Chapter 3: Polynomial Functions 3.1 Complex Numbers 3.2 Quadratic Functions and Graphs 3.3 Quadratic Equations and Inequalities

Complex Numbers Numbers of the form a + bi where a and b are real numbers are called complex numbers in standard form. a is called the real part b is called the imaginary part

Examples of Complex Numbers

Operations with Complex Numbers All properties of real numbers are extended to complex numbers. Example Adding and Subtracting Complex Numbers (a) (3 − 4i) + (−2 + 6i) (b) (−4 + 3i) − (6 − 7i)

Multiplying Complex Numbers Example Multiply (5 − 4i)(7 − 2i).

Complex Conjugates The conjugate of the complex number a + bi is a − bi. Number Conjugate Product 3 − i 3 + i (3 − i)(3 + i) = 9 + 1 = 10 All products are real numbers. 2 + 7i 2 −7i (2 + 7i)(2 − 7i) = 53 −6i 6i (−6i)(6i) = 36

Dividing Complex Numbers Procedure: Multiply the numerator and denominator by the complex conjugate of the denominator.

Forms of Quadratic Equations P(x) = a(x − h)2 + k - Vertex Form P(x) = ax2 + bx + c - Standard Form P(x) = a(x − d) (x – e) - Factored Form

Graphs of Quadratic Functions Find the vertex: Put the equation in Vertex Form: y-intercept: Axis of symmetry: Domain: Range: increasing: decreasing: Graph:

Vertex Formula Example Use the vertex formula to find the coordinates of the vertex of the graph of P(x) = 3x2 − 2x − 1.

Extreme Values The vertex of the graph of P(x) = ax2 + bx + c is the lowest point on the graph if a > 0, or highest point on the graph if a < 0. Such points are called extreme points (also extrema, singular: extremum). For the quadratic function defined by if a > 0, the vertex (h, k) is called the minimum point of the graph. The minimum value of the function is P(h) = k. if a < 0, the vertex (h, k) is called the maximum point of the graph. The maximum value of the function is P(h) = k.

Identifying Extreme Points and Extreme Values Example Give the coordinates of the extreme point and the corresponding maximum or minimum value for each function. (a) P(x) = 2x2 + 4x − 16 (b) P(x) = − x2 − 6x − 8

Height of a Propelled Object If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height s0 feet with initial velocity v0 feet per second is s(t) = −16t2 + v0t + s0 where t is the number of seconds after the object is projected. The coefficient of t² (that is, −16), is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon or the other planets.

Solving a Problem Involving Projectile Motion A ball is thrown directly upward from an initial height of 100 feet with an initial velocity of 80 feet per second. Give the function that describes height in terms of time t. Graph this function. Give the point (4.8, 115.36) is on the graph. What does this mean in reference to the problem?

3.2 Solving a Problem Involving Projectile Motion d) After how many seconds does the projectile reach its maximum height?

Solving a Problem Involving Projectile Motion For what interval of time is the height of the ball greater than 160 feet?

Solving a Problem Involving Projectile Motion f) After how many seconds will the ball fall to the ground?

Quadratic Equations and Inequalities Quadratic Equation in One Variable An equation that can be written in the form ax2 + bx + c = 0 where a, b, and c are real numbers with a  0, is a quadratic equation in standard form. Consider the zeros of P(x) = 2x2 + 4x − 16 x-intercepts of P(x) = 2x2 + 4x −16 solution set of 2x2 + 4x − 16 = 0. Each is solved by finding the numbers that make 2x2 + 4x − 16 = 0 true.

Factoring and Zero-Product Property One way to solve a quadratic equation P(x) = 0 is to use factoring and the zero-product property. Zero-Product Property If a and b are complex numbers and ab = 0, then a = 0 or b = 0 or both.

Factoring and Zero-Product Property Graphically Example Solve 2x2 + 4x − 16 = 0.

Factoring and Zero-Product Property Graphically Example Solve x2 − 6x + 9 = 0.

Factoring and Zero-Product Property Graphically Example Solve x2 − 6x + 9 = 0. Solution x2 − 6x + 9 = 0 (x − 3)2 = 0 x – 3 = 0 or x – 3 = 0 x = 3 or x = 3 There is one distinct zero, 3. It is sometimes called a double zero, or double solution (root) of the equation. Graphing Calculator Solution

Square Root Property and Graphically

Square Root Property and Graphically Graphing Calculator Solution

The Quadratic Formula Complete the square on P(x) = ax2 + bx + c and rewrite P in the form P(x) = a(x − h)2 + k. To find the zeros of P, use the square root property to solve for x.

The Quadratic Formula and the Discriminant The expression under the radical, b2 − 4ac, is called the discriminant. The discriminant determines whether the quadratic equation has two real solutions if b2 − 4ac > 0, one real (double) solution if b2 − 4ac = 0, no real solutions if b2 − 4ac < 0.

Solving Equations with the Quadratic Formula and Graphically Solve x(x − 2) = 2x − 2.

Solving Equations with the Quadratic Formula Graphical Solution

3.3 Solving a Quadratic Equation

Graphical Interpretations for Quadratic Equations and Inequalities (FYI) Solution Set of Is P(x) = 0 {a, b} P(x) < 0 The interval (a, b) P(x) > 0 (-∞, a) U (b, ∞) Solution Set of Is P(x) = 0 {a, b} P(x) < 0 (-∞, a) U (b, ∞) P(x) > 0 The interval (a, b)

Graphical Interpretations for Quadratic Equations and Inequalities (FYI) Solution Set of Is P(x) = 0 {a} P(x) < 0 No Solution P(x) > 0 (-∞, a) U (b, ∞) Solution Set of Is P(x) = 0 {a} P(x) < 0 (-∞, a) U (b, ∞) P(x) > 0 No Solution

Graphical Interpretations for Quadratic Equations and Inequalities (FYI) Solution Set of Is P(x) = 0 No Solution P(x) < 0 P(x) > 0 (-∞, ∞) Solution Set of Is P(x) = 0 No Solution P(x) < 0 (-∞, ∞) P(x) > 0

Solving a Quadratic Inequality Step 1 Solve the corresponding quadratic equation. Step 2 Identify the intervals determined by the solutions of the equation. Step 3 Use a test value from each interval to determine which intervals form the solution set

Solving a Quadratic Inequality Analytically

Solving a Quadratic Inequality Analytically The two numbers −3 and 4 separate a number line into the three intervals shown below. If a value in Interval A, for example, makes the polynomial x2 − x − 12 negative, then all values in Interval A will make the polynomial negative.

Solving a Quadratic Inequality Analytically Solve x2 − x − 12 < 0. Pick a test value x in each interval to see if it satisfies x2 − x − 12 < 0. If it makes the inequality true, then the entire interval belongs to the solution set. Interval Test Value x Is x2 − x − 12 < 0 True or False? A: (−∞, −3) −4 (− 4)2 − (−4) − 12 < 0 ? 8 > 0 False B: (−3, 4) 02 − 0 − 12 < 0 ? − 12 > 0 True A: (4, ∞) 5 52 − 5 − 12 < 0 ? 8 > 0 False

Solving a Quadratic Inequality Graphically Verify the analytic solution to x2 − x − 12 < 0: Y1 < 0 in the interval (−3, 4).