4.1 Objective: Students will look at polynomial functions of degree greater than 2, approximate the zeros, and interpret graphs.

Functions of degree greater than 2 4.1 Polynomial Functions of degree greater than 2

A polynomial function is a function of the form: Remember integers are … –2, -1, 0, 1, 2 … (no decimals or fractions) so positive integers would be 0, 1, 2 … A polynomial function is a function of the form: n must be a positive integer 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 Determine which of the following are polynomial functions. If the function is a polynomial, state its degree. A polynomial of degree 4. We can write in an x0 since this = 1. x 0 A polynomial of degree 0. Not a polynomial because of the square root since the power is NOT an integer Not a polynomial because of the x in the denominator since the power is negative

Graphs of polynomials are smooth and continuous. No gaps or holes, can be drawn without lifting pencil from paper No sharp corners or cusps This IS the graph of a polynomial This IS NOT the graph of a polynomial

Let’s look at the graph of where n is an even integer. and grows steeper on either side Notice each graph looks similar to x2 but is wider and flatter near the origin between –1 and 1 The higher the power, the flatter and steeper

Let’s look at the graph of where n is an odd integer. Notice each graph looks similar to x3 but is wider and flatter near the origin between –1 and 1 and grows steeper on either side The higher the power, the flatter and steeper

Let’s graph Translates up 2 Reflects about the x-axis 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.

HAND BEHAVIOUR OF A GRAPH RIGHT LEFT HAND BEHAVIOUR 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 behaviour 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 behaviour determined by the highest powered term. left hand behaviour: rises right hand behaviour: rises

Even degree polynomials fall on both the left and right hand sides of the graph (like - x2) if the coefficient is negative. turning points in the middle left hand behaviour: falls right hand behaviour: 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. turning Points in the middle right hand behaviour: rises left hand behaviour: falls

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

How do we determine what it looks like near the middle? A polynomial of degree n can have at most n-1 turning points (so whatever the degree is, subtract 1 to get the most times the graph could turn). doesn’t mean it has that many turning points but that’s the most it can have Let’s determine left and right hand behaviour for the graph of the function: 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. How do we determine what it looks like near the middle? The graph can have at most 3 turning points

To find the x intercept we put 0 in for y. 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. To find the x intercept we put 0 in for y. 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)

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.

4.2 Dividing Polynomials © 2002 by Shawna Haider

4.2 Properties of Division Objective: We will learn how to use synthetic division which will help us in finding imaginary roots of polynomial functions.

3x3 4x2 x 2 1 1 1 1 Dividing by a Monomial If the divisor only has one term, split the polynomial up into a fraction for each term. divisor Now reduce each fraction. 3x3 4x2 x 2 1 1 1 1

Subtract (which changes the sign of each term in the polynomial) Long Division If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials and copy the steps. Subtract (which changes the sign of each term in the polynomial) Now multiply by the divisor and put the answer below. x + 11 2 1 Bring down the next number or term Multiply and put below Now divide 3 into 5 or x into 11x First divide 3 into 6 or x into x2 32 698 x - 3 x2 + 8x - 5 Remainder added here over divisor 64 x2 – 3x subtract 5 8 11x - 5 32 11x - 33 This is the remainder 26 28 So we found the answer to the problem x2 + 8x – 5  x – 3 or the problem written another way:

Remainder added here over divisor Let's Try Another One If any powers of terms are missing you should write them in with zeros in front to keep all of your columns straight. Subtract (which changes the sign of each term in the polynomial) y - 2 Write out with long division including 0y for missing term Multiply and put below Bring down the next term Multiply and put below y + 2 y2 + 0y + 8 Divide y into y2 Divide y into -2y Remainder added here over divisor y2 + 2y subtract -2y + 8 - 2y - 4 This is the remainder 12

- 3 1 6 8 -2 - 3 - 9 3 1 x2 + x 3 - 1 1 1 Synthetic Division There is a shortcut for long division as long as the divisor is x – k where k is some number. (Can't have any powers on x). Set divisor = 0 and solve. Put answer here. x + 3 = 0 so x = - 3 1 - 3 1 6 8 -2 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line - 3 Add these up - 9 3 Add these up Add these up 1 x2 + x 3 - 1 1 This is the remainder Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.

Let's try another Synthetic Division 0 x3 0 x Set divisor = 0 and solve. Put answer here. x - 4 = 0 so x = 4 1 4 1 0 - 4 0 6 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line 4 Add these up 16 48 192 Add these up Add these up Add these up 1 x3 + x2 + x + 4 12 48 198 This is the remainder Now put variables back in (remember one x was divided out in process so first number is one less power than original problem so x3). So the answer is: List all coefficients (numbers in front of x's) and the constant along the top. Don't forget the 0's for missing terms.

Let's try a problem where we factor the polynomial completely given one of its factors. You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. - 2 4 8 -25 -50 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line - 8 Add these up 50 Add these up Add these up No remainder so x + 2 IS a factor because it divided in evenly 4 x2 + x - 25 Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is the divisor times the quotient: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0. You could check this by multiplying them out and getting original polynomial

4.3 Zeros of Polynomials Objective: Students will find the zeros of a polynomial and be able to discuss how many zeros are real & imaginary, positive & negative, rational & irrational and large or small in value.

4.3 Zeros of Polynomials Can you find the zeros of the polynomial? 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

What would the left and right hand behavior be? 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 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? 1 of mult. 1 (so crosses axis at 1) -2 of mult. 2 (so touches at 2) (0, 4) Find the zeros and their multiplicity

Steps for Graphing a Polynomial Determine left and right hand behaviour by looking at the highest power on x and the sign of that term. Determine maximum number of turning points 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.

Find and plot y intercept by putting 0 in for x Let’s graph: Determine left and right hand behavior by looking at the highest power on x and the sign of that term. Find and plot y intercept by putting 0 in for x Determine maximum number of turns in graph by subtracting 1 from the degree. Find the zeros (x intercepts) by setting polynomial = 0 and solving. Join the points together in a smooth curve touching or crossing zeros depending on multiplicity and using left and right hand behaviour as a guide. Determine multiplicity of zeros. 0 multiplicity 2 (touches) 3 multiplicity 1 (crosses) -4 multiplicity 1 (crosses) Zeros are: 0, 3, -4 Multiplying out, highest power would be x4 Degree is 4 so maximum number of turns is 3 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 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? Multiply this out to get the polynomial. FOIL two of them and then multiply by the third one.

4.4 Complex and Rational Zeros of Polynomials Objective: In this lesson, we will: Learn that imaginary roots always come in pairs and those pairs are conjugates Find the possible and actual rational zeros of a polynomial function

4.4 Complex and Rational Zeros of Polynomials Using the Rational Root Theorem to Predict the Rational Roots of a Polynomial

Find the Roots of a Polynomial For higher degree polynomials, finding the complex roots (real and imaginary) is easier if we know one of the roots. The table below can help get you started. Complete the table below:

The Rational Root Theorem The Rational Root Theorem gives us a tool to predict the Values of Rational Roots:

List the Possible Rational Roots For the polynomial: All possible values of: All possible Rational Roots of the form p/q:

Narrow the List of Possible Roots For the polynomial: Graph the polynomial: # + real zeros: 1 # - real zeros: 0 # imag. zeros: 2 All possible Rational Roots of the form p/q:

Find a Root That Works For the polynomial: Substitute each of our possible rational roots into f(x). If a value, a, is a root, then f(a) = 0. (Roots are solutions to an equation set equal to zero!)

Find the Other Roots Now that we know one root is x = 3, do the other two roots have to be imaginary? What other category have we left out? To find the other roots, divide the factor that we know into the original polynomial:

Find the Other Roots (con’t) The resulting polynomial is a quadratic, but it doesn’t have real factors. Solve the quadratic set equal to zero by either using the quadratic formula, or by isolating the x and taking the square root of both sides.

Find the Other Roots (con’t) The solutions to the quadratic equation: For the polynomial: The three complex roots of the polynomial are:

4.5 Rational Functions Objective: In this lesson, we will: Sketch graphs of rational functions Look more in depth at the vertical & horizontal asymptotes Find equations of rational functions

4.5 Rational Functions and Their Graphs

Example Find the Domain of this Function. Solution: The domain of this function is the set of all real numbers not equal to 3.

Arrow Notation Symbol Meaning x  a + x approaches a from the right. x  a - x approaches a from the left. x   x approaches infinity; that is, x increases without bound. x  -  x approaches negative infinity; that is, x decreases without bound.

Definition of a Vertical Asymptote The line x = a is a vertical asymptote of the graph of a function f if f (x) increases or decreases without bound as x approaches a. f (x)   as x  a + f (x)   as x  a - f a y x x = a f a y x x = a Thus, f (x)   or f(x)  -  as x approaches a from either the left or the right.

Definition of a Vertical Asymptote The line x = a is a vertical asymptote of the graph of a function f if f (x) increases or decreases without bound as x approaches a. f (x)  as x  a + f (x)  -  as x  a - f a y x x = a x = a f a y x Thus, f (x)   or f(x)  -  as x approaches a from either the left or the right.

Locating Vertical Asymptotes If is a rational function in which p(x) and q(x) have no common factors and a is a zero of q(x), the denominator, then x = a is a vertical asymptote of the graph of f.

Definition of a Horizontal Asymptote The line y = b is a horizontal asymptote of the graph of a function f if f (x) approaches b as x increases or decreases without bound. f y x y = b x y f y = b f y x y = b f (x)  b as x   f (x)  b as x   f (x)  b as x 

Locating Horizontal Asymptotes Let f be the rational function given by The degree of the numerator is n. The degree of the denominator is m. If n<m, the x-axis, or y=0, is the horizontal asymptote of the graph of f. If n=m, the line y = an/bm is the horizontal asymptote of the graph of f. If n>m,t he graph of f has no horizontal asymptote.

Strategy for Graphing a Rational Function Suppose that where p(x) and q(x) are polynomial functions with no common factors. 1. Find any vertical asymptote(s) by solving the equation q (x) = 0. Find the horizontal asymptote (if there is one) using the rule for determining the horizontal asymptote of a rational function. Use the information obtained from the calculators graph and sketch the graph labeling the asymptopes.

Sketch the graph of

The vertical asymptote is x = -2 The horizontal asymptote is y = 2/5

Section 4.5 Rational Functions Find an equation of a rational function f that satisfies x-intercept : 4 Vertical asymptote: x=-2 Horizontal asymptote: y= -3/5 hole at x=1

Section 4.5 Rational Functions X-intercept -- x-4 must be factor in numerator (x-4) Vertical asymptote-- x+2 is factor in denominator x-4 x+2 Horizontal asymptote--Mult. Numerator by -3 and denominator by 5 -3(x-4) 5(x+2) Hole--x-1 must be factor in both num. & den. f(x)= -3(x-4)(x-1) 5(x+2)(x-1)

Ch. 4 Review Answers