Pc Polynomials

Remember integers are … –2, -1, 0, 1, 2 … (no decimals or fractions) so nonnegative integers would be 0, 1, 2 …

A polynomial function is a function of the form: n must be a nonnegative integer

f ( x ) = an x + an −1 x n

n −1

+  + a1 x + ao

All of these coefficients are real numbers

The degree of the polynomial is the largest power on any x term in the polynomial.

Determine which of the following are polynomial functions. If the function is a polynomial, state its degree.

f ( x) = 2x − x 4

g ( x) = 2 x 0 h( x ) = 2 x + 1 3 2 F ( x) = + x x

A polynomial of degree 4.

We can write in an x0 since this = 1.

A polynomial of degree 0. Not a polynomial because of the square root since the power is NOT 1 an integer

x = x2

Not a polynomial because of the x in the denominator since the power is 1 −1 NOT nonnegative

x

=x

Graphs of polynomials are smooth and continuous. No sharp corners or cusps No gaps or holes, can be drawn without lifting pencil from paper

This IS the graph of a polynomial

This IS NOT the graph of a polynomial

Let’s look at the graph of even integer.

g ( x) = x4

f ( x) = x

2

Notice each graph looks similar to x2 but is wider and flatter near the origin between –1 and 1

f ( x) = x

n

where n is an

h( x ) = x 6 and grows steeper on either side The higher the power, the flatter and steeper

Let’s look at the graph of odd integer. Notice each graph looks similar to x3 but is wider and flatter near the origin between –1 and 1

f ( x) = x

g ( x) = x5

f ( x) = x3

n

where n is an

and grows steeper on either side

h( x ) = x 7

The higher the power, the flatter and steeper

Let’s graph

f ( x) = − x + 2

Reflects over x-axis

4

Looks like x2 but wider near origin and steeper after 1 and -1

So as long as the function is a transformation of xn, we can graph it, but what if it’s not? We’ll learn some techniques to help us determine what the graph looks like in the next slides.

Moves up 2

LEFT

and

RIGHT

HAND BEHAVIOR OF A GRAPH

The degree of the polynomial along with the sign of the coefficient of the term with the highest power will tell us about the left and right hand behavior of a graph.

Even degree polynomials rise on both the left and right hand sides of the graph (like x2) if the coefficient is positive. The additional terms may cause the graph to have some turns near the center but will always have the same left and right hand behavior determined by the highest powered term. left hand behavior: rises

right hand behavior: rises

Even degree polynomials fall on both the left and right hand sides of the graph (like - x2) if the coefficient is negative. Can turn in the middle

left hand behavior: falls

right hand behavior: falls

Odd degree polynomials fall on the left and rise on the right hand sides of the graph (like x3) if the coefficient is positive. Can turn in the middle

left hand behavior: falls

right hand behavior: rises

Odd degree polynomials rise on the left and fall on the right hand sides of the graph (like x3) if the coefficient is negative. Can turn in the middle left hand behavior: rises right hand behavior: falls

A polynomial of degree n can have at most n-1 turns (so whatever the degree is, subtract 1 to get the most times the graph could turn). Let’s determine left and right hand behavior for the graph of the function: doesn’t mean it does turn that 4 3 many times 2 but that’s the f x = x − 3x −most 15 xit can + 19 x + 30 turn

( )

degree is 4 which is even and the coefficient is positive so the graph will look like x2 looks off to the left and off to the right. The graph can turn at most 3 times

How do we determine what it looks like near the middle?

)( xx ++119)( xx +− 30 0f (=x )( =x −x 2−)(3xx +−315 5) 4

3

2

x and y intercepts would be useful and we know how to find those. To find the y intercept we put 0 in for x.

f ( 0 ) = 0 − 3( 0 ) − 15( 0 ) + 19( 0 ) + 30 = 30 4

3

2

To find the x intercept we put 0 in for y. We don’t know how to solve this yet so let me give you the factored version of this polynomial. Later in the chapter we’ll learn how to factor. Finally we need a smooth curve through the intercepts that has the correct left and right hand behavior. To pass through these points, it will have 3 turns (one less than the degree so that’s okay)

(0,30)

)( xx ++119)( xx −+ 30 0f (=x )( =x −x 2−)(3xx +−315 5) 4

3

2

We found the x intercept by putting 0 in for f(x) or y (they are the same thing remember). So we call the x intercepts the zeros of the polynomial since it is where it = 0. These are also called the roots of the polynomial. Can you find the zeros of the polynomial?

g ( x) = ( x − 1) ( x + 2 ) ( x − 3) 3

2

There are repeated factors. (x-1) is to the 3rd power so it is repeated 3 times. If we set this equal to zero and solve we get 1. We then say that 1 is a zero of multiplicity 3 (since it showed up as a factor 3 times). What are the other zeros and their multiplicities?

-2 is a zero of multiplicity 2 3 is a zero of multiplicity 1

So knowing the zeros of a polynomial we can plot them on the graph. If we know the multiplicity of the zero, it tells us whether the graph crosses the x axis at this point (odd multiplicities CROSS) or whether it just touches the axis and turns and heads back the other way (even multiplicities TOUCH). Let’s try to graph: What would the left and 2 f x = − x −1 x + 2 right hand behavior be?

( )

(

)(

)

You don’t need to multiply this out but figure out what the highest power on an x would be if multiplied out. In this case it would be an x3. Notice the negative out in front. What would the y intercept be? Find the zeros and their multiplicity

(0, 4)

1 of mult. 1 (so crosses axis at 1) -2 of mult. 2 (so touches at 2)

Steps for Graphing a Polynomial •Determine left and right hand behavior by looking at the highest power on x and the sign of that term. •Determine maximum number of turns in graph by subtracting 1 from the degree. •Find and plot y intercept by putting 0 in for x •Find the zeros (x intercepts) by setting polynomial = 0 and solving. •Determine multiplicity of zeros. •Join the points together in a smooth curve touching or crossing zeros depending on multiplicity and using left and right hand behavior as a guide.

Let’s graph:

f ( x ) = x ( x − 3)( x + 4) 2

••Determine left and right hand behavior by looking atand ••Find and plot y intercept by putting 0 in for x Determine maximum number of turns in graph by Find Join the the points zeros (x together intercepts) in a smooth by setting curve polynomial touching = or 0 •Determine multiplicity of zeros. 0 multiplicity 2 (touches) the highest on xdegree. and the sign ofand thatusing term.left and subtracting 1depending the solving. crossing zerospower on multiplicity 2from 2 3 multiplicity 1 (crosses) 0 = x x − 3 x + 4 Zeros are: 0, 4 3, -4 right Multiplying hand behavior as a guide. out, highest power would be x -4 multiplicity Degree is 4 so maximum number of turns1is(crosses) 3

f ( 0 ) = 0 ( ( 0 − )(3)( 0 +) 4 ) = 0

Here is the actual graph. We did pretty good. If we’d wanted to be more accurate on how low to go before turning we could have plugged in an x value somewhere between the zeros and found the y value. We are not going to be picky about this though since there is a great method in calculus for finding these maxima and minima.

What is we thought backwards? Given the zeros and the degree can you come up with a polynomial? Find a polynomial of degree 3 that has zeros –1, 2 and 3. What would the function look like in factored form to have the zeros given above?

f ( x ) = ( x + 1)( x − 2 )( x − 3)

Multiply this out to get the polynomial. FOIL two of them and then multiply by the third one.

f ( x) = x − 4x + x + 6 3

2