Week 8- Pre Calc 11

November 6, 2017November 6, 2017 katelynw2015

For week 8 in Pre-calc 11 we learned to analize Standard form (vertex form). In my opinion this is the easiest form to graph becasue you are already given the vertexand with it you can find the x and y intercepts, the line of symmetry, and the domain and range. Each number in the standard form tells you something different about how it needs to be graphed.

Standard form: $y=a(x-p)^2+q$

a- determins the width of the graph. If a<1 then the graph will become wider and if a>1 then the graph will become thinner. A also determines if the graph will be opening up or down. If a is positive then the graph will open up, but if it is negative then it will open down.

p- determins the horizontal translation of the graph. If p is positive then the graph will slide left to the negative side. If p is negative then the graph will slide right to the postivive side. Basically you do the opposite of what p really is, so if you have p=-2 then you go +2 along the x axis.

q- determins the vertical translation of the graph. If q is negative then you slide the graph down the y axis into the negative side. If q is a positive then you slide the graph up the y axis into the positive side

This form is also known as vertex form because it gives you the vertex, which will we (p,q)

Example: $y=3(x-2)^2+1$

Just by looking at this I can tell that the graph is going to be skinnier, move 2 to the left (positive) and will also go up one along the y-axis.

There is no x-intercepts because y>1 and the graph is opening up which means the graph will not touch the x-axis

Week 7- Pre Calc 11

November 6, 2017 katelynw2015

In week 7 we started looking over properties of a quadratic function. In this lesson we determined the characteristics of a quadratic function and how to solve it.

We determined if a table of values represented a quadratic function. You could tell this by seeing how much y went up.

Example: 5,0,-7,-16,-27

to get from 5-0 you -5, to get from 0–7 you -7, and to get from -7–16 you -9. The pattern here is you are adding 2 to how much you subtract each time

A few of the first things I learned include; finding the vertex, the x and y intercepts, the line of symmetry, and the domain and range.

Example: $2x^2+8x+6$

Vertex: (-2,2)

x-intercepts: (-3,-1)

y-intercept: (6)

Line of symmetry (-2)

Domain: (XER)

Range: (y>-2)

Week 6-Pre Calc 11

October 18, 2017 katelynw2015

This week in Pre Calc 11we learned the 3 ways to solve quadratic equations. The 3 ways are;

1. Factoring

2. Completing the squares

3. Using quadratic equation

Factoring: this is the easiest way to solve quadratic equations

Example:

$x^2-9x-22=0$ > Start off with your equation

$x^2-11x+2x-22=0$ > find your 2 numbers that multiply to 22 and add to -9 (-11,2)

$(x-11)(x+2)$ > factor by grouping common term (look at week 5 post)

$x=11 x=-2$ > when you have your 2 bracketsyou take the opposite of both numbers and those would be what x equals, in this case the opposite of -11 is 11 and the opposite of 2 is -2

Completing the square: $(b/2)^2$

Example:

$x^2+4x+1=0$ > start with your equation

$(x^2+4x+4)+1-4=0$ > separate $x^2+4x$ and +1 by using brackets, from there you take whatever number is b (4), divide it by 2 and then square it. For this equation our number will be 4, you then put that number beside 4x and you also put the opposite of that number (-4) and put it beside 1

$(x+2)^2-3=0$ > you foil what’s in the bracket and take the copy bracket turn it into $(x+2)$ which is the same as (x+2)(x+2). You also solve outside the bracket so 1-4=-3

$\sqrt{(x+2)^2}=\sqrt{3}> you move -3 to the other side of the = sign by makeing it 3 and the � you’ve squareboth$ latex (x+2)^2$ and 3

$x+2=\sqrt{3}$ > by squaring $(x+2)^2$ it just becomes x+2

$x=-2\sqrt{3}$ > you then isolate x so you need to move 2 to the other side by making it -2

Week 5- Pre Calc 11

October 18, 2017 katelynw2015

This week we started a brand new unit, Factoring polynomials. I’m very happy we stared this unit because this was one of my favourite units to do last year in math 10. We spent t most of the day on Friday when we started the unit doing review. Once I did the first question everything came back to me. In my opinion the easiest factoring to do is when you have perfect square binomials like $x^2-100$ All you need to do is square root both numbers and put them in brackets. Example: $x^2-100$

$(x-10)(x+10)$

i got this answer because I put the square root of $x^2$ (x) at the beginning of each bracket. I then took the square root of 100 which is (10) and I put that at the end of each bracket. $x^2-100$ has a negative sign, that means that one of the 10 will be positive and the other one will be a negative becasue 10x(-10)=-100

When you are factoring trinomials like $x^2+7x+12$ This trinomial is an easy trinomial because it has $larex x^2 at the front$ In order Govett solve this trinomial you need to find 2 numbers that multiply to 12amd add to 7. In this case we would use the numbers 3 and 4 because 3×4=12 and 3+4=7. These are the steps I take to get the answer

$x^2+7x+12$ > we start with the original equation

&$atex x^2+3x+4x+12$ > we take our 2 numbers that multiply to 12 and add to 7 and we put them into our equation in replace of the 7 in order to expand

$x(x+3) 4(x+3)$ > we the night take the common factor out of each equation and put whatever is left over into brackets, in the step we should get the same numbers in both the bracket, I got (x+3)

$(x-3)(x+4)$ > the 2 brackets turn into one and you take the common factor from each equation and put them into a bracket (x+4)

Week 4-Pre Calc

October 5, 2017 katelynw2015

This week was filled with many new and challenging lessons. We went over Absolute Values, Roots and Radicals, Adding and subtracting Radicals, Multiplying and dividing Radicals, and solving Radical equations. From all these lessons the one I found the most challenging was Multiplying and dividing Radicals.

Generally the idea is to:

1. Multiply coefficients

2. Multiply radicands

3. Simplify

one questin that I struggled with INV the beginning of learning multiplication was

$2\sqrt{5}(3\sqrt{45}-8\sqrt{5}+\sqrt{20})$

I have to admit, when I first saw the question I had no idea where to start.

I soon learned that you need to multiple the Radical that’s on the outside of the bracket ( $2\sqrt{5}$ and multiply it with the Radicals inside the bracket.

after this step is complete you should have: $6\sqrt{5×45}-16\sqrt{5×5}+2(10)$

From the last equation all you need to do is solve everything inside the square root sign so you will end up with: 6(15)-16(5)+2(10)

Then all you need to do for the rest of the equation is solve so you should get:

90-80+20

=30

This was one of the first equations that we learned when started our multiply and dividing Radicals lesson. At the start it really confused me and looked very intimmidating, now when I see these types of questions I have no worries becasue I know how to solve them.

Week 2- Pre Calc 11

September 21, 2017September 21, 2017 katelynw2015

This week we had our Arithmetic and Geometric sequences and series test. One thing that I was still unsure about before the test was finding “r” when you are given $t_1$ and $S_\infty$ Once I asked questions to my classmates to get a better understanding I understood how to do it.

Line thig that I have really enjoyed so far is how we can rely on the people in our table group to help us when we need it. I was really struggling with this question and when I asked for help with my table group everyone was happy to help. With the he’ll hey gave me I now understand how to do questions like these and I also know that I can count on them for any other questions I may have.

Week 2- Pre Calc 11 (arithmetic and geometric sequences and series

September 20, 2017 katelynw2015

This past week we have learned many new things about geometric and arithmetic sequences and series. One thing that I have struggled with throughout the week has finding a term when you don’t have the difference.

Steps: The first thing you need to do is find the difference (d)

1. First I wrote out my formula

2. I then substituted the numbers in where they should be. I got “x” by subtracting the subscripts under “t” (8-3=5)

3. I then solved, trying to isolate the variable (d). I had to move 4 to the other side so I had to subtract it from one side which means I had to do the same to the other side so I did 34-4=30

4. Once I had 30 = 5d, I just had to divide 30 by 5 and 5d by 5 to isolate “d”

5. Once I did this I got 6=d which means I have my difference

Once I have my difference I can use it to solve and find any other term in the sequence