# 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.

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.

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

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):

# Solve for x (complicated)

Q:  Solve for x:

-2[8 – 5(2-3x) – 7x] = 4(x – |-9|)

A:

There are many ways to start with this, so I’m just going to start on the right side of the equation first:

-2[8 – 5(2-3x) – 7x] = 4(x – |-9|)

Starting on the right (piece at a time)

We know that |-9| = 9 (by definition of absolute value), so:

4(x – |-9|)

4(x – 9)

Now use the distributive property to distribute the 4 through (multiply the 4 to the x and the 9):

4x – 36

This is the best we can do on the right side.  So the right side (for now) is 4x – 36.

Now let’s look at the left side:

-2[8 – 5(2 – 3x) – 7x]

We need to get rid of the inner most parentheses, so we should deal with the -5(2 – 3x) part.  Distribute the -5 through:

-2[8 – 10 + 15x – 7x] <– that is what happens on the left when the -5 distributed through.

Now, clean up inside the brackets and combine like terms:

-2[-2 +8x]  <— I combined the 8-10 and the 15x-7x

Now distribute the -2 through the brackets to get:

4 – 16x  <– this is as far as the left side can be simplified.  So, combining the left side = right side we get:

4 – 16x = 4x – 36

I’m going to add 16x to both sides (to get rid of the x on the left side):

4 = 20x – 36

Now add 36 to both sides:

40 = 20x

Divide both sides by 20 to get x by itself:

# Logs

Q:  Solve:  log(5x − 7) − log(4x − 5) = 0

# Finding the equation of a line

Q:  What is the equation of the line containing the points (3,5) & (6,-2)?

# Evaluating Functions

Q:  Find the following values of the function: f(x) = 3x + 1

a. f(0)       b. f(3)

c. f(-2)      d.f(-x)

e. -f(x)      f. f(x+2)

g. f(2x)     h. f(x+h)