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!

Graphing a Quadratic Function

Q:  Graph the Quadratic Function:  f(x) = 2x2 – 3x + 1

Answer:

There are two main methods to do this.  I will do both methods (depending on what you have learned in class, you will want to pick either method 1 or method 2 that I show).

Method 1:  Plotting Points

Because this is a quadratic equation, we know the graph will be in the shape of a parabola.  So, I am going to pick a few values for x, then find the y values.  Then, I will plot the points on the graph to make a parabola.  You can pick any values for x when you do this method.

Let’s pick x = 0.  So, plug x into the equation to find y:

2(0)2 – 3(0) + 1 = 0 – 0 + 1 = 1

Point 1:  When x = 0, y = 1:  (0, 1)


Let’s pick x = 1 now:

2(1)2 – 3(1) + 1= 2(1) – 3 + 1 = 2 – 3 + 1 = 0

Point 2:  When x = 1, y = 0:  (1, 0)


Let’s pick x = -1 now:

2(-1)2 – 3(-1) + 1= 2(1) + 3 + 1 = 2 + 3 + 1 = 6

Point 3:  When x = -1, y = 6:  (-1, 6)

Now, plot the points (0, 1) and (1, 0) and (-1, 6) on a graph.  Connect the dots to make a parabola!

Keep in mind:  You can get more than 3 points for more accuracy.

Method 2:  Finding the y-intercept and the vertex:

The y-intercept of any type of graph occurs when you plug in 0 for x.  So, plug in 0 for x into the equation and solve for y:

2(0)2 – 3(0) + 1 = 0 – 0 + 1 = 1

So, the y-intercept is the point (0, 1).

Now, there is a formula to find the vertex of a quadratic equation.  It goes as follows:

Step 1 to find the vertex:  x = -b / 2a (this will give us the x coordinate of the vertex.

Recall, in our equation, a = 2, b = -3 and c = 1

So, the x part of the vertex is x = -(-3) / 2(2) = 3/4

Now, to get the y part, you need to plug the x part into the equation and solve:

2(3/4)2 – 3(3/4) + 1 = -1/8


The vertex is the point (3/4 , -1/8)

So, on your graph, plot the vertex and the y-intercept.  Draw a parabola through the y-intercept until you hit the vertex.  You know the vertex is either the high or the low point.  Then, “bounce off” the vertex and continue drawing the other half of the parabola.