 3 is if x < 0 this means that x is less then zero and x cannot equal zero, so on the y-axis for x = 0 you put a empty circle because x cannot equal zero. For f(x) = 2x, if x __>__ 0 how ever, x is greater than or equal to zero so you put a full circle because x can equal zero for the y-axis.


 * f(0) = (0) - 3**
 * y = -3**

Ok, at the point f(0) = -3 it will be an empty circle because it does not start at zero so the line is discontinuous.


 * f(0) = 2(0)**
 * y = 0**

Ok, at the point f(0) = 0 it will be a full circle because it starts at zero so the line is continuous.

**Graphing Piecewise Function:**
Time to graph the piecewise function up above. But we are going to need points to do so. So we are going to follow the statements at the end of each function like so:

For the First Statement: **x = -1, -2, -3**
 * f(x) = x - 3, if x < 0**


 * f(-1) = (-1) - 3**
 * y = -4**


 * f(-2) = (-2) - 3**
 * y = -5**


 * f(-3) = (-3) - 3**
 * y = -6**

For the Second Statement: **x = 1, 2, 3**
 * f(x) = 2x, if x < 0**


 * f(1) = 2(1)**
 * y = 2**


 * f(2) = 2(2)**
 * y = 4**


 * f(3) = 2(3)**
 * y = 6**

Now plot the point on the graph for each statement.

**Example of a Piecewise Function:**

 * f(x) = x^2 - 1, if 1 __>__ x**


 * 2x - 4, if 1 < x**


 * f(x) = x^2 - 1, if 1 __>__ x**
 * f(1) = (1)^2 - 1**
 * f(1) =** **2 - 1**
 * y = 1**


 * f(1) = 2x - 4, if 1 < x**
 * f(1) = 2(1) - 4**
 * f(1) = 2 - 4**
 * y = -2**


 * f(x) = x^2 - 1, if 1 __>__ x**
 * x = 3, 5**


 * f(3) = (3)^2 - 1**
 * f(3) = 9 - 1**
 * y = 8**


 * f(5) = (5)^2 - 1**
 * f(5) = 25 - 1**
 * y = 24**


 * f(x) = 2x - 4, if 1 < x**
 * x = -1, -3**


 * f(-1) = 2(-1) - 4**
 * f(-1) = -2 - 4**
 * y = -6**


 * f(-3) = 2(-3) - 4**
 * f(-3) = -6 - 4**
 * y = -10**

=7. Composite Functions: Ryan=

Composite Function is a function whose values are found from two given functions by applying one function to an independent variable and then applying the second function to the result and whose domain consists of those values of the independent variable for which the result yielded by the first function lies in the domain of the second.

<span style="font-family: Arial,Helvetica,;">If we have two functions f ( x )  and  g ( x )  we can define a composite function  h ( x )  f ( g ( x ))   Thus if  f ( x )= x 3  and  g ( x )=2 x − 1  we have  h ( x )=(2 x − 1) 3 =8 x 3 − 12 x 2 +6 x − 1 <span style="font-family: Arial,Helvetica,;">On the other hand if we define the composite function k ( x )  g ( f ( x ))  we then have  k ( x )=2( x 3 ) − 1   Notice that  h ( x )  and  k ( x )  are different functions: when composing functions, the order matters. (If you think of functions as //procedures//, this makes intuitive sense: putting on your socks and then your shoes has a different result from doing this the other way round\dots). <span style="font-family: Arial,Helvetica,;">In compounding functions such as h ( x )= f ( g ( x ))  you have to be a bit careful to ensure that the range of  g  is in the domain of  f  : that is, that  g  doesn't throw up any outputs for which  f  is undefined.

[|Composite Functions plus example] [| Patrick JMT - composition of functions tutorial.]