Week 8 – Analyzing the Standard Form

April 8, 2018April 9, 2018 keishan2015

This week in Pre-calculus 11 I learned a lot about graphing quadratic functions and how to analyze them. Something that stuck with me was how to analyze quadratic functions using the standard form: $y=a{(x-p)}^2+q$ .

Things to know:

General Form: $y=ax^2+bx+c$ , when using the general form, c tells you what the y-intercept is, and a tells if the parabola will open up or down (if a is negative it will open down, if a is positive it will open up) as well as if the parabola will be stretched or compressed.
Standard Form: $y=a{(x-p)}^2+q$ , when using the standard form, a will tell you if the parabola will open up or down, if the parabola will be stretched or compress and the scale. In the equation, p will tell you the horizontal translation and q will tell you the vertical translation. The vertex will always be represented by (p,q).
Scale: The scale of a parabola can be found by looking at the coefficient in front of the $x^2$ . If the coefficient in front is equal to one then the parabola will have a scale of 1-3-5 (to check this: Start at vertex go up 1 over 1, up 3 over one, up 5 over one etc. if it matches then the parabola has a scale of 1-3-5), having a scale of 1-3-5 also means that it is congruent to the parent function $y=x^2$ , it can also have a scale of 2,6,10.
Vertex: The highest or lowest point of the parabola.
Axis of Symmetry: intersects the parabola at the vertex. Splits the parabola perfectly in half.
Minimum & Maximum: used to indicate if the vertex is at the highest or lowest possible point. If the parabola is opened up it will be at its lowest height (minimum), if the parabola is opened down the parabola is at its highest point (maximum).
Domain: All possible values of x, will always be $x\in\Re$
Range: All possible values of y

Example:

Firstly since our equation is written in the proper format ( $y=a{(x-p)}^2+q$ ) we should write down what we know and put it into the formula. We know: a =1, p =2, and q =6 [ $y=1(x-2)^2+6$ ].
To start off it is best to find what the vertex is since it tells you a lot about a quadratic function. To find the quadratic function we use what we know: (p,q) –> (-2,6). If we know that the vertex is (-2,6) then we know that the axis of symmetry is: x= -2 (the axis of symmetry is located where the vertex is). If we then take a look at the value of a we know that the parabola of this equation will open up and have a scale of 1-3-5 (congruent to parent function: $y=x^2$ ). We know that it will open up because the value of a is positive and we know it will have a scale of 1-3-5 because the value of a is 1. If we know that the parabola opens up then we know that it is at its lowest point (minimum) where y =6. Finally based on what we know we can write down the domain and range. Since this is a parabola the domain will always be $x\in\Re$ . By looking at the function we know that the range will be y≥6 because the lowest that the parabola goes on the y axis is 6, and since it is opening up, all possible values for y would be greater than or equal to 6.

Example:

Firstly since our equation is written in the proper format ( $y=a{(x-p)}^2+q$ ) we should write down what we know and put it into the formula. We know: a =-2, p =-3, and q =5 [ $y=-2(x+3)^2+5$ ].
Since we have put what we know into the formula we can write what the vertex is: (p,q) –> (3,5)
If the vertex is (3,5) then we know that the Axis of symmetry is x = 3
Since the value of a is -2 we know that the parabola will open downwards and will be at its highest point (maximum). We also know that it will have a scale of 2-6-10 (because it has a value of 2 and not 1, you double the scale of 1-3-5 –> 2-6-10) and not congruent to the parent function.
We also know that the domain is $x\in\Re$ and the range is y≤5 (highest point on the y axis is 5, and the parabola is facing downwards, there would be no greater value than 5).

Core Competencies

April 3, 2018 keishan2015

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Week 7 – Discriminants

March 18, 2018March 19, 2018 keishan2015

This week in Pre-Calculus 11 we learned about discriminant’s and how they can be useful. A discriminant is a portion of the quadratic formula that you can use to predict how many solutions you will have, based on if the value is positive, negative, or equal to zero. When graphing the discriminant represents how many times that the parabola touches the x axis. You should also note that finding the discriminant is NOT a way of solving quadratic equations.

Note, if the value of the discriminant is:

Positive: then that means that you will have to solutions
Negative: then that means that there are no solutions
Equal to zero: then that means that there is one solution

The formula for finding the discriminant:

*highlighted in yellow*

Example:

Touches the x axis in two spots

Firstly, since our equation is already written in the proper format of ( $ax^{2}+bx+c=0$ ), we need to write down the values for a, b, and c. (a = 2, b = 9, c = -4)
Next, we input our values into the formula.
Use BEDMAS to simplify the equation until you are left with a value.
Since the number we are left with is 113, and it is positive, then we know that this quadratic equation will have two possible solutions.

Example:

Doesn’t touch the x axis

First we need to identify the values of a, b, and c. (a = 6, b = -2, c = 7)
Next we input the values that we know into the formula.
Use BEDMAS to simplify to a value.
Since the value that we are left with is 164, and it is negative, we know that that this quadratic equation will have no solutions.

Example:

Touches the x axis in one spot

Firstly, since our equation is not in the proper format we need to rearrange it so that it is.
Now that the equation is in the proper format we can write down the values for a, b, and c. (a = 1, b = -8, c = 16)
Input the values that you know into the formula.
Use BEDMAS to simplify to a value.
Since what we are left with is zero, we know that this quadratic formula will have one solution.

Week 6 – Quadratic Formula

March 11, 2018March 12, 2018 keishan2015

This week in Pre-calculus 11 we learned a number of different ways to solve quadratic equations. One method that I liked was the quadratic formula. The quadratic formula can be used to solve any equation (rational, irrational, etc.), even ones involving fractions!

The quadratic formula is based off of the values in a general quadratic equation:

$ax^{2}+bx+c=0$
In this equation a is not equal to zero, a≠0

The quadratic formula:

Example:

Since the equation is set up in the pattern that we need it in we can find the values of a, b, and c. (a=2, b=-10, c=12)
since we have the values for a, b and c we can input them into the formula. (the two negatives in front of the 10 cancel each other out and make it a positive).
After writing in the values you need to use BEDMAS to simplify the radicand in the numerator.
You then need to check if you can simplify the radical. We know that four is a perfect square, so we can write it as a whole number ( $\sqrt{4}=2$ ).
after simplifying the radical you check to see if any of the coefficients have anything in common so that we can simplify the fraction. In this case all of the coefficients are divisible by 2.
Since our answer can’t be further simplified we leave it as is.

Example:

Since our equation isn’t written in the proper format we need to rearrange it. We do this by moving the 5 over to the left side.
Now that it is written in the proper format we can find the values of a, b and c. (a=1, b=3, c=-5).
Next you need to input what you know into the formula.
Simplify the radical using BEDMAS.
Check to see if the radical can be simplified. $\sqrt{29}$ can’t be written as a mixed radical.
Check if the coefficients have a common factor. (-3, 1, and 2 have nothing in common).
Since we can’t further simplify the answer we just leave it as is.

Week 5 – Solving Radical Expressions

March 4, 2018March 5, 2018 keishan2015

This week in Pre-calculus 11 we learned about radical expressions and how you would solve them. We learned that in order for you to solve a radical expression you must understand that square roots and squaring are inverse operations. Inverse operations are opposite operations. ex. addition is the opposite of subtraction.

Before solving a radical expression, you must recall:

What you do to one side you must do to the other
You need to isolate in order to get a variable by itself
When you divide by a negative number in a inequality you have to flip the sign (this is only important when you are writing your restrictions)

Note: when writing your restrictions you must write whatever is under the square root sign and isolate to find the restrictions for x.

ex. $\sqrt{2x+6}$ = 2

Restriction:

2x+6 ≥ 0
2x ≥ -6
x ≥ -3

Example:

First, we need to get rid of the $\sqrt{}$ sign. In order to do this we put brackets around it and square it. Since we square the left side we need to square the right side as well.
In the expression x is already isolated we don’t need to move anything around; x = 36.
Since all expressions involve variables, after finding a value for x we need to write a restriction. Since it is a square root x needs to be greater than or equal to zero, x ≥ 0. We then double check that our value for x fits into our restriction.
After solving for x and making a restriction you then need to do a check to make sure that you got the correct value for x. All you do is input your value for x into the expression and solve it. Both sides need to equal the same number, if not, you need to go back and double check that you solved the expression correctly.

The steps for solving an expression are:

Solve for x
Write a restriction
Check

Example:

Since there is a coefficient of 4 in front of the $\sqrt{x}$ ,we divide both the left and right side by 4. By doing this it makes the expression easier for us to solve. $\frac{8}{4}=2$ .
We then square both sides to get rid of the $\sqrt{}$ .
Since we have found the value of x, we need to write a restriction. Since it is a square root x needs to be greater than or equal to zero, x ≥ 0. Check that your value for x fits into the restriction.
Check.

Similar Example:

Example:

Sometimes there will be instances where there is No Solution to the expression. These expressions are referred to as Extraneous Solutions. You will be able to tell if an expression is an extraneous solution if the square root in a question is equal to a negative number. The reason that this type of expression will have no solution is because it isn’t a true root. No square root should be equal to a negative number just like how you can’t have a negative number as the radicand.
If you come across an expression where the square root is equal to a negative number, simply write no solution.

Week 4 – Multiplying and Dividing Radicals

February 25, 2018February 26, 2018 keishan2015

This week in Pre-Calculus 11 we learned how to simplify expressions involving radicals. We learned how to add/subtract and multiply/divide radicals.

Multiplication:

When multiplying radicals all you need to do is multiply the coefficients of the radicals by each other and the radicands by each other. After doing that you check to see if the radicand can be simplified, and simplify if possible.

Example:

First, in this expression you just multiply the coefficients and radicands together.
You then simplify the radicand if possible. Since the radicand is 32, we look for any perfect squares that could be a factor of it. In this case $3\sqrt{32}$ is the same as ( $\sqrt{16}$ )( $\sqrt{2}$ )
If we know that 16 goes into 32 we take the square root of 16 (4) and then multiply it by the coefficient in front of the radicand. 12(4) = 48.

Example:

First, you need to simplify the expression that is in the brackets before multiplying. To do this all the radicands need to be the same.
Both $3\sqrt{24}$ and $\sqrt{54}$ can be simplified. $3\sqrt{24}$ can be simplified to $6\sqrt{6}$ (the perfect square 4 goes into 24 six times). $\sqrt{54}$ can be simplified to $3\sqrt{6}$ (the perfect square 9 goes into 54 six times).
Since all the radicands in the brackets are the same to simplify all we do is add the like terms. We add or subtract the coefficients and keep the radicands the same (6+3-8 = 1).
We then multiply the two radicals together and simplify.

Division:

When dividing radicals, the only rule is that you CAN’T leave a radical in the denominator as well as a negative.

Conjugate: A conjugate is formed by changing the sign in between two numbers. For example, the conjugate of x + y would be x – y.

Example:

First, we need to understand that there is a radical in the denominator.
To get rid of it we multiply both the numerator and the denominator by the conjugate which in this case would be $\sqrt{5}$ + $\sqrt{3}$ which we write as a fraction. For the fraction to be accurate it should be equivalent to 1.
We the multiply the radicals together and add like terms. In the denominator when you multiply the radicals together there will be some like terms that cancel out ie. $+\sqrt{15}$ and $-\sqrt{15}$ .
After we are left with a fraction that has no radical in its denominator. Though the expression may look like it can’t be further simplified it still can be. The coefficients can still be simplified since they are both divisible by the denominator.

Note: if the denominator was a negative number and the coefficients weren’t divisible by it, then we would divide everything by -1 to get rid of the negative sign.

Week 3 – Absolute Value of a Real Number

February 18, 2018February 19, 2018 keishan2015

This week in Pre-calculus 11 I learned about the absolute values of real numbers and how to solve for them. I learned that the absolute value of a real number is defined as the principal square root of the square of a number.

Principle square root: $\sqrt{36}=6$

The symbol used for the absolute value of a number is two vertical lines: | |

When solving for the absolute value of a number you are saying how far it is away from zero. Whenever you are trying to find the absolute value of a negative number the value will always be positive because, you cant have a negative distance.

ex.

|6| = 6 (6 is 6 numbers away from 0)
|-12| = 12 (-12 is 12 numbers away from 0)
|45| = 45 (45 is 45 numbers away from 0)

If there is an equation inside the absolute value symbol you must solve the equation and find the absolute value of the product.

ex.

|1 – 5| = |-4| = 4
|4 + 6| = |10| = 10
|3 – 9(-2)|= |3 + 18| = |21| = 21

If there is a coefficient in front of the absolute value symbol then you have to first solve for the absolute value and then multiply it by the coefficient. You do not distribute the coefficient inside the symbols.

ex.

5 |2 + 4(-3)|
5 |2 – 12|
5 |-10|
5(10)
50

ex.

– |- $\sqrt{64}$ |
– |-8|
-(8)
-8

Week 2 – Geometric Sequences

February 12, 2018 keishan2015

This week in Pre-calculus 11 I learned about geometric sequences and how to use the formulas to isolate variables. In a geometric sequence each term is multiplied by a constant, known as a common ratio.

My geometric sequence:

3, 12, 48, … 49152

from this sequence we know:

$t_1=3$
$t_n=49152$
r = 4

To find r we use the formula, $r=\frac{t_n}{t_{n-1}}$ , so:

$r=\frac{t_n}{t_{n-1}}$
$r=\frac{12}{3}$
r = 4

In this sequence we are trying to find what n is. To do that we would use the general formula used to find $t_n$ :

$t_n=(t_1)(r)^{n-1}$ , we then input all information that we already know.

$t_n=(t_1)(r)^{n-1}$
$49152=(3)(4)^{n-1}$

we then need to divide everything by 3 to make it easier for us to isolate n.

$[49152=(3)(4)^{n-1}]/3$
$16384=4^{n-1}$

To then find n you keep multiplying 4 by different powers until it matches the value of $t_n$ divided by $t_1$ . In this case we would multiply 4 by different powers until we get 16384.

$16384=4^{7}$

If we know tat the power is 7 then that means that we know, 7 = n -1. We now know that n = 8 because 8 – 1 = 7.

Week 1 – My Arthmetic Sequence (Updated)

February 4, 2018February 8, 2018 keishan2015

This week in Pre-Calculus 11 I learned about sequences and series and the differences between them.

My Arthmetic sequence is:

16, 4, -8, -20, -32, …

In this sequence we know,

$t_1=16$
d = -12 (the sequence decreases by 12 every time)
n = 5 (five is the last known term of the sequence)
$t_n=-32$

Pt. 1

To find the last number of a sequence ( $t_n$ ) a general formula ( $t_n=t_1+(n-1)d$ ) can be altered and simplified to suit my sequence:

$t_n=16+(n-1)(-12)$

$t_n=-12n+28$

So, if we were trying to find the 50th term ( $t_{50}$ ) we would just use the above formula and input all of the information that we know: n = 50 / d = -12/ $t_1=16$