# End Behavior for linear and Quadratic Functions

**1 End Behavior for linear and Quadratic Functions**

A linear function like f(x) = 2x − 3 or a quadratic
function f(x) = x^{2} + 5x +3 are pretty generic. We have

the tools to determine what the graphs look like just by looking at the
functions. Today, I want to start

looking at more general aspects of these functions that carry through to the
more complicated polynomial

functions (e.g., f(x) = 2x^{4} − 3x^{3}+ 7x^{2} − x + 11). Specifically, I want to look
at what the graphs look like

on the ends, what they look like near the x-axis, and distinguishing aspects of
the graph like bumps and

wiggles.

Let’s start with end behavior. We’ve seen this so far as the ends of the curves
we’ve drawn that point up or

down and signify that the functions run off to positive or negative infinity
(±∞). In particular, we want to

describe how the functions behave for very large values of x.

First, let’s look at the function f(x) = x^{2} + 5x + 3 at a somewhat large number,
like x = 1,000,000 for

example. We get

f(1,000,000) = (1,000,000)^{2} + 5(1,000,000) + 3.

If we look at each term separately, we get the numbers

1,000,000,000,000

5,000,000

3

Imagine graphing the point (1,000,000 , 1,000,005,000,003) (Good luck!). It
wouldn’t look much different

from (1,000,000 , 1,000,000,000,000).

We will say that x^{2} dominates x, when x is very large. Similarly, x dominates
any constant. Notice that

the same thing happens for large negative numbers like x = −1,000,000. Out
towards the ends of the graph,

therefore, f(x) = x^{2}+5x+3 doesn’t look a whole lot different from f(x) = x^{2}, and
when we’re just sketching

graphs, they don’t look different at all. Similarly, the function f(x) = 2x − 3
looks a lot like f(x) = 2x for

large values of x.

Figure 1:

As another example, consider the linear function f(x) = −3x+11. Since the x-term
dominates the constant

term, the end behavior is the same as the function f(x) = −3x. For large
positive values of x, f(x) is large

and negative, so the graph will point down on the right. Similarly, the graph
will point up on the left, as o

n the left of Figure 1.

On the other hand, if we have the function f(x) = x^{2}+5x+3, this has the same end
behavior as f(x) = x^{2},

so the ends will go up on both sides, as on the right side of Figure ??. I’ve
given these a little curve upwards

to indicate that x^{2} gets bigger faster than x does.

1.1 Quiz 04-A

What is the end behavior of the following functions?

2 FACTORINGS OF QUADRATIC FUNCTIONS 2

1. f(x) = 2x − 4

2. f(x) = −x

3. f(x) = −2x^{2} + 11x + 4

**2 Factorings of Quadratic Functions
**

I’m going to assume that you can factor quadratic expressions, at least in the simpler cases. I want to focus

on what information we can draw from the factorings.

Given the quadratic function f(x) = x

^{2}− 4x + 3, we can factor it as follows.

f(x) = (x − 1)(x − 3).

Since anything times zero is zero, we can see that if x = 1 or if x = 3, we get f(1) = 0 and f(3) = 0. These

are the places where the graph crosses the x-axis, as can be seen in Figure 2.

Figure 2:

For very large values of x (both positive and negative), the magnitude of x^{2} is
much larger than x, so it will

dominate to the right and left. The sign on the x^{2}-term, therefore, determines
whether the parabola will

open upwards or downwards.

Figure 3:

For example, consider the function

g(x) = −2x^{2} + 8x − 6,

which is the first function multiplied by −2. It will open
downwards. In fact, we can factor as follows.

g(x) = −2x^{2} + 8x − 6 = −2(x^{2} − 4x + 3) = −2(x − 1)(x − 3).

The x-intercepts are the same, x = 1, 3, but now everything is multiplied by a
negative number, and that

turns things upside-down. Look at Figure 3. The 2 stretches everything
vertically, so the graph also looks

skinnier.

The basic factorings give us three possibilities. If we shift the function
function f up one unit, we get the

following.

h(x) = x^{2} − 4x + 4 = (x − 2)(x − 2).

Since both factors are the same, only x = 2 is an x-intercept. The x^{2}-term is
positive, so the parabola opens

upwards. The graph must look as it does in Figure 4, therefore.

Figure 4:

If we shift the function up any higher, it won’t intersect the x-axis at all.
This corresponds to the fact that

the function

j(x) = x^{2} − 4x + 5

does not factor over the real numbers. In fact, if we try to solve the equation
x^{2} − 4x + 5 = 0 using the

quadratic formula, we get

We have to use imaginery numbers to find square roots of
negative numbers. Anyway, the graph is shown

in Figure 5

We can also multiply by constants to stretch and compress the graphs vertically,
and multiplying by negative

numbers makes the parabolas open downwards.

**2.1 Quiz 02-B (Note: We didn’t do this in class.)
**

Sketch the graphs of the following quadratic functions. For the answers, give the x-intercepts and whether

the graph opens up or down.

**f(x) = (x + 3)(x − 1).**

QB1.

QB1.

**QB2.**f(x) = (x + 1)

^{2}. (The list of answers has been changed as of 1/17/05. The correct answer should now

get credit in Blackboard.)

**QB3.**f(x) = x

^{2}+ 1. This does not factor over the reals, and the vertex is at x = 0.

**QB4.**f(x) = −3(x + 3)(x − 1).

**3 Homework 04
**

For each of the given functions, find the x-intercept(s) and the end behavior.

1. f(x) = x − 4.

2. f(x) = (x + 4)(x − 2).

3. f(x) = (x − 3)

^{2}.

4. f(x) = x

^{2}+ x + 1. This does not factor.

5. f(x) = 2(x − 3)(x − 5).

6. f(x) = −2(x + 1)(x + 1).

7. f(x) = −x

^{2}− x − 1. Compare this to problem 4.