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## Determining the End Behavior of a Function

How do you determine the end behavior of a function?  And, what does this mean?

When looking at a graph, the “end behavior” is referring to what is happening all the way to the far left of the graph and all the way to the far right of the graph.  Your goal is to analyze the y-value (height or function value) of the function when x is really large and negative, and then again when x is really large and positive.  What is the pattern on each end?  What is the “end behavior”?

Notationally, we are thinking:

1. As x → -∞, y → ?
2. As x → +∞, y → ?

OK, so let’s try this on a polynomial example:

Q:  What is the end behavior of the function y=5x3+7x2-2x-1

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## Integration: Sines, Cosines and U-Substitution

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:

∫cos5(x)sin4(x)dx

∫cos(x)*cos4(x)*sin4(x)dx

∫cos(x)*(cos2(x))2*sin4(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx

Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

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## 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.

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## 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
Now we need to clean up the expression on the right:

-9m*2m2 + -9m*6m + -9m*-1 ….

-9m*2m2 can be thought of as -9*m*2*m*m = -18m3

(multiply the coefficients together, and multiply the m’s together)

-9m*6m = -54m2

-9m*-1 = 9m

So, final answer cleans up like:

-9m(2m2 + 6m – 1) = -18m3 – 54m2 + 9m

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## 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!

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

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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!

(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

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Q:  Solve for x:  (x + 3)(x – 4) < 0

A:

Step 1:  Find the zeroes

Since the quadratic is already factored, this isn’t too tricky.  If it wasn’t factored, you’d have to factor first! (always make sure there is a 0 on one side of the equation / inequality before proceeding).

OK, so what are the zeroes?

x + 3 = 0 or x – 4 = 0

The zeroes are x = -3 or x = 4

So, draw a number line and plot the zeroes on the number line:

Now, you have to test each “interval” that is separated by the “zeroes”.  There are three intervals to test.

Interval 1:  The numbers to the left of -3  –> in interval notation this is (-infinity, -3)

Interval 2:  The numbers between -3 and 4  –> in interval notation this is (-3, 4)

Interval 3:  The numbers to the right of 4  –> in interval notation this is (4, infinity)

Step 2:  Test each interval

Pick any number on interval 1 and test it into the original inequality.  I’ll pick -5:

(x + 3)(x – 4) < 0

(-5 + 3)(-5 – 4) < 0

(-2)(-9) < 0

18 < 0      <—  this is false, so numbers on this interval [interval 1] are not part of the solution.

Pick any number on interval 2 and test it into the original inequality.  I’ll pick 0:

(x + 3)(x – 4) < 0

(0 + 3)(0 – 4) < 0

(3)(-4) < 0

-12 < 0      <—  this is true, so numbers on this interval [interval 2] are a part of the solution.

Pick any number on interval 3 and test it into the original inequality.  I’ll pick 5:

(x + 3)(x – 4) < 0

(5 + 3)(5 – 4) < 0

(8)(1) < 0

8 < 0      <—  this is false, so numbers on this interval [interval 3] are not part of the solution.

SO:  The only interval that “worked” was interval 2.  Therefore, the solution is all number between -3 and 4.

In interval notation, we write that like: (-3, 4)

In inequality notation, we write that like: -3 < x < 4

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Q:  Solve  (a/2)2 – a > 0

Solving Algebraically:

Step 1:  Simplify

We can simplify the equation by squaring the fraction to get:

a2/4 – a > 0

OR in different form:

(1/4) a2 – a > 0

Now, to get rid of the fraction and clean things up, I am going to multiply everything by 4 (since the fraction is 1/4 — you don’t have to do this, just a “cleaning up” step”):

4*(1/4) a2 – 4*a > 4*0

a2 – 4a > 0

Normally, when you solve inequalities, you isolate the variable by moving things around to the left / right side.  When it is a quadratic, you don’t want to do that.  You want all of the numbers and variables on one side and zero on the other side.  We have that, so we are good to go.

Step 2:  Factor

Now, factor the left side:

a2 – 4a > 0

a (a – 4) > 0

Step 3:  Identify the zeroes

As we are used to doing with quadratics, we need to find what values you plug in to make “zeroes”

So, take each factor and set it equal to zero like so:

a = 0 and a – 4 = 0

Solve to get:

a = 0 and a = 4

The zeroes are:  0 and 4.

So, this parabola (quadratic) crosses the x-axis at 0 and 4.  Now, we want to find where the quadratic is greater than 0.

With analysis, we know the quadratic is concave up in shape (a smiley face).  On your paper, draw an x-y graph with a parabola that crosses the x-axis and 0 and 4 and is concave up.  You should get:

So, where is the parabola > 0 (above the x-axis?)

When a < 0 and when a > 4

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## Graphing Lines by Plotting Points

Q:  Graph x + 2y = 4

A:  We are going to graph this by making an x-y table of values.  A typical (though not necessary) thing to do is solve for y first — this just means “get y by itself”

x + 2y = 4

Subtract x from both sides (to try to get y alone)

2y = 4 – x

Now, it’s standard for to list the x first, so we may write this as:

2y = -x + 4

To get y by itself, divide everything by 2:

y = -x/2 + 4/2

This simplifies to:

y = -1/2 x + 2

So, to make a table of values, pick some numbers for x.  It is standard to pick “0, 1, 2” or something similar.  So:

x  |  0  |  1  |  2  |

y  |      |      |      |

Plug in each value for x to find y:

When x = 0, y = -1/2 (0) + 2 = 0 + 2 = 2

When x = 1,  y = -1/2 (1) + 2 = -1/2 + 2 = 1.5

When x = 2, y = -1/2 (2) + 2 = -1 + 2 = 1

Then the table is complete:

x  |  0  |  1    |  2  |

y  |  2  | 1.5  |  1  |

These are just points you need to plot on a graph:  (0, 2)  (1, 1.5)  and (2,1)

Plot the points then connect the dots to make a line!

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## Graphing Quadratics by creating a table

Q:  Sketch graphs of these three quadratics relations on the same set of axes

a) y=-3x2

b) y=1/4 x2
A:  There are many ways to sketch graphs (quadratics).  The most basic way is to:  (1) make a table of values, (2) plot the points, (3) connect the dots.

So, let’s start with the first equation (a) y=-3x2 and make a table of values:

STEP 1

Start by picking values for x.  I will pick -2, -1, 0, 1, 2 for x:

x  | -2 | -1 | 0 | 1 | 2

y  |     |     |      |    |

Now we need to calculate the y values.

When x = -2, y=-3(-2)2 = -3(4) = -12

When x = -1, y=-3(-1)2 = -3(1) = -3

When x = 0, y=-3(0)2 = -3(0) = 0

When x = 1, y=-3(1)2 = -3(1) = -3

When x = 2, y=-3(2)2 = -3(4) = -12

So, the table is complete:

x  | -2   |  -1  | 0   | 1   | 2

y  | -12 |  -3 |  0  | -3 |  -12

STEP 2:

Plot the points: (-2, -12)  (-1, -3)  (0, 0)  (1, -3)  (2, -12)  on the axes

STEP 3

Connect the dots to make a parabola.  You should have a picture like so:

Now we need to do the same thing for equation (b) y=1/4 x2

STEP 1

Start by picking values for x.  I will pick -2, -1, 0, 1, 2 for x:

x  | -2 | -1 | 0 | 1 | 2

y  |     |     |      |    |

Now we need to calculate the y values.

When x = -2, y=1/4 (-2)2 = 1/4 (4) = 1

When x = -1, y=1/4 (-1)2 = 1/4 (1) = 1/4

When x = 0, y=1/4 (0)2 = 1/4 (0) = 0

When x = 1, y=1/4 (1)2 = 1/4 (1) = 1/4

When x = 2, y=1/4 (2)2 = 1/4 (4) = 1

So, the table is complete:

x  | -2   |  -1    | 0   |   1   | 2

y  |  1    |  1/4 |  0  | 1/4 |  1

STEP 2:

Plot the points: (-2, 1)  (-1, 1/4)  (0, 0)  (1, 1/4)  (2, 1)  on the axes

STEP 3

Connect the dots to make a parabola.  You should have a picture like so:

SO…. If we wanted to put these two graphs on the same axes, it would look something like (depending on your scale on the y-axis):

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## Find solutions to a Trig Function

Q:  Find all solutions for X where 0 ≤ X < 360 degrees and 9*cos(X) – 5 = 0.

Okay, the 0 ≤ X < 360 just tells you we are looking for answers in one complete circle (no more, no less).

So, we first need to solve for X like so:

9*cos(X) – 5 = 0

9*cos(X) = 5

cos(X) = 5/9

X = cos-1(5/9)

Plug this into your calculator and you get:

X = 56.25 degrees

This is part of your answer, but not all of it.  The calculator will only give one answer, and will not consider answers in multiple quadrants.  We will use this answer to find any other solutions for X.

From our problem, we saw that:

cos(X) = 5/9

this means that the cosine value is positive.  In what quadrants are the cosine values positive??

So, X = 56.25 degrees is the answer for the angle in Quadrant 1.

To get the angle in Quadrant 4 that also works, you gotta do 360 – 56.25 = 303.75

Therefore, 303.75 degrees is another answer for X.

What this is saying is that:

cos(56.25)= 5/9 and cos(303.75) = 5/9 also.

So, X = 56.25 and 303.75

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## Simplifying Trig Expressions Using Identities

Q:  Simplify cos4t – sin4t

A:

First thing we gotta notice is that this is a difference of squares.  cos4t is the same as (cos2t)2.  So, we need to use what we know about factoring from algebra to factor this:

cos4t – sin4t can be factored using the difference of squares formula/concept (this just takes practice to see this and realize it):

(cos2t – sin2t)(cos2t + sin2t)

But, now we can use a trig identity because cos2t + sin2t = 1, so plug that in:

(cos2t – sin2t)(cos2t + sin2t)

(cos2t – sin2t)(1) = (cos2t – sin2t)

Now we gotta notice that even though it is simplified a bunch, we have another identity:

(cos2t – sin2t) = cos(2t)

And now we are as simplified as we get.