Monday, February 27, 2017

Week of February 27, 2017

Algebra 1

Function Notation and Evaluating Function

You've been playing with "y=" sorts of equations for some time now. And you've seen that the "nice" equations (straight lines, say, rather than ellipses) are the ones that you can solve for "y=" and then plug into your graphing calculator. These "y=" equations are functions. But the question you are facing at the moment is "Why do I need this function notation, especially when I've got perfectly nice 'y=', and how does this notation work?"
Think back to when you were in elementary school. Your teacher gave you worksheets containing statements like "[  ] + 2 = 4" and told you to fill in the box. Once you got older, your teacher started giving you worksheets containing statements like "x + 2 = 4" and told you to "solve for x".
Why did your teachers switch from boxes to variables? Well, think about it: How many shapes would you have to use for formulas like the one for the area A of a trapezoid with upper base a, lower base b, and height h? The formula is as follows:
A = \dfrac{h}{2}(a + b)
If you try to express the above, or something more complicated, using variously-shaped boxes, you'd quickly run out of shapes. Besides, you know from experience that "A" stands for "area", "h" stands for "height", and "a" and "b" stand for the lengths of the parallel top and bottom sides. Heaven only knows what a square box or a triangular box might stand for!
Why did your teachers switch from boxes to variables? Well, think about it: How many shapes would you have to use for formulas like the one for the area A of a trapezoid with upper base a, lower base b, and height h? The formula is as follows:
A = \dfrac{h}{2}(a + b)
If you try to express the above, or something more complicated, using variously-shaped boxes, you'd quickly run out of shapes. Besides, you know from experience that "A" stands for "area", "h" stands for "height", and "a" and "b" stand for the lengths of the parallel top and bottom sides. Heaven only knows what a square box or a triangular box might stand for!
Algebra 2
Assessments: Quiz on Friday, unit test next Thursday, the projected test date.

Welcome back! Today, we start the class with some applications of polynomials, to ease back into the routine. Then, we move to long division with polynomials, you will do great once you see the pattern. Many of you know I will review the long division from elementary school for the steps.
HW; None

We review long division again and look at synthetic division. It is  tough at the beginning, yet you will master the process very quickly. We also review factoring today for the rest of the unit needs.
HW; None

It is finding roots of polynomials today with synthetic division and factoring. We will review our factoring skills too. I know all of you will love the lesson today.
HW; None

It is finding roots day three, more advanced with the type of factoring
HW; Study for quiz tomorrow

Yes, we made it to Friday! It is quiz day, a day for you to shine. Also, we take notes on 4th degree polynomials.
HW; None

Have a great weekend!

Monday, February 13, 2017

Week of February 13, 2017


Systems of Linear Equations:
  Solving by Addition / Elimination

The addition method of solving systems of equations is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations.
If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution.
    x + 6 = 11
        –6    –6
    x       =   5
You'll do something similar with the addition method.
  • Solve the following system using addition.
    • 2x + y = 9
      x – y = 16
    Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:
      2x + y = 9
      3x – y = 165x      = 25
    Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll back-solve in that one:
      2(5) + y = 9
        10 + y = 9
                y = –1

      Then the solution is (xy) = (5, –1).
It doesn't matter which equation you use for the backsolving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten:
    3(5) – y = 16
      15 – y = 16
            –y = 1
              y = –1
...which is the same result as before.
  • Solve the following system using addition.
    • x – 2y = –9x + 3y = 16
    Note that the x-terms would cancel out if only they'd had opposite signs. I can create this cancellation by multiplying either one of the equations by –1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the –1 through the entire equation. (That means both sides of the "equals" sign!)
    I'll multiply the second equation.
      [x - 2y = -9] + [-x - 3y = -16] = [-5y = -25]
    The "–1R2" notation over the arrow indicates that I multiplied row 2 by –1. Now I can solve the equation "–5y = –25" to get y = 5. Back-solving in the first equation, I get:
      x – 2(5) = –9
      x – 10 = –9
      x = 1

      Then the solution is (xy) = (1, 5).
A very common temptation is to write the solution in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the y-value first and then the x-value second, and of course in points the x-value comes first. So just be careful to write the coordinates for your solutions correctly. 


You can also evaluate compositions symbolically. It is simpler to evaluate a composition at a point because you can simplify as you go, since you'll always just be plugging in numbers and simplifying. Evaluating a symbolic compositon, where you're first plugging x into some function and then plugging that function into some other function, can be much messier. But the process works just as the at-a-number composition does, and using parentheses to be carefully explicit at each step will be even more helpful.
  • Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o g)(x).
  • In this case, I am not trying to find a certain numerical value. Instead, I am trying to find the formula that results from plugging the formula for g(x) into the formula for f(x). I will write the formulas at each step, using parentheses to indicate where the inputs should go:
      ( f o g)(x) = f (g(x))     = f (–x2 + 5)     = 2(             ) + 3     ... setting up to insert the input formula     = 2(–x2 + 5) + 3     = –2x2 + 10 + 3     = –2x2 + 13
If you plug in "1" for the x in the above, you will get ( f o g)(1) = –2(1)2 + 13 = –2 + 13 = 11, which is the same answer we got before. Previously, we'd plugged a number into g(x), found a new value, plugged that value into f(x), and simplified the result. This time, we plugged a formula into f(x), simplified the formula, plugged the same number in as before, and simplified the result. The final numerical answers were the same. If you've done the symbolic composition (the composition with the formulas) correctly, you'll get the same values either way, regardless of the value you pick for x. This can be a handy way of checking your work.
Here's another symbolic example:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
  • Given f(x) = 2x + 3 and g(x) = –x2 + 5, find (g o f )(x).
    • (g o f )(x) = g( f(x))     g(2x + 3)     = –(           )2 + 5    ... setting up to insert the input     = –(2x + 3)2 + 5     = –(4x2 + 12x + 9) + 5     = –4x2 – 12x – 9 + 5     –4x2 – 12x – 4

There is something you should note from these two symbolic examples. Look at the results I got:
    ( f o g)(x) = –2x2 + 13 (g o f )(x) = –4x2 – 12x – 4
That is, ( f o g)(xis not the same as (g o f )(x). This is true in general; you should assume that the compositions ( f o g)(x) and (g o f )(x) are going to be different. In particular, composition is not the same thing as multiplication. The open dot "o" is not the same as a multiplication dot "•", nor does it mean the same thing. While the following is true:
    f(x) • g(x) = g(x) • f(x)            [always true for multiplication] cannot say that:
    ( f o g)(x) = (g o f )(x)           [generally false for composition]
That is, you cannot reverse the order in composition and expect to end up with the correct result. Composition is not flexible like multiplication, and is an entirely different process. Do not try to multiply functions when you are supposed to be plugging them into each other.

Usually composition is used to combine two functions. But sometimes you are asked to go backwards. That is, they will give you a function, and they'll ask you to come up with the two original functions that they composed. For example: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
  • Given h(x) = (x + 1)2 + 2(x + 1) – 3, determine two functions f (x) and g(x) which, when composed, generate h(x).
  • This is asking you to notice patterns and to figure out what is "inside" something else. In this case, this looks similar to the quadratic x2 + 2x – 3, except that, instead of squaring x, they're squaring x + 1. In other words, this is a quadratic into which they've plugged x + 1. So let's make g(x) = x + 1, and then plug this function into f (x) = x2 + 2x – 3:
      ( f o g)(x) = f (g(x))     = f (x + 1)     = (         )2 + 2(         ) – 3     = (x + 1)2 + 2(x + 1) – 3
    Then h(x) may be stated as the composition of  f (x) = x2 + 2x – 3 and g(x) = x + 1.
  • Given h(x) = sqrt(4x + 1), determine two functions f (x) and g(x) which, when composed, generate h(x).
  • Since the square root is "on" (or "around") the "4x + 1", then the 4x + 1 is put inside the square root. I need to take x, do "4x + 1" to it, and then take the square root of the result:
      g(x) = 4x + 1,  f(x) = sqrt(x), and h(x) = ( f o g)(x).

© Elizabeth Stapel 2003-2011 All Rights Reserved

Monday, February 6, 2017

Week of February 6,2017


Solve systems of equations by graphing

A system of linear equations contains two or more equations e.g. y=0.5x+2 and y=x-2. The solution of such a system is the ordered pair that is a solution to both equations. To solve a system of linear equations graphically we graph both equations in the same coordinate system. The solution to the system will be in the point where the two lines intersect.


Graph the equations in a coordinate plane
The two lines intersect in (-3, -4) which is the solution to this system of equations.

Video lesson

Algebra 2

Operations on Polynomials

   The definition of a polynomial is not easily explained because it involves several special terms. Knowing these terms is crucial.
TermDefinitionExample in Red
coefficientIt is a constant that is either alone or being multiplied by an expression.3x5 and -7x2y
exponentIt is the power to which a number or expression is being raised.62 and (-2c)3
integerIt is a number that contains no fractional or decimal part.5 and -300
   Now we are ready to understand the definition of a polynomial.
   Definition: A polynomial is an expression composed of coefficients and variables under addition, subtraction and multiplication and exponents on those variables must be non-negative integers.
   This table will help you discern the difference between polynomials and non-polynomials.

ExamplePolynomial or Non-Polynomial
2x2 + 3x - 5Polynomial

   When adding polynomials, like terms must be combined. For instance, 3c and 5c can be added to get 8c. Likewise, 3x2y and -7x2y can be added to get -4x2y. However, 5x3y and 10x2y5 cannot be added together because they do not have the same exact variables and the exact powers on those variables.   Let those examples guide us regarding the following problem.

   The best way to handle this is to perform the task vertically, instead of horizontally, while aligning like terms.

   With this arrangement of polynomials, it's easier to determine which terms to combine together.

   Consequently, here is the solution.

      ideo: Adding Polynomials
      uiz: Polynomial Addition

   On this next example, care has to be taken.

   The reason for care is due to the first polynomial. It is missing an x-squared term and an x-term. This is why place-holder terms must be included.

   The vertical placement below emphasizes the correct alignment of like terms to be added.

   Consequently, the solution is...

      ideo: Adding Polynomials
      uiz: Polynomial Addition

   When subtracting numbers, it is possible to change the problem to addition. Here is a case in point.
   This problem can be changed to an addition problem. All we have to do is switch the subtraction to addition and then change the second number to its opposite, like this.

   When problems are converted into addition, they are usually done more successfully. The answer is -9, which is harder to obtain as a subtraction problem. When dealing with polynomial subtraction, we can do the exact same process. Here is an example of a subtraction problem with polynomials.

   We can also change this problem to addition. Change the subtraction to addition and then switch the last polynomial to its opposite. Our new example would then be…

   Notice how the second polynomial changed. The -6 changed to 6. The 7 changed to -7 and the 4 changed to -4. Now, the problem is a polynomial addition problem, which is best accomplished vertically.

   The answer can be gained by adding like terms. The like terms are those that have the same variables and powers on those variables. This vertical form makes it easier to find and add those like terms.

Combining those highlighted like terms gives us the following solution.

      ideo: Subtracting Polynomials
      uiz: Polynomial Subtraction

   Here is another problem, but this one is in vertical form.

   We still have to change this problem to addition. We first have to realize it means we are subtracting the bottom polynomial from the top polynomial. So, we have to take the opposite of the second (the bottom) polynomial. This would give us this addition problem.

   Notice that the last polynomial, the bottom polynomial, was changed to its opposite. Now, we simply add like terms and the like terms have already been aligned.

   After we add like terms, we will get this solution.

      ideo: Subtracting Polynomials
      uiz: Polynomial Subtraction

   The best way to multiply polynomials is to do so using a visual organizer. We used a visual organizer in grammar school, called a multiplication table. That is exactly what needs to be used for polynomial multiplication. For example, we will multiply these binomials. [A binomial is a polynomial that has two-terms, bi–nomial.]
   To do this, we will place the first polynomial on the top of our table and the second on the side of our table. [Some people use the method called “FOIL” instead. However, “FOIL” only works for binomials and is useless for other types of polynomial multiplication problems.]

   Now, we will multiply a column times a row. We will multiply “x” times “x” and get x2 and then place that in the table.

   Likewise, we will multiply “x” times “-5” to get -5x. We will put that in the table.

   For our last column, we will begin by multiplying "3" times "x." The result is this.

   The last item of the table is gained by multiplying "-5" times "3."

   This leaves us with the following table of elements.

   This process has left us with four terms inside the table.

   There are two elements within the table that are like terms and they need to be combined together. These are the two like terms.

   If we write out these four elements horizontally, we get this polynomial.

x^2 + 3x - 5x - 15
   We need to combine the inner-most terms, which are the like terms. After adding the "3x" and the "-5x," we get this solution.

x^2 - 2x - 15
      ideo: Multiplying Polynomials
      uiz: Polynomial Multiplication

   For our next example, we will look at a much more difficult problem. We will multiply a binomial times a trinomial.

   Again, we will place these polynomials on the outside of a table. Place the first along the top and the second along the side.

   Now, we will begin the process of multiplying columns times rows to gain elements inside the table. Let's start with "x2" times "x" to get "x3."

   Multiply "x" times "4x" to get "4x2."

   Multiply "x" times "-7" to get "-7x."

   Multiply "2" times "x2" to get "2x2."

   Multiply "2" times "4x" to get "8x."

   Multiply "2" times "-7" to get the final element in the table, which is "-14."

   Alright. We have a completed table! We are almost done.

   Notice there are six elements within our table. These elements will be used to gain our solution.

   The six elements will be simplified to produce the solution. Notice, however, there are like terms that need to be combined. These like terms have like powers and have been color-coded below.

   Taking the elements of the table out and writing them horizontally, we get this unsimplified polynomial. Like terms have been color-coded.

   We will combine the like terms by adding "2" with "4" and then adding "8" with "-7." This is the solution.