MTH 209 Week 4 Third.

MTH 209 Week 4 Third

Final Exam logistics Here is what I've found out about the final exam in MyMathLab (running from the end of class this week (week 4 at 10pm) to 11:59pm five days after the last day of class. . Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Final Exam logistics There will be 50 questions.
You have only one attempt to complete the exam. Once you start the exam, it must be completed in that sitting. (Don't start until you have time to complete it that day or evening.) You may skip and get back to a question BUT return to it before you hit submit. You must be in the same session to return to a question. There is no time limit to the exam (except for 11:59pm five nights after the last class). You will not have the following help that exists in homework: Online sections of the textbook Animated help Step-by-step instructions Video explanations Links to similar exercises You will be logged out of the exam automatically after 3 hours of inactivity. Your session will end. IMPORTANT! You will also be logged out of the exam if you use your back button on your browser. You session will end. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Equations by Factoring I (Quadratics)
Section 6.6 Solving Equations by Factoring I (Quadratics) Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives The Zero-Product Property Solving Quadratic Equations
Applications

Zero-Product Property
To solve equations we often use the zero-product property, which states that if the product of two numbers is 0, then at least one of the numbers must be 0.

Example Try Q’s pg ,15,19,23 Solve each equation. a. b. Solution

Solving Quadratic Equations
Any quadratic polynomial in the variable x can be written as ax2 + bx + c with a ≠ 0. Any quadratic equation in the variable x can be written as ax2 + bx + c = 0, with a ≠ 0. This form of quadratic equation is called the standard form of a quadratic equation.

Example Solve each quadratic equation. Check your answers. a x2 = –12x b. x2 − 49 = 0 Solution a x2 = –12x x2 + 12x + 36 = 0 (x + 6)(x + 6) = 0 x = −6 36 + x2 = –12x 36 + (−6)2 = –12(−6) To check this value, substitute −6 for x in the given equation. 72 = 72 The only solution is −6.

Example (cont) b. x2 − 49 = 0 (x + 7)(x − 7) = 0 x + 7 = 0 x − 7 = 0
Try Q’s pg ,35,43,49 b. x2 − 49 = 0 (x + 7)(x − 7) = 0 x + 7 = 0 x − 7 = 0 x = −7 x = 7 To check these values, substitute 7 and −7 for x in the given equation. 72 − 49 = 0 (−7)2 − 49 = 0 0 = 0 0 = 0 The solutions are −7 and 7.

Example Solve 2x2 − 7x = −5 Solution 2x2 − 7x = −5 2x2 − 7x + 5 = 0
Try Q’s pg Solve 2x2 − 7x = −5 Solution 2x2 − 7x = −5 2x2 − 7x + 5 = 0 (2x – 5)(x − 1) = 0 2x – 5 = 0 or x – 1 = 0 or x = 1 The solutions are

Example Try Q’s pg a If a model rocket is launched at 48 feet per second, then its height, h, after t seconds is h = 48t – 16t2. After how long does the rocket strike the ground? Solution The rocket strikes the ground when the height is 0. 48t – 16t2 = 0 16t(3 – t) = 0 The rocket strikes the ground after 3 seconds. 16t = 0 3 – t = 0 t = 0 t = 3

Example A frame surrounding a picture is 2 inches wide. The picture inside the frame is 7 inches longer than it is wide. If the overall area of the picture and frame is 198 square inches, find the dimensions of the picture inside the frame. Solution Let x be the width of the picture and x + 7 be its length. x + 4 2 x + 7 x 2 x + 11

Example (cont) Try Q’s pg (x + 4)(x + 11) =198 x2 + 15x + 44 = 198 x2 + 15x − 154 = 0 (x − 7)(x + 22) = 0 x − 7 = 0 or x + 22 = 0 x = 7 or x = −22 The only valid solution for x is 7 inches. Because the length is 7 inches more than the width, the dimensions are 7 inches and 14 inches.

Other Functions and Their Properties
Section 8.4 Other Functions and Their Properties Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives Expressing Domain and Range in Interval Notation
Absolute Value Function Polynomial Functions Rational Functions (Optional) Operations on Functions

Expressing Domain and Range in Interval Notation
The set of all valid inputs for a function is called the domain, and the set of all outputs from a function is called the range. Rather than writing “the set of all real numbers” for the domain of f, we can use interval notation to express the domain as (−∞, ∞).

Example Try Q’s pg ,21,25 Write the domain for each function in interval notation. a. f(x) = 3x b. Solution a. The expression 3x is defined for all real numbers x. Thus the domain of f is b. The expression is defined except when x – 4 = 0 or x = 4. Thus the domain of f includes all real numbers except 4 and can be written

Absolute Value Function
We can define the absolute value function by f(x) = |x|. To graph y = |x|, we begin by making a table of values. x |x| –2 2 –1 1

Example Try Q’s pg Sketch the graph of f(x) = |x – 3|. Write its domain and range in interval notation. Solution Start by making a table of values. x y 3 2 1 4 6 The domain of f is The range of f is

Polynomial Functions The following expressions are examples of polynomials of one variable. As a result, we say that the following are symbolic representations of polynomial functions of one variable.

Example Try Q’s pg ,45,47,51 Determine whether f(x) represents a polynomial function. If possible, identify the type of polynomial function and its degree. a. b. c. cubic polynomial, of degree 3 not a polynomial function because the exponent on the variable is negative not a polynomial

Example Try Q’s pg ,73 A graph of is shown. Evaluate f(1) graphically and check your result symbolically. Solution To calculate f(–1) graphically find –1 on the x-axis and move down until the graph of f is reached. Then move horizontally to the y-axis. f(1) = –4

Example Evaluate f(x) at the given value of x. Solution
Try Q’s pg Evaluate f(x) at the given value of x. Solution

Example Use and to evaluate each of the following. Solution

Example (cont) Use and to evaluate each of the following. Solution

Example (cont) Use and to evaluate each of the following. Solution
Try Q’s pg Use and to evaluate each of the following. Solution

Quadratic Functions and Their Graphs
Section 11.1 Quadratic Functions and Their Graphs Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives Graphs of Quadratic Functions Min-Max Applications
Basic Transformations of Graphs More About Graphing Quadratic Functions (Optional)

The graph of any quadratic function is a parabola.
The vertex is the lowest point on the graph of a parabola that opens upward and the highest point on the graph of a parabola that opens downward.

The graph is symmetric with respect to the y-axis
The graph is symmetric with respect to the y-axis. In this case the y-axis is the axis of symmetry for the graph.

Example Try Q’s pg ,31 Use the graph of the quadratic function to identify the vertex, axis of symmetry, and whether the parabola opens upward or downward. a. b. Vertex (0, 2) Axis of symmetry: x = –2 Open: up Vertex (0, 4) Axis of symmetry: x = 0 Open: down

Example Try Q’s pg Find the vertex for the graph of Support your answer graphically. Solution a = 2 and b = 8 Substitute into the equation to find the y-value. The vertex is (2, 11), which is supported by the graph.

Example Identify the vertex, and the axis of symmetry on the graph, then graph. Solution Begin by making a table of values. Plot the points and sketch a smooth curve. The vertex is (0, –2) axis of symmetry x = 0 x f(x) = x2 – 2 3 7 2 2 1 1 3

Example Identify the vertex, and the axis of symmetry on the graph, then graph. Solution Begin by making a table of values. Plot the points and sketch a smooth curve. The vertex is (2, 0) axis of symmetry x = 2 x g(x) = (x – 2)2 4 1 1 2 3 4

Example Identify the vertex, and the axis of symmetry on the graph, then graph. Solution Begin by making a table of values. Plot the points and sketch a smooth curve. The vertex is (2, 0) axis of symmetry x = 2 x h(x) = x2 – 2x – 3 2 5 1 3 1 4 2 3 4

Example Try Q’s pg Find the maximum y-value of the graph of Solution The graph is a parabola that opens downward because a < 0. The highest point on the graph is the vertex. a = 1 and b = 2

Example A baseball is hit into the air and its height h in feet after t seconds can be calculated by a. What is the height of the baseball when it is hit? b. Determine the maximum height of the baseball. Solution a. The baseball is hit when t = 0. b. The graph opens downward because a < 0. The maximum height occurs at the vertex. a = –16 and b = 64.

Example (cont) The maximum height is 66 feet. Try Q’s pg

Basic Transformations of Graphs
The graph of y = ax2, a > 0. As a increases, the resulting parabola becomes narrower. When a > 0, the graph of y = ax2 never lies below the x-axis.

Example Try Q’s pg Compare the graph of g(x) = –4x2 to the graph of f(x) = x2. Then graph both functions on the same coordinate axes. Solution Both graphs are parabolas. The graph of g opens downward and is narrower than the graph of f.

Parabolas and Modeling
Section 11.2 Parabolas and Modeling Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives Vertical and Horizontal Translations Vertex Form
Modeling with Quadratic Functions (Optional)

Vertical and Horizontal Translations
The graph of y = x2 is a parabola opening upward with vertex (0, 0). All three graphs have the same shape. y = x2 y = x2 + 1 shifted upward 1 unit y = x2 – 2 shifted downward 2 units Such shifts are called translations because they do not change the shape of the graph only its position

The graph of y = x2 is a parabola opening upward with vertex (0, 0). y = x2 y = (x – 1)2 Horizontal shift to the right 1 unit

The graph of y = x2 is a parabola opening upward with vertex (0, 0). y = x2 y = (x + 2)2 Horizontal shift to the left 2 units

Example Try Q’s pg ,19,27 Sketch the graph of the equation and identify the vertex. Solution The graph is similar to y = x2 except it has been translated 3 units down. The vertex is (0, 3).

Example Sketch the graph of the equation and identify the vertex. Solution The graph is similar to y = x2 except it has been translated left 4 units. The vertex is (4, 0).

Example Try Q’s pg ,37 Sketch the graph of the equation and identify the vertex. Solution The graph is similar to y = x2 except it has been translated down 2 units and right 1 unit. The vertex is (1, 2).

Example Try Q’s pg Compare the graph of y = f(x) to the graph of y = x2. Then sketch a graph of y = f(x) and y = x2 in the same xy-plane. Solution The graph is translated to the right 2 units and upward 3 units. The vertex for f(x) is (2, 3) and the vertex of y = x2 is (0, 0). The graph opens upward and is wider.

Example Try Q’s pg ,68 Write the vertex form of the parabola with a = 3 and vertex (2, 1). Then express the equation in the form y = ax2 + bx + c. Solution The vertex form of the parabola is where the vertex is (h, k). a = 3, h = 2 and k = 1 To write the equation in y = ax2 + bx + c, do the following:

Example Write each equation in vertex form. Identify the vertex. a. b. Solution a. Because , add and subtract 16 on the right.

Example (cont) b. This equation is slightly different because the leading coefficient is 2 rather than 1. Start by factoring 2 from the first two terms on the right side. Try Q’s pg

Section 11.3 Quadratic Equations
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives Basics of Quadratic Equations The Square Root Property
Completing the Square Solving an Equation for a Variable Applications of Quadratic Equations

Basics of Quadratic Equations
Any quadratic function f can be represented by f(x) = ax2 + bx + c with a  0. Examples:

Basics The different types of solutions to a quadratic equation.

Example Solve each quadratic equation. Support your results numerically and graphically. a. b. c. Solution a. Symbolic: Numerical: Graphical: x y 1 5 2 1 The equation has no real solutions because x2 ≥ 0 for all real numbers x.

Example (cont) b. The equation has one real solution. x y 5 4 4 1 3
2 1 b. The equation has one real solution.

Example (cont) c. The equation has two real solutions.
Try Q’s pg ,37,39 c. x y 4 2 8 1 9 2 The equation has two real solutions.

The Square Root Property
The square root property is used to solve quadratic equations that have no x-terms.

Example Solve each equation. a. b. c. Solution a. b. c.
Try Q’s pg ,53,57 Solve each equation. a. b. c. Solution a. b. c.

Real World Connection If an object is dropped from a height of h feet, its distance d above the ground after t seconds is given by

Example A toy falls 40 feet from a window. How long does the toy take to hit the ground? Solution

Example Try Q’s pg Find the term that should be added to to form a perfect square trinomial. Solution Coefficient of x-term is –8, so we let b = –8. To complete the square we divide by 2 and then square the result.

Example Try Q’s pg Solve the equation Solution Write the equation in x2 + bx = d form.

Example Solve the equation for the specified variable. Solution
Try Q’s pg ,107 Solve the equation for the specified variable. Solution

Example Use of the Internet in Western Europe has increased dramatically. The figure shows a scatter plot of online users in Western Europe, together with a graph of a function f that models the data. The function f is given by: where the output is in millions of users. In this formula x = 6 corresponds to 1996, x = 7 to 1997, and so on, until x = 12 represents a. Evaluate f(10) and interpret the result. b. Graph f and estimate the year when the number of Internet users reached 85 million. c. Solve part (b) numerically.

Example (cont) Solution a. Evaluate f(10) and interpret the result. Because x = 10 corresponds to 2000, there were about 51.4 million users in 2000.

Objectives Solving Quadratic Equations The Discriminant
Quadratic Equations Having Complex Solutions

QUADRATIC FORMULA The solutions to ax2 + bx + c = 0 with a ≠ 0 are given by

Example Solve the equation 4x2 + 3x – 8 = 0. Support your results graphically. Solution Symbolic Solution Let a = 4, b = 3 and c = − 8. or or

Example (cont) Try Q’s pg 4x2 + 3x – 8 = 0 Graphical Solution

Example Try Q’s pg Solve the equation 3x2 − 6x + 3 = 0. Support your result graphically. Solution Let a = 3, b = −6 and c = 3.

Example Try Q’s pg Solve the equation 2x2 + 4x + 5 = 0. Support your result graphically. Solution Let a = 2, b = 4 and c = 5. There are no real solutions for this equation because is not a real number.

THE DISCRIMINANT AND QUADRATIC
EQUATIONS To determine the number of solutions to the quadratic equation ax2 + bx + c = 0, evaluate the discriminant b2 – 4ac. 1. If b2 – 4ac > 0, there are two real solutions. 2. If b2 – 4ac = 0, there is one real solution. 3. If b2 – 4ac < 0, there are no real solutions; there are two complex solutions.

Example Try Q’s pg a, b Use the discriminant to determine the number of solutions to −2x2 + 5x = 3. Then solve the equation using the quadratic formula. Solution −2x2 + 5x − 3 = 0 Let a = −2, b = 5 and c = −3. b2 – 4ac = (5)2 – 4(−2)(−3) = 1 or Thus, there are two solutions.

THE EQUATION x2 + k = 0 If k > 0, the solution to x2 + k = 0 are given by

Example Solve x2 + 17 = 0. Solution The solutions are
Try Q’s pg Solve x = 0. Solution The solutions are

Example Try Q’s pg Solve 3x2 – 7x + 5 = 0. Write your answer in standard form: a + bi. Solution Let a = 3, b = −7 and c = 5. and

Example Solve Write your answer in standard form: a + bi. Solution Begin by adding 2x to each side of the equation and then multiply by 5 to clear fractions. Let a = −2, b = 10 and c = −15.

Example (cont) Let a = −2, b = 10 and c = −15. Try Q’s pg

Example Try Q’s pg Solve by completing the square. Solution After applying the distributive property, the equation becomes Since b = −4 ,add to each side of the equation. The solutions are 2 + i and 2 − i.

Objectives Basic Concepts Graphical and Numerical Solutions
Symbolic Solutions

Example Try Q’s pg ,11 Determine whether the inequality is quadratic. a. 6x + 2x2 – x3 ≥ 0 b x2 + 2 < 6x2 + x Solution a. The inequality 6x + 2x2 – x3 ≥ 0 is not quadratic because it has an x3-term. b. Write the inequality as follows. 8 + 7x2 + 2 < 6x2 + x 10 + 7x2 < 6x2 + x x2 – x + 10 < 0 The inequality is quadratic because it can be written in the form ax2 + bx + c > 0 with a = 1, b = −1, and c = 10.

Example Make a table of values for x2 − 2x – 15 and then sketch the graph. Use the table and graph to solve x2 – 2x – 15 ≤ 0. Write your answer in interval notation. Solution Interval notation [−3, 5] x y = x2 – 2x – 15 −3 −2 −7 −1 −12 −15 1 −16 2 3 4 5

Example (cont) Try Q’s pg ,27 x2 − 2x – 15

Example Try Q’s pg Solve x2 > 4. Write your answer in interval notation. Solution The graph of y = x2 – 4 is shown with intercepts −2 and 2. Thus the solution set is given by x < −2 or x > 2, which can be written in interval notation as

Example Solve each of the inequalities graphically. a. x2 + 3 > 0 b. x2 + 3 < 0 c. (x − 3)2 ≤ 0 Solution a. Because the graph is always above the x-axis, x2 + 3 is always greater than 0. The solution set includes all real numbers, or (−∞,∞). b. Because the graph never goes below the x-axis, x2 + 3 is never less than 0. Thus there are no real solutions.

Example (cont) Try Q’s pg ,55,57 Solve each of the inequalities graphically. a. x2 + 3 > 0 b. x2 + 3 < 0 c. (x − 3)2 ≤ 0 Solution a. Because the graph never goes below the x-axis, (x − 3)2 ≤ 0 is never less than 0. When x = 3, y = 0, so 3 is the only solution to the inequality (x − 3)2 ≤ 0.

Example Solve each inequality symbolically. Write your answer in interval notation. a. x2 – 4x – 12 ≥ 0 b. 10 – x2 > 9x Solution a. x2 – 4x – 12 = 0 The solutions lie outside the values. In interval notation the solution set is (x + 2)(x – 6) = 0 x + 2 = 0 x – 6 = 0 x = −2 x = 6

Example (cont) Try Q’s pg ,37 Solve each inequality symbolically. Write your answer in interval notation. a. x2 – 4x – 12 ≥ 0 b. 10 – x2 > 9x Solution b. – x2 − 9x + 10 < 0 The solutions lie between these two values. In interval notation the solution set is (−10, 1). – x2 − 9x + 10 = 0 x2 + 9x − 10 = 0 (x + 10)(x – 1) = 0 x + 10 = 0 x – 1 = 0 x = −10 x = 1

Example Try Q’s pg A rectangular building needs to be 9 feet longer than it is wide. The area of the building must be at least 532 square feet. What widths x are possible for this building? Solution x(x + 9) ≥ 532 x2 + 9x = 532 x2 + 9x – 532 = 0 The width is positive, so the building must be 19 feet or more, x ≥ 19 feet.

Due for this week… Homework 4 (on MyMathLab – via the Materials Link)  The fifth night after class at 11:59pm. Read Chapter 12.1, 12.3 and Do the MyMathLab Self-Check for week 4. Learning team hardest problem assignment. Complete the Week 4 study plan after submitting week 4 homework. Participate in the Chat Discussions in the OLS Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

End of week 4 You again have the answers to those problems not assigned Practice is SOOO important in this course. Work as much as you can with MyMathLab, the materials in the text, and on my Webpage. Do everything you can scrape time up for, first the hardest topics then the easiest. You are building a skill like typing, skiing, playing a game, solving puzzles. Next Week: Composite Functions, Inverse Functions, Logarithmic functions, Arithmetic/Geometric sequences/series.

MTH 209 Week 4 Third.

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