# 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) 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

A:  OK.  Let’s look at the left end behavior first:

As  x approaches -∞, what is the function (y-value) doing?

Imagine x=-1000000 (some super large and super negative number, like the idea of -∞), we have:

y=5(-1000000)3+7(-1000000)2-2(-1000000)-1

Don’t do the actual math.  Just think:

Is this number large or small?

Is it positive or negative?

I can look at the x3 term and see that it dominates this function. x2 and x are small peanuts compared to x3. So, in reaity, in polynomials, I can focus on the term of the largest degree:

y=5(-1000000)3+7(-1000000)2-2(-1000000)-1

y=5(-1000000)3

This number gives y = negative and super large.

So, I can jump to conclusions here…

As x → -∞, y → -∞

(As x approaches negative infinity, y approaches negative infinity).

Now, let’s look at the right end behavior:

As  x approaches +∞, what is the function (y-value) doing?

Imagine x=+1000000 (some super large and super positive number, like the concept of +∞), we have:

y=5(+1000000)3+7(+1000000)2-2(+1000000)-1

And, by the same reasoning, we can focus on the term of largest degree:

y=5(+1000000)3+7(+1000000)2-2(+1000000)-1

y=5(+1000000)3 = super large and super positive

So, as x → +∞, y → +∞

(As x approaches positive infinity, y approaches positive infinity)

Note: in this example, y behavior mimicked x behavior, this isn’t always the case!

# 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

# 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

# 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!

(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

# 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.

# Solve the system of equations (conics)

Q:  Solve the system of equations (both of which are conic sections)
1:   x2 + y2 – 20x + 8y + 7 = 0
2:  9x2 + y2 + 4x + 8y + 7 = 0

A:  I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)

I am going to multiply equation 1 by (-1) and then add it to equation 2:

2:  9x2 + y2 + 4x + 8y + 7 = 0

1: -x2 – y2 + 20x – 8y – 7 = 0   [after it has been multiplied by -1]

Now, add them together to get equation 3:

3:  8x2 +24x = 0

Now, equation 3 only has x’s so we can solve for x (by factoring):

(8x)(x + 3) = 0

So, x = 0 or -3

But, we aren’t done… Find the solution to this system of conics means that we are finding the point(s) of intersection — and it turns out there are two points of intersection since there were two solutions for x.  So, our points (solutions) are:

(0, ?)  and (-3, ?)

We need to find the y-coordinate of each solution separately.  We do this by plugging in the x value to either of the original equations (both will lead us to the same correct solution):

First x-value:  (0, ?)

x2 + y2 – 20x + 8y + 7 = 0

(0)2 + y2 – 20(0) + 8y + 7 = 0

y2  + 8y + 7 = 0

Solve for y by factoring:

(y + 7)(y + 1)=0

y = -7, -1

OH!  So this really means that there are more than two solutions…. When x = 0, we found two answers for y.

(0, -7) and (0, -1)

Second x-value:  (-3, ?)

x2 + y2 – 20x + 8y + 7 = 0

(-3)2 + y2 – 20(-3) + 8y + 7 = 0

9 + y2 +60 + 8y + 7 = 0

y2 + 8y + 76 = 0

Since this does not factor, we need to solve for y by using the quadratic formula.  I did not show my work here, but we end up with a negative under the square root, so there are no real solutions for when x = -3.

Therefore, there are two real solutions to this equation (2 points of intersection) and they are:

(0, -7) and (0, -1)

Here is a visual of what we are solving for….. We had the equations for the two graphed conic sections and we found the two points of intersection shown below:

Q:  Simplify:

A:  There are a few ways to do this, but first I will rationalize the denominator (that means get rid of the square root in the denominator) and then I will simplify the square root (simplest radical form):

Step 1:  Rationalize the denominator:

Multiply the top and bottom by sqrt(75):

Now, simplify algebraically:

Simplify the denominator to get 150… for some reason, I can’t make my picture keep it in a fully fraction with 150 in the bottom, so the image shows the 150 pulled out like so (either way is fine):

Now reduce the 15 and the 150 to get:

The denominator has been rationalized (no square roots in the denominator) and the fraction has been reduced.

Step 2:  Put the numerator into simplest radical form:

To go to simplest radical form, you want to see what “perfect squares” divide into the square root.  So, what perfect squares divide into 75?  25 is a perfect square that divides into 75:  3*25=75.

So, we can break up the square root like so (again, the square roots can be on top of the fraction, just having image problems):

Now, what is the square root of 25?  That’s 5, so simplify:

And, now reduce the fraction part of the 5 and the 10 in the denominator to get:

You are done!

# Graph the system of equations

Q:  Graph the system of linear equations, determine the number of solutions, and find the solution:

y = 4x + 2 and y =- 2x – 3

A: Okay, for the sake of talking about the lines I will name them:

Line 1:  y = 4x + 2

Line 2:  y =- 2x – 3

A “solution” to a system of linear equations is the point where the lines intersect.  So, we need to graph both of these lines on the same grid and see if and where they intersect.

Step 1:  Graph Line 1 and Line 2 on the same grid:

Tricks for graphing line 1:  Since line 1 is in the form y = mx + b, we can use the method of finding the slope and the y-intercept to graph the line.

The slope of line 1 is 4 and the y-intercept of line 1 is 2.

On your x-y graph, go up 2 on the y-axis and put a point [this is the y-intercept, or the “b” value]

Now, starting from the y-intercept point 2:  go up 4 and right 1 (this is the slope 4/1).  Create another point.  Connect the dots to make a line.  You should have:

Now, we need the same strategy to graph line 2 on the same grid.

Line 2:  y =- 2x – 3

The y-intercept is -3 and the slope is -2.  Plot the y-intercept of -3 on the y-axis.  From that point:  go down 2, right 1 to create another point.  Connect the dots to make a line.  Your graph should now look like:

So, the question now is:  How many solutions are there and what is the solution.  Recall, the solution is (x, y) point of intersection.
Clearly, there is 1 solution.  The lines do cross at 1 point.

What is that point?  Well, you have to estimate that point from the graph.  This is one problem with solving a system by graphing — it can be inaccurate.  Depending on the “neatness” of your graph, your answer will vary.

To me, the answer appears to be around the point (-.8, -1.3)

*To get the accurate solution, we solve these types of problems algebraically.