Precalculus Notes: Unit 2 – Polynomial Functions
Objective: The student will sketch the graph of a quadratic equation. The student will write the equation of a quadratic function.
Quadratic Function: a polynomial function of degree 2
· Graph is u-shaped, called a parabola
· Parabolas are symmetric with respect to a line called the axis of symmetry
· Parabolas have a vertex, which is the point on the parabola where it intersects the axis of symmetry, and is the maximum or minimum of the quadratic function
· General Form: Axis of Symmetry:
o x-intercepts can be found using the quadratic formula:
o x-intercepts can be found by factoring.
o x-intercepts can be found by completing the square.
· Vertex Form (standard form):
· Vertex: Axis of Symmetry:
o In both forms, if , the parabola opens UP; if , the parabola opens DOWN
Graphing a Parabola
Ex1a: Write the equation in vertex form. Find the vertex and axis of symmetry. Then graph the function.
Ex. 1b. You Try: Write in vertex form. Find vertex & axis of symmetry:
f(x) = -6x + x2 – 3
Ex. 1c. Identify the x-intercepts of the quadratic function
Writing the Equation of a Parabola
· Substitute the vertex for in vertex form.
· Substitute the given point for and solve for a.
Ex2a: Write an equation for the parabola with vertex that passes through the point .
Ex.2b You try: Write the equation for a parabola with a vertex (1,2)
that passes through the point (0, 5).
Average Rate of Change (recall):
Ex3: Find the average rate of change of from to .
Ave Rate of Change =
Vertical Free-Fall Motion: formula for the position of an object in free fall (on Earth)
s = position g = acceleration due to gravity = initial velocity = initial position
Ex4: A flare is shot straight up from a ship’s bridge 75 feet above the water with an initial velocity of 76 ft/sec. How long does it take to reach its maximum height? How long does it take to reach 163 feet?
The flare reaches it max height at the vertex of the parabola.
It takes for the flare to reach its maximum height.
Note: This is NOT the graph of the path of the flare. The graph relates the position of the flare to time. The flare reaches 163 feet when .
It reaches 163 feet at
Algebraic Method:
Ex. 5 A soda pop company’s sales average 26,000 cans per month when the cans sell for 50 cents each. For each nickel increase in the price, the sales per month drop by 1000 cans. Find the maximum revenue.
You Try: Write the function in vertex form. Find the vertex and axis of symmetry. Then graph.
Reflection: Using a table of values, how can you determine whether it is a linear or quadratic function?
Syllabus Objective: 2.9 – The student will sketch the graph of a polynomial, radical, or rational function.
a function that can be written in the form
Note: This is called Standard Form – when the exponents of x descend
Degree: the largest exponent, , of x
Leading Coefficient: the coefficient of the first term when written in standard form
Constant Term: the numerical term,
Classifying Polynomial Functions
I. Number of Terms
· 1 term = monomial
· 2 terms = binomial
· 3 terms = trinomial
· 4 or more terms = polynomial
II. Degree
Degree / Name (Degree) / Standard Form / Example / Classification of Example0 / Constant / / / Constant Monomial
1 / Linear / / / Linear Monomial
2 / Quadratic / / / Quadratic Trinomial
3 / Cubic / / / Cubic Binomial
4 / Quartic / / / Quartic Polynomial
5 / Quintic / / / Quintic Monomial
Ex1: Which of the following are polynomial functions? State the degree and the leading coefficient or explain why it is not a function.
a)
b)
Ex2 Describe how to transform the graph. Name the y-intercept.
a.)
b.) g(x) = 2(6 – 3x)4 – 1
The Leading Coefficient Test to Predict End Behavior of Polynomial Functions
Odd Degree:
If the leading coefficient is positive, then .
· If the leading coefficient is negative, then .
Even Degree:
· If the leading coefficient is positive, then .
· LIf the leading coefficient is negative, then .
Ex2: Indicate if the degree of the polynomial function shown in the graph is odd or even and indicate the sign of the leading coefficient. (see pg. 105 exploration)
a.) b.)
Zeros (x-intercepts, Roots) of Polynomial Functions: nth-degree polynomials have at most extrema (relative minima or maxima) and n zeros.
Recall that a zero of a function f is a number x for which f(x)=0.
1. x=a is a zero of function f.
2. x=a is solution of the polynomial function f(x)=0
3. (x-a) is a factor of the polynomial f(x).
4. (a,0) is an x-intercept of the graph of f.
Zeros of Polynomial Functions
Multiplicity: “repeated” zeros; If a polynomial function f has a factor of , and not , then c is a zero of multiplicity m of f.
· Odd Multiplicity: f crosses the x-axis at c; changes signs
· Even Multiplicity: f “bounces” or is tangent to the x-axis at c; doesn’t change signs
Ex. 3 Sketch the graph of . Describe the multiplicity of the zeros.
End behavior (degree of ______, odd or even, ______leading coefficient):
;
Domain: Range:
Continuity:
Symmetry:
Boundedness:
Local Extrema:
Asymptotes:
Ex5: Graph the function and list its characteristics.
Domain: Range: Increasing: Decreasing:
Symmetry:
Boundedness: Extrema: Local Max: Local Min:
End Behavior: ;
Zeros:
Ex6: Find the zeros and extrema of the function . Graph:
Zeros: Extrema:
Newton’s Law of Cooling
Ex7: The rate at which an object cools varies as the difference between its temperature and the temperature of the surrounding air. When a 270° C steel plate is placed in air that is at 20° C, it is cooling at 50° C per minute. How fast is it cooling when its temperature is 100° C?
R = rate of cooling, = initial temp, = surrounding temp:
Solve for k first:
Watch for hidden behavior:
Ex8: Graph the function in a Standard window and find the zeros.
is degree 3, so the end behavior is , so we know it must cross the x-axis again to the right.
New window: Zeros are
Intermediate Value Theorem (Location Principle): If a and b are real numbers with , and if f is continuous on , then f takes on every value between and .
Therefore, if and have opposite signs, then for some number c in .
Ex9: Use the Intermediate Value Theorem to show has zeros in these intervals: and
f changes sign (negative to positive) in the interval , so f must have at least one zero in .
f changes sign (positive to negative) in the interval , so f must have at least one zero in .
Application of Polynomial Functions
Ex10: You cut equal squares from the corners of a 22 by 30 inch sheet of cardboard to make a box with no top. What size squares would need to be cut for the volume to be 300 cubic inches?
Define x as the length of the sides of the squares. Write a formula for the volume of the box.
Solve the equation ______by graphing.
Squares with lengths of approximately ______inches should be cut.
Ex. 11
Find polynomial functions with the following zeros. (There are many correct solutions.)
a.)
b.)
You Try: Sketch the graph of the polynomial function.
g(x) = -(x-2)3(x + 1)2 f(x) = .2(x +1) 2(x – 3)(x + 5)
Reflection: How many zeros can a function of degree n have? Explain your answer.
Syllabus Objectives: 2.2 – The student will calculate the intercepts of the graph of a given relation.
Review: Long Division
Long Division of Polynomials (same process!)
Ex1a: Find the quotient.
Note: Every term of the polynomial in the dividend must be represented. Since this polynomial is missing an term, we must include the term .
Solution:
Remainder Theorem: If is divided by , then the remainder, .
Ex2a: Find the remainder without using division for divided by .
Note: Compare your answer to the long division problem above.
Ex. 2b. You try: (2x3 – 3x2 – 5x + 6) ÷ (x2 – 3)
Factor Theorem: is a factor of if and only if .
Ex3: Determine if is a factor of without dividing.
Show that :
so by the Factor Theorem, is a factor of .
Synthetic Substitution:
Ex4: Find if using synthetic substitution.
· Using the polynomial in standard form, write the coefficients in a row.
· Put the x-value to the upper left.
· Bring down the first coefficient, then multiply by the x-value.
· Add straight down the columns, and repeat.
The number in the bottom right is the value of .
Ex 5: Find if using synthetic substitution.
This polynomial function is in standard form, however it is missing two terms. We can rewrite the function as to fill in the missing terms.
Synthetic Division:
Ex.6 a) (2x3 – 3x2 – 5x – 12) ÷ (x – 3)
b) Divide x4 – 8x3 + 11x – 6 by x + 3.
Recall:
Possible Rational Zeros of a Polynomial Function =
Ex7: Find the possible rational zeros of
Step 1: The leading coefficient is _____. _____ is the only factor of ______.
Step 2: The constant is ____. All of the factors of _____ are
Step 3: List the possible factors -
*If we tested for actual zeros using synthetic substitution
we would find thatare zeros.
Descartes’ Rule of Signs: The number of real zeros equals the number of sign changes in or that number less a multiple of 2. The number of negative real zeros equals the number of sign changes in or that number less a multiple of 2.
Ex8: Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of the function .
Number of sign changes of : 1 sign change
Number of sign changes of : 2 sign changes
There is 1 possible positive real zero and 2 or 0 possible negative real zeros.
Upper and Lower Bound Rules: used to determine if there is no zero larger or smaller than a number c, if the leading coefficient is POSITIVE , using synthetic division for
· Upper Bound: c is an upper bound for the real zeros of f if and every number in the last line of using synthetic division has signs that are all nonnegative.
· Lower Bound: c is a lower bound for the real zeros of f if and every number in the last line of using synthetic division has signs that are alternately nonnegative and nonpositive.
Ex9: Verify that all of the real zeros of the function lie on the interval .
:
and signs are all nonnegative, so is an upper bound of the zeros
:
and signs alternate, so is an lower bound of the zeros
Conclusion: ______
Finding the Real Zeros of a Polynomial:
Ex10: Find all real solutions of the polynomial equation.
Note: There is one sign change, so______
Possible Rational Zeros:
Synthetic Substitution (try 1):
1 is a zero, so we can write .
Factor the quadratic:
Solutions:
Reflect: How is the Factor Theorem connected to the Remainder Theorem?
Objective: Students will use imaginary unit i to write complex numbers. Add, subtract and multiply complex numbers, use complex conjugates to write the quotient of two complex numbers in standard form, plot complex numbers in the complex plane.
I. Number Sets
Transcendental numbers are not the roots of any algebraic equations. All other complex numbers are algebraic.
Powers of i: Cont to i7
i0 = 1
*
i2 = -1
i3 =
Ex.1 a) (3 + 7i) – (2 – 4i) b) (3 + 7i)(2 – 4i)
Ex.2 a) b) (2 – i)3
Complex Conjugate of z: (a + bi) is : (a – bi)
Ex.3 a) Multiply 3 – 4i by its conjugate.
b) Find x & y. (2 + 3i)(x – i) = 13 + yi
Ex.4
Complex Solutions to Quadratic Equations
Ex.5 5x2 + 2x + 1 = 0
Reflect:
Syllabus Objectives: 1.7 – The student will solve problems using the Fundamental Theorem of Algebra. 1.8 – The student will calculate the complex roots of a given function.
Fundamental Theorem of Algebra
· A polynomial function of degree n has n complex zeros (some zeros may repeat)
· Complex zeros come in conjugate pairs
Review: conjugate
· A polynomial function with odd degree and real coefficients has at least one real zero
Calculator Exercise: State how many complex zeros the function has. Use
a graphing calculator for b & c to determine how
many of the zeros are real.
a) f(x) = 2x2 – 12x + 19
b) f(x) = 5x3 + 4x2 + 3x – 4
c) f(x) = x5 + 2x4 – 22x2 + 2x3 – 18x + 55
Ex1: Find all the zeros of the function and write the polynomial as a product of linear factors.
Step One: Check for any rational zeros.
Possible rational zeros: