Quadratics: Standard Form to Vertex Form

Q:  Write the equation of y = -x2 + 2x + 2 in the form of y = a(x-h)2 + k

A:  To do this, we are going to use a strategy called “completing the square” — a fairly complicated algebra 2 concept.

We are currently in the form y = ax2+bx+c and we want to get to y = a(x-h)2 + k

First, let’s establish a, b, and c in our equation: y = -x2 + 2x + 2

a = -1, b = 2, c = 2

I will walk you through the steps on how to do the problem.  At the end, I will provide some explanation behind the concept and the “why”.

Step 1)  Divide the entire equation by “a”

So, divide everything by -1:

y = -x2 + 2x + 2

y/ -1 = (-x2 + 2x + 2) / -1

And simplify to get:

-y = x2 – 2x – 2

Step 2)  “Complete the Square”

This step involves finding what number to add (or subtract) into the equation that will make the x terms factor nicely.  To find this number, we follow a simple pattern:

Take 1/2 of the coefficient in front of the x term and then square it.

The coefficient in front of x is -2.

1/2 * -2 = -1

Square -1 to get +1.

The number we need to “Complete the Square” is +1.

Add this number to both sides of the equation (remember to do it to BOTH SIDES to keep the equation balanced).  Also, add in by the x’s just for ease, like so:

-y + 1 = x2 – 2x + 1 – 2

So, our modified equation is:

-y + 1 = x2 – 2x + 1 – 2

Step 3)  Factor the x terms:

We are now going to factor the part of the equation I’ve highlighted.  The part we factor involves the x’s and the “complete the square” number that we added to the equation:

-y + 1 = x2 – 2x + 1 – 2

Factoring x2 – 2x + 1 gives (x – 1)(x – 1) OR (x – 1)2

So, we now have:

-y + 1 = (x – 1)2 – 2

Step 4) Isolate y

We have finished the hardest part (completing the square)!  Now, we just solve for y and we are done:

-y + 1 = (x – 1)2 – 2

Subtract 1 from both sides:

-y = (x – 1)2 – 3

Multiply everything by -1:

y = -(x – 1)2 + 3

And there it is!  We have taken the original equation: y = -x2 + 2x + 2 and re-written it in a different form: y = -(x – 1)2 + 3


Some logic behind the process:

Completing the square is the process of figuring out “what number” is needed to add (or subtract) in the equation so that it will factor easily into something like (x-h)2.  Once we determine the number needed, we add it to both sides of the equation to maintain balance.  Remember, we aren’t altering the actual equation, we are just changing its appearance.  Once we found the correct number, the equation will factor the way we need it to.

 

Multiplying Polynomials

Q: Multiply -9m(2m2 + 6m – 1)

A:  When adding and subtracting polynomials, you can only combined like terms.  This is not the case with multiplication.  You can multiply unlike terms together.

-9m(2m2 + 6m – 1)

In this problem the whole quantity in the parentheses is being multiplied by -9m.  So, -9m needs to multiply each term:

-9m(2m2 + 6m – 1) = -9m*2m2 + -9m*6m + -9m*-1

Read More »

Subtracting Polynomials

1:  Subtract: (-4y2-3y+8) – (2y2-6y+2)

A:  Here we are just asked to clean things up and combine like terms: basically, add apples with apples and oranges with oranges, but don’t accidentally add apples with oranges!

y2 terms get combined with y2 terms
y terms get combined with y terms
constants (lone numbers) get combined with constants

So, let’s use color to highlight the like terms:

(-4y2-3y+8) – (2y2-6y+2)

How many total y2 terms do you have? -4y2 – 2y2 = -6y2

How many total y terms do you have? -3y – -6y = -3y+6y = 3y

How many total constant terms do you have? 8+2 = 8 – 2 = 6

Did we forget anyone?  Did anyone not have a match that we need to bring along?  Nope! Everyone from both sets of parentheses is accounted for!

So, final answer:

(-4y2-3y+8) – (2y2-6y+2) = -6y2 + 3y + 6

 

Adding Polynomial Expressions

Q:  Add: (3x3+2x2-5x) + (-4x3-x2-8x)

A:  Here we are just asked to clean things up and combine like terms: basically, add apples with apples and oranges with oranges, but don’t accidentally add apples with oranges!

x3 terms get added with x3 terms
x2 terms get added with x2 terms
x terms get added with other x terms

So, let’s use color to highlight the like terms:

(3x3+2x2-5x) + (-4x3x2-8x)

How many total x3 terms do you have? 3x3+ -4x3 = -1x3

How many total x2 terms do you have? 2x2+x2 = 1x2

How many total x terms do you have? -5x+ -8x = -13x

Did we forget anyone?  Did anyone not have a match that we need to bring along?  Nope! Everyone from both sets of parentheses is accounted for!

So, final answer:

(3x3+2x2-5x) + (-4x3x2-8x) = -1x3 + 1x2 13x

Since we tend not to write the “1”, you might see the answer displayed as:

x3 + x2 13x

 

Betting and Errors in Poker

If we break down betting to the very basics, there are two types of bet, or two reasons you bet:

  • Value Bet: A bet with the goal of getting worse hand to call
  • Bluff: A bet with the goal of getting a better hand to fold

To determine the appropriate type of bet to make, you need to ask yourself “What type of errors are my opponents (or is this particular opponent) making?”

If poker were a game like chess, we would analyze which types of bets are best based on your hand, your position, your chip stack, the board, etc…  In chess, moves can be analyzed regardless of your opponent.  In poker, this is not the case.  Whether you should be value betting or bluffing hinges on the errors being made by your current opponent(s).

Are you opponents putting chips in the pot with worse hands?  Are your opponents folding too often?  These questions help determine your reasons for betting.

Types of Errors

First, let’s look at the definitions of Type 1 Errors and Type 2 Errors and what these errors look like in the world of statistics.

Example:  You want to know if you are pregnant and you take a pregnancy test.  There are 4 potential outcomes:

  • You are not pregnant and the test comes back negative.
  • You are not pregnant and the test comes back positive.
  • You are pregnant and the test comes back negative.
  • You are pregnant and the test comes back positive.

Two of the above outcomes lead to correct results.  Let’s view this in a table:

Not Pregnant Pregnant
Negative Result Correct Result
Positive Test Correct Result

The other two outcomes, which are blank in the table above, lead to errors (incorrect results). These errors are defined as:

  • Type 1 Error: You are not pregnant and the test comes back positive – a false positive.  A Type 1 Error is the assertion of something that is absent.
  • Type 2 Error: You are pregnant and the test comes back negative – a false negative.  A Type 2 Error is the failure to assert something that is present.

In table format:

Not Pregnant Pregnant
Negative Result Correct Result Type 2 Error – False Negative
Positive Test Type 1 Error – False Positive Correct Result

In the pregnancy example, think about the consequences of each error type as it pertains to an individual (or a society). What are the implications of Type 1 Errors? What are the implications of Type 2 Errors?

Betting Errors

Looking at the basics of betting, we can create a similar table to model a simplified situation.  In this situation, you bet.  Your opponent either has a better hand or a worse hand, and either calls or folds.  See the errors that can be made by your opponent below:

Opponent has a better hand Opponent has a worse hand
Opponent Calls No Error Type 2 Error
Opponent Folds Type 1 Error No Error

Bluff

When you bluff, you are counting on your opponent to make a Type 1 Error.  You are asserting something that is absent (a strong hand) and your opponent falsely believes you and folds.

If your opponents are waiting for strong hands and folding too often, they are making Type 1 Errors.

Value Bet

When you value bet, you are counting on your opponent to make a Type 2 Error.  You fail to assert strength (making your opponent sense weakness) and your opponent falsely believes you and calls.

If your opponents are calling too much or seeing too many rivers, they are making Type 2 Errors.

Are you helping your opponents correct their errors?

You need to take advantage of the types of errors your opponents are making.  As well, your bets should not encourage your opponents to correct their errors.

For example: drastic over-bets encourage players who make Type 2 Errors to play correctly and fold when otherwise they would have called a standard-sized bet with a weaker hand, giving you lots of value.

And, a player who makes Type 1 Errors will fold to standard-sized or over-sized river bets, but be encouraged to play correctly by calling smaller-sized bets.

Figure out which type of errors your opponents are making and then bet accordingly.

Get Math Help from Me

My name is Stacey and I live in Corvallis, Oregon with my two kids.  I am passionate about math tutoring. Here are ways you can get help from me:

family

  • If you have a quick question, I can help you on Twitter.
  • I tutor people privately both in person and online.  See my Facebook tutoring profile page to learn more about me.
  • I can tutor you on Skype — if you learn well from me, you can hire me.
  • Browse through examples and useful tips here, or post a new question for me to answer.

You can contact me via Twitter or Facebook if you are interested in setting up a tutoring session:

Solving logarithmic equations

Q:  Solve
log16x + log4x + log2x = 7

A:  It is easier (though not always necessary) to have all of the logs in the same base to proceed.  Since that is not the case, we need to use the change of base formula to put all of the logs into the same base.

Change of base formula says:
logab = logcb / logcwhere c can be any base of your choosing

What base should we go to?  Since 16, 4 and 2 can all be formed by powers of 2, let’s go to base 2:

log16x + log4x + log2x = 7

log2x / log216 + log2x / log24 + log2x / log22 = 7

Now, we can simplify:

log2x / 4 + log2x / 2 + log2x / 1 = 7

1/4 log2x + 1/2 log2x + log2x = 7

7/4 log2x = 7

log2x = 7 *(4/7)

log2x = 4  (remember, this reads:  the power you put on 2 to get x is 4)

x = 16

Solving Exponential Equations by Factoring

Q:  Solve for x:
22x + 2x+2 – 12 = 0

A:  The first thing we need to notice (from practice and experience) is that we can re-write this like so using our knowledge / rules of exponents:

(2x)2 + 22*2x – 12 = 0
Now simplify a little:

(2x)2 + 4*2x – 12 = 0

So, look at this in a new light.  What if we substitute each 2x with y?

(2x)2 + 4*2x – 12 = 0   turns into   y2 + 4*y – 12 = 0

(this isn’t necessary, but is helpful for visualization)

We see this is in the form of a quadratic and can be factored:

(y + 6)(y – 2) = 0

So, y = -6 or y = 2

Remember, y was a substitution for 2x. So, we really have:
2x = -6   or   2x = 2

Solve each equation separately.  Let’s start with:

2x = -6

Solve for x by taking the log of both sides (you can use the log of any base: log base 10, log base 2, natural log):

log(2x)= log(-6)
We can stop right here because you cannot take the log of a negative number. This equations yields no solutions.

So, the second equation:

2x = 2
log(2x) = log(2)

Logarithm rules say that the x exponent can come down as a multiplier like so:

x*log(2) = log(2)

Divide both sides by log(2) to get:

x = 1.

Solving Percent Problems using Proportions

Q:  How do you solve percent problems using proportions?

A:  Generally, there are 3 main parts to a percent problem:  the percent, the part and the total.

The set up is the same for all problems:

part/total = percent/100

(remember, percent means per 100, so it is always out of 100)

Example 1:  Finding the percent:

What percent is 75 of 85?

Set-up

75/85 = x/100

Cross multiply:

75*100 = 85*x

7500 = 85x

7500/85 = x

88.235% = x

So, 75 is 88.235% of the number 85.

Example 2:  Finding the “part”:

What is 76% of 50?

Set-up

x/50 = 76/100

Cross multiply:

100x = 76*50

100x = 3800

x = 3800/100

x = 38

So, 76% of 50 is 38.

Example 3:  Finding the “total”:

60 is 24% of what number?

Set-up

60/x = 24/100

Cross multiply:

24x = 60*100

24x = 6000

x = 6000/24

x = 250

So, 60 is 24% of the number 250.

Graphing Conic Sections

Q:  Identify the center, foci, vertices, and co vertices then graph the following conic section:
x2 + 9y2 = 36

A:  I am going to refer to the Conic Section guide that I had previously put together for students who I tutor:

Click to access conic_sections.pdf

This is a must have for those studying conic sections.  Click on it — it’s free 🙂

So, we look at our given equation:  x2 + 9y2 = 36 and recognize by its form that it is an ellipse.

To match the standard form of an ellipse, we need it equal, so divide both sides by 36 to get:

x2/36 + 9y2/36 = 36/36

Reduce to get:

x2/36 + y2/4 = 1

Refer to the Conic Section guide I linked to from above to follow along!

This is a horizontal ellipse (since the number under the x is larger than the number under the y)

The center is (0,0) since there are no values being subtracted from the x or the y.

The major axis:  a = sqrt(36) = 6  (in the x-direction)

The minor axis: b = sqrt(4) = 2  (in the y-direction)

The distance from the center to to foci can be found as follows:

c2 = a2 – b2

c2 = 36 – 4

c2 = 32

c = sqrt(32)

And, in simplest radical form, that is:

c = 4*sqrt(2)

So, the distance from the center to each foci (in the x-direction) is 4*sqrt(2)

Now, take all of the information from a, b and c found above to get the points:

Center: (0, 0)

Vertices:  (-6, 0) and (6, 0)

Co-Vertices:  (0, -2) and (0, 2)

Foci:  (-4*sqrt(2), 0) and (4*sqrt(2), 0)

Plot these points on a graph and then you’ve got it!