# Week 9: How to convert from General form to Standard form

This week in Pre Calc 11 I learned how to convert general form into standard form. I will show you how it uses completing the square but it a little different from last unit of solving quadratic equations. In this unit it helps us graph what the function will look like (this should look like a parabola)

Converting from General Form to Standard form.

ex/ $2x^2 -11x +4$

Step 1: Place brackets around $2x^2-11x$ , then divide by 2 to get$x^2$ instead of $2x^2$

• $(2x^2 -11x) +4$
• $2(x^2 - \frac{11}{2}x ) +4$

Step 2/3: Divide $\frac{11}{2}x$ by 2 and square it; place it in two blank spaces beside $\frac{11}{2}x$ one being added and one being subtracted.

• $2(x^2 - \frac{11}{2}x +{blank} - {blank}) +4$
• $2(x^2 - \frac{11}{2}x +\frac{121}{16} - \frac{121}{16}) +4$

Step 4: Multiply the 2 you factored out in step 1 to $\frac{121}{16}$ to remove it from the brackets.

• $2(x^2 - \frac{11}{2}x +\frac{121}{16}) +4 -\frac{242}{16}$

Step 5: Factor the inside of the brackets.

$\star$ Hint: What you squared in step 2/3 is you factor.

• $2(x^2 - \frac{11}{2}x +\frac{121}{16}) +4 -2\frac{242}{16}$
• $2(x - \frac{11}{4})^2 -\frac{242}{16} + 4$

Step 6: Simplify

• $2(x - \frac{11}{4})^2 -\frac{242}{16} + 4$
• $2(x - \frac{11}{4})^2 -\frac{121}{8} + 4$
• $2(x - \frac{11}{4})^2 -\frac{121}{8} + \frac{32}{8}$
• $2(x - \frac{11}{4})^2 -\frac{89}{8}$

Final Answer: $2(x - \frac{11}{4})^2 -\frac{89}{8}$

You can check if you have done your calculations correctly by using desmos.com

$\star$ notice how both equations line up perfectly, this means that both are equivalent.

Standard form can tell us:

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

a = 2

p$\frac{11}{4}$

q$-\frac{89}{8}$

Horizontal translation: $\frac{11}{4}$ units right

Vertical translation: $-\frac{89}{8}$

Vertex : ($\frac{11}{4}$ , $-\frac{89}{8}$)

Axis of symmetry$\frac{11}{4}$

This is how you convert General form into standard form in the unit of Quadratic equations. The standard form can tell you a lot about what it looks like and how to graph it. This is my favorite form in this unit because it tells me so much information and it is very useful. 🙂

# Week 8: General Form, Quadratics

This week I learned many cool and fascinating facts about the general form of a quadratic equation ( $ax^2 + bx +c$ ) as well as analyzing $y=a (x-p)^2 + q$

General form

$x^2$ :

• Positive – Open upward
• Negative – Opens downwards

bx:

• is a mixture of both $ax^2$ and bx so it can differ

c:

• Is the y axis

Analyzing equation:  $y=a (x-p)^2 + q$

using this formula can tell us a lot of things that will help us graph the equation. In fact it will tell us 8 helpful clues.

ex. $latex -2(x+2)^2 – 1 1. Vertex: the vertex will be p and q … (-2,-1) * p is always backwards from the formula because in the original equation it is (x – p) therefore it has to be a negative number for it to become positive in the new equation. 2. Axis of symmetry: This will be p because it is the x value on a graph …-2 3. Opens up or down: This is determined if $x^2$ is negative or positive. If it is negative it will be opening down, and if it is positive it will be opening upwards… Opens down because it is -2 4. If it is congruent to $y = x^2$. This means that the pattern of the graph would be 1, 3, 5. Meaning 1 over 1 up, 1 over 3 up, 1 over 5 up etc. If it does not follow the 1,3,5 rule it means that a has a value different from -1 or 1. Say a = 4, multiply 4 to each number from the original pattern. ex/ 1,3,5 is now equal to 4, 12 , 20 etc. This pattern will also tell us if the parabola will be stretched or compressed. If 1> a < 0 ( fraction ) it means that the parabola will be compressed. If 1 < a > 1 it means that the parabola will be stretched…. this equation is congruent to y = $-2x^2$, it is being stretched because 2 > 1, it is negative so it will be flipped upside down but the structure is the same. (congruent) 5. Minimum or maximum: If a is positive, it will have a minimum. If a is negative it will have a maximum… It has a maximum of -1 6. Domain. We know the XER because the parabola never ends. 7. Range: If a is positive y will be greater than the minimum. If a is negative, y will be smaller than the maximum… y < -1, yER IMPORTANT$latex y = 3x^2 (x – 3)^2 + 4

The horizontal translation will be 3 units to the right even though it says -3.

I also learned that general form can be turned into the analyzing equation of  $y=a (x-p)^2 + q$ by completing the square.

Here is a video that helped me understand how to complete the square

=

# Week 7: The discriminant

A few weeks ago, before spring break we learned about the discriminant. This is used in the unit of quadratics and it is a portion of the quadratic formula. The discriminant is able to tell us the amount of solutions it will have.

# of solutions:

• positive number : 2 solutions
• negative number: no solution
• equal to zero : 1 solution

. ex. $5x^2 - 9x + 4 = 0$

1. determine a, b, and c from the equation:    a = 5    b= -9    c = 4
2. enter into the discriminant formula:
• $b^2 - 4ac$
• $-9^2 - 4(5)(4)$
• $81 - 4(20)$
• $81 - 80$ = 1

This number is positive, therefore, it has two solutions

ex2/ $latex x^2 + 8x + 16 = 0 1. a = 1, b = 8, c = +16 2. $b^2 - 4ac$ 3. $8^2 - 4(1)(16)$ 4. $64 - 4(16)$ 5. $64 - 64)$ = 0 6. (equal to zero) 1 solution ex3/$latex -2x^2 + 3x – 10 = 0

1. a= -2,   b = 3,    c= -10
2. $b^2 - 4ac$
3. $3^2 - 4(-2)(-10)$
4. $9 - 4(+20)$
5. $9 - 80$ = -71
6. (negative number) No solution

This is how you would find the discriminant, as well as finding how many solutions it has. Using this can also help determine if the answer will be a rational or irrational number.

# Week 6: Solving using the Quadratic Formula

This week in Pre Calculus 11 I learned a few things that really helped me. I learned how to recognize perfect square trinomials, how to use the box method to factor, and how to use the quadratic formula when solving.

I will quickly go over each of these things.

Perfect Square Trinomials

ex. $x^2 + 10x + 25$

Always notice the first and third terms

• 1st term: $x^2$
• 2nd term: 10x
• 3rd term: 25

Notice that $x^2$ and 25 are perfect squares

$\sqrt{x^2} = x$    and    $\sqrt{25} = 5$

The 2nd term will always be $2\sqrt{1st}\cdot\sqrt{3rd}$

ex. 10x:

• $2\sqrt{x^2}\cdot\sqrt{25}$
• $2\cdot5\cdot{x}$
• 10x

This is how you recognize perfect square trinomials

Box Method (Factoring)

This box method is typically used with ugly polynomials but can be used with any expression. You want to be careful about using this method because the numbers could get big.

ex. $5x^2 + 9x + 4$

Step 1: Insert the 1st and 3rd term into the box,

Top left box: This is where you would put the first part of the expression. ex/ $5x^2$

The bottom right box: This is where the last part of your expression would go. ex/+ 4

Step 2: Multiply the 1st and 3rd term and find the products

Step 3: Chose the pair that would +. – to the 2nd term

Step 4: Insert them into the box.

* Doesn’t matter which one empty box they’re put in

*take out what is common horizontally and vertically

Step 5: Insert into the brackets and expand to check

This is how you use the box method

The quadratic formula can be used to solve any quadratic equation. This can be a really useful and quick way to solve an equation.

ex/ \$latex 3x^2 – 4x -1 = 0 %

Step 1: Determine a, b, and c and substitute into the equation.

• a = $3x^2$
• b = -4x
• c = -1

Step 3: Simplify the radicand further

Step 4: Divide by the common denominator to get it into simplest form

This is how you use the quadratic formula

Today we learned how to recognize perfect squares trinomials, how to use the box method, and how to use the quadratic formula.

# Week 5: Solving Radical Equations

This week in Pre Calculus 11, I learned how to solve radical equations. It was a little bit difficult for me at some points, because I didn’t understand squaring. Today I will teach you how to deal with these kinds of expressions.

Steps:

1. Solving
2. Restrictions
3. Checking

Example

$\sqrt{x}$ = 6

SOLVING

• remove the radical sign by squaring both sides of the equation.

$(\sqrt{x})^2$ = $6^2$

• Simplify

x = 36

Restrictions

Note: a number under a  radical sign can not be negative because a number times itself is positive weather it be -x or x. The only exception is if it has a odd power.

In the example above x needs to be a positive number so its restriction would be x > 0 or x = 0.

Check It Out

• $\sqrt{36}$ = 6
• 6 = 6

This is how to solve a basic radical equation. Now I will move on to a more difficult equation to maximize your understanding.

Example 2

1 $-3\sqrt{2x}$ = -3 $-2\sqrt{2x}$

Note: When solving eq, what you do to one side you have to do to the other.

SOLVING

1. Place the constants on side of the equation
• 1 + 3 $-3\sqrt{2x}$ = -3 + 3 $-2\sqrt{2x}$
•  4 $-3\sqrt{2x}$ = $-2\sqrt{2x}$

Note: The radicans are the same so they can be combined together. Place them on one side of the equation.

• 4 $-3\sqrt{2x}$ = $-2\sqrt{2x}$
• 4 $-3\sqrt{2x}$ + $3\sqrt{2x}$ = $-2\sqrt{2x}$ + $3\sqrt{2x}$
• 4 = $1\sqrt{2x}$

3. Square both sides to remove the radical sign.

• $4^2$$(\sqrt{2x})^2$
• $4^2$ = 2x
• 16 = 2x

4. Isolate x

• $\frac{16}{2}$$\frac{2x}{2}$
• $\frac{16}{2}$ = x
• 8 = x

RESTRICTIONS

x > 0   or   x = 0

Check It Out

• 1 – $3\sqrt{2(8)}$ = -3 – $2\sqrt{2(8)}$
• 1 – $3\sqrt{16}$ = -3 – $2\sqrt{16}$
• 1 – $3\cdot4$ = -3 – $2\cdot4$
• 1 – 12 = -3 – 8
• -11 = -11

This is how you solve radical equations.

Comment below if this helped you or if you have any questions. 🙂

# Week 4: Rationalizing the Denominator

Hey, this week in Pre Calc 11 I learned something new that is difficult for me to understand. by me teaching this to you we both have a chance to learn more about rationalizing the denominator.

How to Rationalize the Denominator

ex. $\frac{1}{5\sqrt{3}}$

Step One: Multiply by the denominator

Step Two: Root the Radical Expression in the Denominator

Step Three: Simplify Further if Possible

FINAL ANSWER IS $\frac{\sqrt{3}}{15}$

Now you know how to remove a radical expression from the denominator. 🙂

# Week 3 Absolute Value and Simplifying Radical Expressions

This week in Pre Calculus 11, we learned what an absolute value of a real number was and how to simplify radical expressions.

Absolute Value of a Real Number : The principal square root of a square number.

SOLUTION: The absolute value of a negative number is the opposite number, and the absolute value of a positive number and 0 are the same.

ex/    |-99| = 99         | 12 | = 12

This is because distance is always positive.

The long lines act as brackets but are not and it represents absolute value.

• ex 1/         4 |15-20|
•                     4 |-5|
•                     4 (5)
•                     = 20

• ex 2/         |6 +(-10)| -|5-7|
•                     |6-10| – | -2|
•                     |-4| – |2|
•                     4  –   2
•                     = 2

Here is a video that really helped me learn and understand a little bit more about absolute value.

we also briefly reviewed radical expressions

ex/ $\sqrt{25}$ =   5  and $3\sqrt[2]{5}$ = $\sqrt{45}$

Solving

Step 1: Find perfect squares.

Step 2: Take them out / simplify

Step 3: Do the same with the variables (treat them like numbers)

ex/ $\sqrt[2]{18x}$

• ex/ $\sqrt[2]{18x^2}$
• $\sqrt{9\cdot{x^2}\cdot2}$
• $\sqrt{3\cdot3\cdot{x}\cdot{x}\cdot2}$
• $3x\sqrt[2]{2}$

This is how to work with variables in Radical Expressions

# Week 2 Geometric Series

Week 2: This week in Pre Calculus 11 we learned about infinite and finite geometric series.

GEOMETRIC SEQUENCES – Each term is multiplied by a constant, known as the common ratio.

There are two types of infinite geometric series. (adding the sequence is a series to find a sum).

DIVERGING – A diverging series means the ratio is    r > 1   or    r < -1. This means the number of terms will increase, which means the partial sum increases, so the series does not have a finite sum. (NO SUM) It only has a sum if it is asking for a specific sum of terms.

CONVERGING – A converging series means the ratio is 1 > r > 0 (smaller than 1, bigger than 0) or -1 < r < 0. ( bigger that -1, smaller than 0). The partial sum will appear to get closer to a number, therefore we estimate the finite sum. (HAS SUM).

To find the sum:

1.  Find Ratio
2. Identify if it Diverges or Converges.
3. Diverges, STOP (No Sum)
4. Converges, CONTINUE (Sum)
5. Use the formula $S_{\infty} = \frac {a}{1- r}$

ex. 8 + 16 + 32 + 64 + 128 + 256 …

Ratio: $\frac {16}{8} = 2$

Identify: Diverges. NO SUM

ex. 2)     8 + -7.2 + 6.48 + -5.832 + 5.2488 …

Ratio: $\frac {-7.2}{8} = -0.9$

Identify: Converges, ratio is a decimal bigger that -1 but smaller than 0.

FORMULA :

• $S_{\infty} = \frac {a}{1 - r}$
• $S_{\infty} = \frac {8}{1 - (-0.9)}$
• $S_{\infty} = \frac {8}{1 + 0.9}$
• $S_{\infty} = \frac {8}{1.9}$
• $S_{\infty}$ = 4.21052…
• Estimated sum = 4.2

This is how to find the sum of a geometric series.

If a question asks for the sum of a term the formula to use is:

• $S_{n} = a \frac {(r^n -1)}{r - 1}$

ex. 8 + 16 + 32 + 64 + 128 + 256 …

Find $S_{10}$ :

• $S_{10} = a \frac {(r^n -1)}{r - 1}$
• $S_{10} = 8 \frac {(2^{10} - 1)}{2 - 1}$
• $S_{10} = 8 \frac {(1024-1)}{2-1}$
• $S_{10} = 8 \frac {1023}{1}$
• $S_{10} = 8 (1023)$
• $S_{10} = 8184$

# Week 1 Arithmetic Sequences

This week was the first week of Pre Calculus 11 for me. So far it has been good, as we are learning about series and sequences. Today I am going to teach you the difference between arithmetic sequences and an arithmetic series. I will also show the formula to find the nth-term and how to find the sum of the terms. STAY TUNED 🙂

A SEQUENCE for example, is a set of numbers that are changing in some way.

ex/ 5, 10, 20, 40, 80

5 is called $t_1$ , 10 = $t_2$ , 15 = $t_3$ and so on.    The t stands for term.

An ARITHMETIC SEQUENCE is a sequence that changes by a constant amount. The constant amount is also known as the common difference.

ex/ 5, 10, 15, 20, 25,     d=+5 ;The common difference is +5

to find the common difference the formula is  $t_2$  –  $t_1$

NOT ARITHMETIC,       NO CONSTANT DIFFERENCE

An ARITHMETIC SERIES is the terms of an arithmetic sequence added together. The point is to find the sum of the desired terms.

ex/ 5+10+15+20+25 = 75

How to find the nth-term

Say you wanted to find what $t_{50}$ is in the sequence 12, 9, 6, 3, 0 is but you don’t want to continue writing the whole sequence out. Well your in luck, there is a fast way.

The formula to find the nth-term is  $t_n$ = $t_1$ + d(n-1)

• The nth-term is $t_n$$t_{50}$
• d= -3
• $t_1$ = 12

Now we insert all the known numbers into the formula

1. $t_n$ = $t_1$ + d(n-1)
2. $t_{50}$ = 12 + (-3)(50-1)
3. $t_{50}$ = 12 + (-3)(49)
4. $t_{50}$ = 12 -147
5. $t_{50}$ = -135

There you are, you just figure out how to find the 50th term quick and easy.

How to find the sum of a series

We will take the example from above and determine the sum of  $S_{50}$$S_{50}$ means $t_1$ to $t_{50}$ will be added.

The formula to determine the sum is $S_n = \frac {n}{2}(t_1 + t_n)$

• $S_n$$S_{50}$
• n = 50
• $t_1$ = 12
• $t_n$ = -135

Insert into formula

1. $S_n = \frac {n}{2}(t_1 + t_n)$
2. $S_{50} = \frac {50}{2}( 12 + (-135))$
3. $S_{50} = 25( -123)$
4. $S_{50} = - 3,075$

There you have it. Today you learned what an arithmetic sequence and series were, what the formula to find the nth-term and the sum of a series was, and how to solve. I hope this blog post helps you in your future with practice and studying.