Week 14 – Pre-calculus 11

This week in Math 11 we learned how to simplify rational expressions. We do this by cancelling out what is equivalent. The first thing we do is figure out our non-permissible values, these are values that we cannot use because if we do our denominator will equal 0 and a math rule is that we can’t have a 0 in our denominator. Then we use the different math skills we have learned to first simplify the numerator and denominator, then once everything is simplified then we can cancel out what is the same or something common.


\frac{x^2 +13x + 36}{x^2 - 81}
  • right now due to the way our equation is written we can’t say what our non-permissible values are
  • so first thing we can do is factor
  • now that it is factored we can see that x cannot equal -9 and 9 because our rule is that we cannot have a zero inn our denominator and we we use those numbers as x’s we will get a zero
  • we see that we can something in the numerator and denominator which is the (x+9)
  • so we would cancel both of those out and be left with our final answer because we cannot reduce anymore

And as our final answer we would have our simplified expression and what our non-permissible values are.


x cannot equal -9 and 9

Totem Post Writing

Post Writing:

  • When Canadians see something, they don’t like they just remove to get it out of their sight. Just like in the story they just remove the totem pole and but it in the basement. I think that the totem poles continuing to return in the story is a metaphor for the first people fighting back saying that they’re not going to leave because just like in the story with the totem poles, the First people came to Canada first so why should they have to leave. I feel that even in the end it’s not a true resolution that works for both sides. They just decide to ignore the totem poles and eventually forget about them. So, is this how Canadian history treated the First Peoples? History has shown that we have just ignored the first nations and have just taken or removed from what is theirs.

Week 13 – Pre-calculus 11

This week in Math 11 we learned how to graph reciprocal linear functions. We do this by first reciprocating our function, then we graph the original function, then we find our invariant points on our graph which are always 1 and -1 and we graph our hyperbola that intersects through our invariant points. The hyperbola never touches our asymptotes which is y=0 and it can vary for our vertical asymptote depending on our graph.


y=-3x + 6

  • the first thing we do is reciprocate our graph which is pretty easy, we know that technically it is over 1 so we just flip it.

y = \frac{1}{-3x +6}

now that we have our reciprocal we can graph our original function

  • we see that our y-intercept is 6 and our slope is -3

Now we find our invariant points on our graph which is -1 and 1

Now that we know our invariant points we can graph our hyperbola through those points

  • In this case our vertical asymptote would be x=2 because our vertical asymptote is always right between our invariant points and and horizontal asymptote is always going to be y=0(in grade 11 at least)
  • so our hyperbola intersects our invariant point but never touches our asymptotes

Week 12 – Pre-calculus 11

This week in Math 11 we learned how to solve absolute value equations algebraically. When solving with an absolute value and or a square root you may come across extraneous solutions which are solutions that are not true solutions of the equation given. Algebraically we can find out whether a solution is extraneous by verifying and inserting the “x” value back into the original equation to see if the equation is true or to see whether both sides equal each other.

3 = |2x + 5|

  • first thing we do is separate the equation
  • because they original equation uses absolute value we times everything by a negative to make it to make it positive.

3 = 2x + 5

3 = -(2x + 5)

  • now that we have our equations we solve each of them

3 = 2x +5

3 – 5 = 2x

-2 = 2x

\frac{-2}{2} = \frac{2x}{2}

x = -1

Now for the second equation:

3 = -(2x +5)

3= -2x – 5

3 + 5 = -2x

8 = -2x

\frac{8}{-2} = \frac{-2x}{-2}

x = -4

Now we have to make sure these solutions are not extraneous by verifying:

  • to do this we insert each solution back into the original equation

x = -1      x = -4


X = -1: Verify

3 = |2x + 5|

3 = |2(-1) + 5|

3 = |-2 + 5|

3 = |3|

3 = 3

X = -4: Verify

3 = |2x + 5|

3 = |2(-4) + 5|

3 = |-8 + 5|

3 = |-3|

3 = 3

We see that both solutions make the equation true because both sides are equal.


Week 11 – Pre-calculus 11

This week in Math 11 we learned about how to graph linear inequalities with two variables. By graphing the linear inequality we can use a test point to find out which side of the line has points that will work as a solution for our inequality to make it true. When graphing inequalities we need to know if the line we are graphing is a broken line(dotted line) or a solid line. A broken line tells us that the points on the line are not solutions, so every set of coordinates on the true side excluding the line is a solution. A solid line tells us that the coordinates on the line are also solutions, so every set of coordinates on the true side including the line is a solution. To know when the linear inequality is a solid or broken line we look at the sign of the inequality given; for example, greater than(>) and less than(<) signs are both broken lines. Greater than or equal to(\geq) and less than or equal to(\leq) signs are solid lines. Now to graph out inequality we have to see what we know first just by looking at the inequality without doing anything to it.

Ex.) y \leq 2x + 5

First thing is that we can see that the sign is less than or equal to so we know our line is going to be solid. Just by looking at this we can see that it is linear because the highest degree of our variables is 1, also knowing the information from grade 10 we can see that this inequality is in the form of            “y = mx + b”, where “m” is equal to the slope and “b” is equal to the y-intercept. So knowing this we can graph our y-intercept which would be (0, 5)

Now we would use our “m” which tells us the slope to finish the line which is 2.

We know that the slope = \frac{rise}{run}

So to make 2 and fraction we just put it over 1 and use that as our rise and run.

Now that we have our line we can use a test point to figure out which side has the solutions

  • To make it easy we can use the most easy test point which is (0, 0)
  • Now we substitute the zeros into our original inequality
  • y \leq 2x + 5
  • 0 \leq 2(0) + 5
  • 0 \leq 5
  • Now we see that the solution is true which means that everything including the line on the side of the test point we used is a solution for this inequality.
  • to show this we shade in that side

This is how we show our answer for these types of questions.

Week 10 – Pre-calculus 11

This week in Math 11 we learned about sign charts which help us find our roots by factoring then once we find our roots we use a number line and our roots to figure out different signs of our parabola and to know which numbers we can use to replace the variable and solve to make the equation true.

Example: 4x \leq 2 (x^2 - 3)

  • This is our equation so, the first thing we do is to factor and find our roots.

4x \leq 2 (x^2 - 3)

4x \leq  2x^2 - 6

0 \leq 2x^2 - 4x - 6

0 \leq \frac{2x^2}{2} - \frac{4x}{2} - \frac{6}{2}

0 \leq 2(x^2 - 2x - 3)

0 \leq 2(x – 3)(x + 1)

X_1 = 3          X_2 = -1

  • Now that we know our roots we can use a sign chart to figure out information about which numbers make the equation true.
  • We can find out whether our parabola opens up or down.

 –                                      –                   0                +

(x – 3): x = 3                                   -1                                           3

  • For this one our zero would be 3 because our”x” is equal to zero
  • on a number line everything to the right of “0” is positive and everything to the left is negative.


    –               0                     +                                          +

(x + 1): x = -1                                  -1                                           3

  • For this one our zero would be -1 because our “x” is equal to zero

Now if we put these together we can find out to number to make our equation true.


                                        –                                       –                   0                +

(x – 3): x = 3                                   -1                                           3

                                        –               0                     +                                   +

(x + 1): x = -1                                  -1                                           3

  • Now we look at each column of the signs and multiply them

                                        –                                       –                   0                +

(x – 3): x = 3                                   -1                                           3

                                        –               0                     +                                   +

(x + 1): x = -1                                  -1                                           3

(-)(-) = +                          (-)(+) = –                     (+)(+) = +

Positive(+)                      Negative(-)                   Positive(+)

Now if we look back at our original equation: 4x \leq 2 (x^2 - 3)

  • we see that 4x is less than or equal to $ 2 (x^2 - 3)
  • so to figure out which section makes the equation true we would get a number from each section and plug it in to our original equation to see whether it is true or false

                                        –                                       –                   0                +

(x – 3): x = 3                                   -1                                           3

                                        –               0                     +                                   +

(x + 1): x = -1                                  -1                                           3

(-)(-) = +                          (-)(+) = –                     (+)(+) = +

Positive(+)                      Negative(-)                   Positive(+)

Section 1: use -2               Section 2: use 0            Section 3: use 4

  • NOTE: Always use “0” as a test number if possible.

Section 1: use -2

4x \leq 2 (x^2 - 3)

4(-2) \leq 2 ((-2)^2 - 3)

-8 \leq 2 (4 – 3)

-8 \leq 2 (1)

-8 \leq 2


  • -8 is less than 2

Section 2: use 0

4x \leq 2 (x^2 - 3)

4(0) \leq 2 ((0)^2 - 3)

\leq 2 (0 – 3)

\leq 2 (-3)

\leq -6


  • 0 is NOT less than -6

Section 3: use 4

4x \leq 2 (x^2 - 3)

4(4) \leq 2 ((4)^2 - 3)

16 \leq(16 - 3)

16 \leq(13)

16 \leq 26


  • 16 is less than 26

Now that we know which ones are true and which ones are false we can write our answer.

  • Because our sections that are true are the outer sections we write our answers for “x” separately.
  • if the true section was the middle section and the outer sections were false we would combine our answer for “x”.
  • In our original equation we used the “equal to” sign as well , so we would use this for our answer as well.


\leq -1


\geq 3

Week 9 – Precalculus 11

This week in precalc 11 we learned about how to find maximum values and revenue using quadratic functions, we can also find different variables depending on the word problem for example if there’s a question about area then the question may also ask for the length and width and we have to use the quadratic function to solve for the side lengths along with the maximum values.

Example: A company can sell 56 phones, each one being sold for $300. For every $30 the price increases, the number of phones sold lower by 3. What is the maximum revenue that the company can make by selling the phones?

The formula we use for revenue is: (price)(# of sales)

so knowing this lets see what we know

  • we have 60 phones
  • each one is sold for $300
  • the price can increase by $30
  • But then the number of phones drops by 3
  • Were trying to find the maximum value

know we need to use a variable to find out our maximum revenue

The variables will be how many phones we decrease by and how much we increase the price

So lets plug in the number we know into the formula

(300 + 30x)(60 – 3x)

Now we can solve for the variables by completing the square but first we distribute

(300 + 30x)(60 – 3x)

18,000 – 900x + 1,800x – 90x^2

-90x^2 + 600x + 15,000

Now we can complete the square and find our vertex which will give us our “x” value and maximum revenue.

\frac{-90x^2}{-90} +\frac{900x}{-90} + 18,000

-90(x^2 – 10 + 25 – 25) + 18,000

-90(x - 5)^2 + 18,000

Vertex is : (5, 18,000)

Our “5” shows our value for “x” and the 18,000 shows our maximum value but because in this question it’s asking for revenue it is our maximum revenue.

Sometimes a question may ask, “which price for a phone will maximise the revenue?”

and what you do is go back to our formula with the numbers

(300 + 30x)(60 – 3x)

and in this case the “(300 + 30x)” talks about the price for a single phone so we plug in our “x” value we got from our vertex and solve to find our the price of one phone that would maximise the revenue.

(300 + 30x)

(300 + 30(5))

(300 + 150) = 450

so this means the price of $450 for one phone will maximise the revenue in this question.


Week 8 – Pre-calculus

This week in Precalc 11 we learned about graphing quadratic functions and how to analyse them. We do this to see what we know just by looking at the function. Just by looking at a quadratic function we can find the y-intercept, whether the parabola is opening up or down and knowing whether we have a minimum or maximum vertex. That is what we can find when we only see the function but if we see the function graphed we can find a lot more. Along with what we can find with just the function we can find the axis of symmetry, the coordinates of the vertex, the x-intercept(s), the domain and the range.

Image result for parabola

The way we find out all this information just by looking at the function of this graph is.

y=ax^2 + bx + c

  • we see that the parabola is opening up so that would mean that the coefficient of $ax^2$ has to be positive because it is the leading coefficient of the highest term.
  • The way we find out the y-intercept is that it’s the point on the graph where “x” is zero and the parabola crosses through the y-axis. The point where the parabola crosses is the intercept.
  • so in our equation if we were to make all of our X’s zero then that would just leave us with “c” which would be our y-intercept.

Image result for parabola

Now if we look at this parabola on a graph we can find out a lot more.

  • when we tried to find out the y-intercept we knew that the “x” have to be zero and the parabola had to cross the y-axis but now if we want to find the x-intercept we do the opposite. the “y” value must be zero and the parabola must cross the x-axis and the point where it crosses is our x-intercept.
  • The vertex can be the highest or lowest point on a parabola. The vertex on this graph would have (3, 1) as it’s coordinates. The way we see if it’s a maximum or minimum vertex is whether the parabola is opening up or down. If it opens down it would have a maximum vertex, if it was opening up it would have a minimum vertex.
  • The axis of symmetry is when you could cut the parabola in half and both sides would be symmetrical. On this graph if we cut vertically on the “3” of the x-axis it would cut the parabola perfectly in half and both sides would be symmetrical, so to show this we say that x=3 for the axis of symmetry.
  • The domain and range talk about the restrictions that the x and y axis have. Range is for y and domain is for x. In the graph we see that along the x-axis the parabola has negative points and positive points so we would say that “X” is an element of all real numbers.
  • For the Range we see the parabola never crosses the negative side and is only positive so knowing that we have a minimum vertex and the y for the vertex coordinates is 1 we can say that y is greater or equal to 1.

Week 7 – Precalc 11

This week in Precalc 11 we learned about discriminant’s. A discriminant is a part of a formula that helps you figure out whether your quadratic formula is going to have no real roots, 1 real root or 2 equal real roots, or if it has 2 unequal real roots. We figure this out by plugging in number into the formula and seeing if it is less than 0, equal to zero, or greater than zero.

  • The original formula we get the discriminant from is:

Image result for quadratic formula

  • We take this formula and only use the b^2 - 4ac
  • To know which numbers go in place or the variables we look at the quadratic equation.

ax^2 + bx + c = 0  NOTE: it must always equal zero, this way you know which one is “a” and “b” and “c”. AND MAKE SURE YOU DON”T FORGET YOUR SIGNS.

Example: 5x^2 - 8x + 6 = 0

  • so just by looking we know that it equals zero.
  • and that a=5, b=-8, and c=6
  • so now that we know these things we can plug them into our discriminant formula.


  • b^2 - 4ac


  • 1) (-8)^2 - 4(5)(6)
  • 2) 64 – 120
  • 3) -56 < 0


  • We see that -56 is less than zero which means they’re no real roots.
  • If the discriminant was greater than “0” the quadratic would have 2 distinct real roots
  • If the discriminant was equal to “0” the quadratic would have 1 real root or 2 equal real roots.