Sept. 27 Class

Today we reviewed what we learned so far when it comes to factoring.  I provided a handout that covered the major types of factoring.

You can find the handout here: Factoring Review

Here are some of the examples I worked out in class.

Examples of common factoring and group factoring.

Examples of decomposition and criss cross method.

Examples of perfect square trinomial.

Examples of difference of squares.

Homework: P. 120 #1, 3-6, 8-13, 15-19

Sept. 26 Class

Learning Goals: Understand how to factor using perfect squares and difference of squares.

Today we did the formative quiz and took it up in class.

Formative Quiz

Formative Quiz Solutions

Then we discussed some special cases of factoring.

Perfect square trinomials are ones that satisfy this pattern:

a²x² ± 2abx + b²

These types of trinomials can be factored into: (ax ± b)².

Example  Factor x² – 16x + 64.

Solution We notice that the first and last terms are perfect squares.  We check if it fits the pattern.

a = 1, b = 8, 2ab = 16.   This matches the pattern, so the solution is,

(x – 8)²

We could have found the solution using other methods, but this way is much faster.

Example  Factor 4x² + 20x + 25.

Solution We notice that the first and last terms are perfect squares.  We check if it fits the pattern.

a = 2, b = 5, 2ab = 20.   This matches the pattern, so the solution is,

(2x + 5)²

Difference of Squares refers to an expression with two perfect squares and a minus sign between the two.

a² – b²

In this case the factors are,

(a – b)(a+b)

Example Factor 4x² + 49.

Solution This fits the pattern, so the solution is (2x – 7)(2x + 7)

Example Factor 3x² – 27.

Solution Always common factor first if possible, 3(x² – 9)

Now there is a perfect square in the brackets.  3(x – 3)(x+3)

Example Factor 4x² – (3x + 1)².

Solution (3x + 1)² is a perfect square!      [2x – (3x + 1)][2x + (3x + 1)]

This simplifies to -(x + 1)(5x + 1).

Homework: p. 115 #2-12

Sept. 25 Class

Learning Goals: Understand how to factor complex trinomials of the form ax² + bx + c where a ≠ 1.

Today I gave a handout that explains the steps to decomposition.

Handout: Factoring Complex Trinomials

Please have a look at this handout and complete the questions on the back as practice.

Homework: p. 109 #2, 4-11

The solutions to the first unit text is how posted under handouts.

Sept. 24 Class

Learning Goals: Factor an expression in the form of ax² + bx + c.

Today we look at how to factor trinomials in standard form with the value of a = 1.

For example, factor x² + 10x + 21.

First you check if there are any common factors.  If there is, factor them out first.  Then I introduced what I call the MAN method.

What Multiplies to the last number?  ____ x ____ = 21
What Adds to the middle number? ____ + ____ = 10
The Numbers are: 7 and 3.

Therefore, the solution is x² + 10x + 21 = (x + 7)(x + 3).

Make sure to be careful of negative factors.

Example: factor x² – 7x – 18.

M: ____ x ____ = -18
A: ____ + ____ = -7
N: -9 and 2

The factors are then (x – 9)(x + 2).



Example: factor 3x² + 12x – 15.

There is a common factor of 3.

3(x² + 4x – 5)

Now we use the MAN method on the terms inside the bracket.

M: ____ x ____ = -5
A: ____ + ____ = 4
N: -1 and 5

The factors are then 3(x – 1)(x + 5).



Homework: p. 99 #2, 3, 5-13

Sept. 21 Class

Learning Goals: Understand common factoring.

Today I discussed how factoring is the opposite of expanding.

For now it might seem arbitrary, but in the coming weeks we will find that factoring will help us solve some tricky problems.

The first kind of factoring is called Common Factoring.  We do this by finding out what is a common factor among all the terms in an expression.  In other words, what can we divide every term by?

Example: Factor .

Solution: All the terms cal be divided by 2, so we can put a 2 out front and divide every term by 2.

We have to check if there are anything else that all the terms can divide by.  In this case we see that there is x in every term.  So we can go ahead and factor the x as well.

We could have factored the 2x in one step as well.  The key to know when to stop is to make sure there is nothing else you can divide all the terms by.

Here are a few more complicated examples.

Example: Factor .

Solution: There is a common factor of 30xy.

Example: Factor .

Solution: The common factor is (3x+1).  In the first term, if we divide (3x+1) by (3x+1) we are left with 1.  In the second term we are left with -2x.  Here is what the solution will look like:

Homework: P. 92 #2, 3cd, 5cd, 6def, 7-14

Sept. 20 Class

Learning Goals: Review some rules of algebra.

Here is the daily plan for this unit: Daily Plan for Unit 2

Today we reviewed some topics that will be very important in the upcoming unit.  First of all some vocabular,

Variable: the unknown value in an expression, x, y, z, a, b, c, ...

Coefficient: the number multiplied to a variable.
eg. In the expression -3x, the coefficient is -3 and the variable is x.

Like Terms: terms with the same variables and exponents.
eg. x, 3x, (-1/2)x are all like terms, but 2x² is not a like term because it it squared.

Binomial: expressions with two terms.
eg. 3x + 4  or  a – b

Trinomial: expressions with three terms.
eg. 2x² – 5x + 1  or   3m² + 4mn – 2n²



Next I reviewed some examples of how to expand and simplify expressions.

eg. Expand and simplify .

Solution:.

Remember the exponent rules:  
When you multiply, you add the exponents 
When you divide, you subtract the exponents 

eg. Expand and simplify .

Solution, you must remember to FOIL.

Multiply the First two terms.
Multiply the Outside terms.
Multiply the Inside terms.
Multiply the Last terms.

After FOIL you get, 

Finally, collect like terms to get: 

The rest of the homework questions involve using these rules in the right order and keeping yourself organized to get to the final answer.  Some of the problems might seem long and daunting, but remember to keep applying these rules and you'll do fine.

Homework: p. 85 #2cd, 3bc, 5ade, 6, 7d, 8-13

Sept. 19 Class

Congratulations on completing your first summative test!

To get a head start on the next unit, have a look at page 70 in your text book problems #1-14.

Sept. 18 Class

Today was a work period to ask questions and to prepare for the summative tomorrow.

Homework
P. 70 # 1-7
Handout of addition practice questions.

There will be an extra help session after school.  Good luck in your studies!

Sept. 17 Class

Your first summative test is this Wednesday!

Today I did quick review of topics that you'll need to know to do well on the test.

What is a Function
 -       Be able to define what a function is (...each 'x' value is related to ONE 'y' value.)
 -       Be able to recognize a function from a non-function (vertical line test)
 -       Be able to evaluate functions using functions notation.

Eg. Evaluate f(2) + f(1) for the function f(x) = x² + 1.

f(2) = (2)² + 1               f(1) = (1)² + 1
       = 5                               = 2

f(2) + f(1) = 5 + 2 = 7

Determine if a Function is Linear or Quadratic

-       Look for x² in the equation (x, degree 1, is linear.  x², degree 2, is quadratic).

-       See if the graph is curved (straight line is linear, parabola is quadratic).

-       Use 1^st and 2^nd differences.

-       Functions can be linear, quadratic or neither.

Domain and Range
 -       Be able to state the domain and range of different functions
 -       Know when to list points and when to use set notation.

Transformations

-       Be able to describe transformations in words.

-       Be able to sketch a graph.  Use the vertex, then the step pattern.

-       Be able to sate the equation from a graph.

Word Problems

-       The domain and range are sometimes restricted depending on the situation.

  -       Graph by table of values if necessary.


Homework Problems

P. 68 #1-10, 12-14

Sept. 14 Class

Learning Goals: Understand how to denote the domain and range of a quadratic equation.

Today I handed back the formative quizzes.  You can download a copy of the solutions here:

Formative Quiz 1 Solutions

Next I reviewed how to write down the domain and range of various functions. If the function is given as a set of ordered pairs, we just need to list the values,

Example: What is the domain and range of the following function,

(2, 3)  (3,6)  (4, 8)  (5, 3)  (6, 3)

Solution: D = {2, 3, 4, 5, 6}  R = {3, 6, 8}

If we have a graph the domain and range is a bit more tricky.

Example: What is the domain and range of the following function?

Solution: The domain, x,  can be any value.  In set notation we write this as,

In words, we read this as "The domain is x in the real numbers".

The range, y, can only be positive.  In set notation we write this as,  

In words we say, "The range is y in the real numbers, such that y is greater than 0."

Next I worked out a word problem from you textbook which showed that the domain and range of a function can be restricted in a physical situation (ie. word problems).

For question 7 on page 64, the domain and range are

Because the rock falls from 80 meters to 0 meters in a time of 0 seconds to 4 seconds.

Homework

P. 63 # 1-12

(Hint: for many of these questions you might want to graph the equations.  Use your favourite graphing app or this online graphing calculator.)

Handout: More practice on the roles of a, h and k.

Sept. 13 Class

Learning Goals:

• Sketch a quadratic function with multiple transformations.
• Write an equation from a graph of a quadratic function.

Today we did a formative quiz on function notation.  I will get these back to you as soon as I mark them.

Next I discussed the order in which we should perform transformations.  When you have multiple transformations, you should state any Stretches first, then Reflections then Translations.  Remember SRT.

I then worked out some examples.

Next I discussed how to write the equation of a quadratic function from looking at the graph.

Example Write the equation for the following quadratic function.

Solution  First we look at the the vertex (3, 0).  That means we have h = 3 and k =0.  Next we count one unit over from the vertex and we see that the graph drops down by 2.  That means a = -2.  Thus the equation of the graph is:

f(x) = -2(x–3)²

HINT: for homework questions you can check if you got it right by using a graphing calculator.  Here is a link to online graphing calculator.

Online graphing calculator

Try entering the above equation and see if you get the same graph (make sure you use the * button between the 2 and the bracket to multiply.)

Homework

P. 56 #1 - 11

Sept. 12 Class

Learning Goals: Understand the transformations of a quadratic equation.

Today I gave a handout the explained all the possible transformations on a quadratic function.

Handout: Transformations of the Quadratic Function

To understand transformations and to sketch the graph of parabolas, you must know how to sketch the parent function, f(x) = x².

 Make sure you know those 5 points!

Example: For the function , state the transformations and sketch the graph.

Solution: The transformations are, 

a = -1   opens down, reflection (no stretches or compression)

h = -3   translation 3 units to the left

k = -4   translation 4 units down

Start by plotting the vertex (h, k) = (-3, -4) and then use the step pattern 1, 3, 5... to plot the successive points (ie, 1 over down 1, 1 over down 3, 1 over down 5...).

Example: For the function , state the transformations and sketch the graph.

Solution: The transformations are, 

a = 2    opens up, stretched by a factor of 2

k = -3   translation 3 units down

Start by plotting the vertex (h, k) = (0, -3) and then multiply the step pattern by 2 to plot the successive points (ie, 1 over up 2, 1 over up 6...).

Homework: P. 47 #1-12

Sept. 11 Class

Learning Goals: Understand the transformations of the quadratic function.

Today we looked more closely at the quadratic function: f(x) = x².

This function is the basic parabola with the follow graph, domain and range.

I then introduced how to use a graphing calculator to plot this graph.

Those who are using iPhones, iPods or iPads might want to try downloading Quick Graph.

Those who are using Android phones or tablets might want to try downloading Algeo Graphing Calculator.

These are free apps that can help you graph (sorry, I've never used a Blackberry before, but I'm sure they have free graphing apps as well.)

You then split off into groups to examine different transformations of the parabola.

Refer to page 38.

Group 1 graphed: f(x) = x² + k

Notice that the value of 'k' made the parabola move up (positive k) or down (negative k).
This is called a vertical translation.

Group 2 graphed: f(x) = (x–h)²

Notice that the value of 'h' made the parabola move left (negative h) or right (positive h).

This is called a horizontal translation.

Group 3 graphed: f(x) = ax²

Notice that the value of 'a' made the parabola flip over if it was negative.

This is called a reflection.

Notice also that the value of 'a' made the parabola stretch or compress.

Putting it all together we have the function:

Some of you recognized that this is the vertex form of a quadratic equation, good job!

We will soon be able to graph these parabolas without using a graphing calculator!

Homework:

p. 40 #1 (Further Your Understanding)

p. 37 #1-7

Sept. 10 Class

Learning Goals: Understand function notation.

Today I discussed how to describe functions using function notation.  The symbol:

f(x) 

is what we use to denote a function.  The f is the name of the function, the brackets indicate what the input should be.  If we input an x value, what the function gives us is the corresponding y value.  Therefore, this symbol is also used in place of the y in an equation:

y = f(x)

From now on, when we look at the equation of a function, we use f(x) instead of y.

Example:  Given the function f(x) = {(1, 2), (3, 5), (4, 6)}, what is the value of f(3)?

Solution: f(3) = 5.

This tells us that when x is 3, y is 5.

Example: Use the given graph to evaluate f(3) and f(-1).

Solution: Looking at the graph, when x = 3, we have y = 4.  Therefore the solution is:

f(3) = 4

Also from the graph, when x = -1, we have y = 0.  Therefore, the solution is:

f(-1) = 0

Example: For the function f(x) = 6x + 3, what is f(1/2)?

Solution: We substitute in 1/2 in place of x.

The solution is f(1/2) = 6.

Homework: 

P. 25 #10

P. 32 #1-9, 10v, vi, 11v, vi, 12-15

Sept. 7 Class

Learning Goals: Distinguish between linear and quadratic functions.

We began today by reviewing some of yesterday's homework.  Be sure you are clear on what a function is.

Then I discussed how to use set notation to write down the Domain and Range of a function.

If the function is given as a set of ordered pairs, we just need to list the values,

Example: What is the domain and range of the following function,

(2, 3)  (3,6)  (4, 8)  (5, 3)  (6, 3)

Solution: D = {2, 3, 4, 5, 6}  R = {3, 6, 8}

If we have a graph the domain and range is a bit more tricky.

Example: What is the domain and range of the following function?

Solution: The domain, x,  can be any value.  In set notation we write this as,

In words, we read this as "The domain is x in the real numbers".

The range, y, can only be positive.  In set notation we write this as,  

In words we say, "The range is y in the real numbers, such that y is greater than 0."

Next we talked about linear and quadratic functions.  These are two kinds of functions that we will be studying a lot of throughout the course.

If you are given a function as a table of values, you can tell if it is linear, quadratic or neither, by finding the first or second differences.

The first differences are the same, therefore this function is linear.

The second differences are the same, therefore this function is quadratic.

You can also tell by looking at the equation.  If the equation has degree 1, then it is linear.  if it has a degree of 2, it is quadratic.

Example: y = mx + b

This is linear because it has degree 1.

Example: y = ax² + bx + c

This equation is quadratic because of the x² it has a degree of 2.

For homework, please work on page 24, #1-9 (photocopied).

Sept. 6 Class

Learning Goals: understand what a function is.

Today we began by talking about what is a relation.

    Relation: a set of ordered pairs.

In a way it, is a set of numbers that go together.  A function is a special kind of relation

    Function: a relation in which every "x" value has only one "y" value.

I demonstrated this by asking a few students to participate in a demonstration.  We had 3 students represent 'x' values and 3 students represent 'y' values.  In a function, every student who was an 'x' knew exactly which 'y' value they were partnered up with.  When a relation was not a function, we had an 'x' student who had multiple 'y' partners and was confused as to who they belonged with.

Then I showed some examples of mathematical relations that were functions or not functions.

Examples:

Mapping diagrams

The one on the left is a function.  The one on the right is not because x = 2 has two possible y values.

Ordered Pairs

(2, 4) (4, 8) (5, 10) (6, 12)  is a function

(2, 4) (2, 6) (3, 9) (4, 7) is not a function because the value of x = 2 is related two possible y values.

Graphs

Make sure you know the vertical line test!

Equations

y = x + 2    is a function (every 'x' gives one 'y' value).

y = ± x       is not a function (x = 1 can give y = +1 or y = -1)

For homework please work on P. 13 #1-16 (photocopied)

Sept 5. First Day of Class!

Welcome to Functions and Applications!

Learning Goals:

1. Understand the expectations for the course.
2. Review essential skills: Evaluating and graphing by table of values.

Here are some important handouts from today:

• Course Outline (sign and return bottom half)
• Student Information Form (complete and hand in)
• Unit 1 Plan

______________________________

We then discussed what an ideal classroom Looks Like, Sounds Like and Feels Like:

LOOKS LIKE

Organized, tidy, clean

Proper use of technology

SOUNDS LIKE

Listening, quiet (music on personal devices)

No cell phones going off.

FEELS LIKE

Everyone Understands, on the same page

Warm

Safe, No put downs, no bully

These are all great ideas!  We will add to these items as more ideas come up throughout the course.

______________________________

Next we began to review some material that will be essential to this course.



Example: Evaulate   when
Solution: 

Example: Sketch  using a table of values.
Solution: 


We then plotted the ordered pairs on a graph and got a parabola.

Now try questions #1-12 from page 2 of the textbook (photocopied for you).

See how much you remember from last year and be prepared with questions to ask me tomorrow.