# Precalculus Chapter 2 Polynomial and Rational Functions.

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Precalculus Chapter 2 Polynomial and Rational Functions

Objectives: Identify key characteristics and graph quadratics and other polynomials. Use polynomial division to find both real and complex roots. Graph rational functions and find asymptotes, intercepts and holes associated with their graphs.

Sections 2.1 Quadratic Functions 2.2 Polynomial Functions of Higher Degree 2.3 Real Zeros of Polynomial Functions 2.4 Complex Numbers 2.5 The Fundamental Theorem of Algebra 2.6 Rational Functions and Asymptotes 2.7 Graphs of Rational Functions

Objectives: Graph quadratic functions. Analyze graphs of quadratic functions. Solve quadratic functions. Find minimum and maximum values of quadratic functions in real-life applications.

Definition of Polynomial An algebraic expression of the form Where all coefficients a i are real numbers, The degree n is a integer The degree of a polynomial is the highest power of x in the expression.

Definition of Polynomial Each expression of the form a n x n is called a term. a n is called the coefficient of the term in x n. The coefficient a n of the highest power of x is called the leading coefficient. a 0 is called the constant term.

Skills Which of the following are polynomial expressions? yes no yes no

The Zero Polynomial If P(x) is a polynomial for which P(x)=0 for all x, then P(x) is called the zero polynomial. The zero polynomial has no degree.

Standard Polynomials

Quadratic Equations Standard Form:  One variable  Two variables (function) Turning Point (Vertex Form):  One variable  Two variables (function) vertex (h,k)

Changing the form of Quadratic Equations Standard form ⇒ to ⇒ vertex form: Complete the Square Vertex form ⇒ to ⇒ standard form: Multiple and combine like terms

Completing the Square We work with a quadratic equation to make one side a perfect square

Completing the Square Completing the square Leading coefficient must be 1 to complete the square

Vertex Form Changing from standard form to vertex form π Add something (b/2) 2 in to make a perfect square trinomial Subtract the same amount to keep it even. Now create a binomial squared This gives us the ordered pair (h,k) 11 19

Expressing Quadratic Function in Vertex Form To convert a quadratic equation from standard form, y = ax 2 + bx + c = 0, to vertex form, y = a(x - h) 2 + k, use completing the square. In both forms of the general quadratic equation there is the variable “a”. Is the value of the variable “a” the same in both equations? Yes, “a” is the dilation factor and controls the shape of the parabola, it is the same in both equations. Use “a” to check you have completed the square properly.

Completing the Square Changing from standard form to vertex form

Your Turn: Change to vertex form by completing the square

Solving Quadratic Equations vs. Functions Quadratic equation – single variable  ax 2 +bx+c=0  Set equal to zero and then solve for x. Quadratic function – two variables  f(x)=ax 2 +bx+c or y=ax 2 +bx+c  Set f(x)=0 (to find the x-intercepts, zeros, roots or solutions of the function) and then solve for x.

Solving Quadratic Functions (Equations) 1.Factoring & use the zero product property 2.Completing the square 3.Quadratic formula 4.Graphing

Zero-Product Property The Zero-Product Property If the product of two factors is zero, then at least one of the factors is 0. That is, if ab = 0, then a = 0 or b = 0 or both a and b are 0.

Zero Factor Property Example: Solve: (x – 5)(x + 4) = 0 x – 5 = 0 or x + 4 = 0 Set each expression equal to 0. x = 5 x =  4

Solving Quadratic Equations Steps for Solving a Quadratic Equation by Factoring Step 1: Write the equation in standard form, ax 2 + bx + c = 0. Step 2: Factor the polynomial on the left side of the equation. Step 3: Set each factor found in Step 2 equal to zero using the Zero-Product Property Step 4: Solve each first-degree equation for the variable. Step 5: Check: Check your answers by substituting the values of the variable into the original equation.

Solving Quadratic Equations Solve x 2 – 5x = 24. First write the polynomial equation in standard form. x 2 – 5x – 24 = 0 Now we factor the polynomial using techniques from the previous sections. x 2 – 5x – 24 = (x – 8)(x + 3) = 0 We set each factor equal to 0. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = -3 Example:

Solving Quadratic Equations Example: Solve: 6x + 18x 2 = 0 18x 2 + 6x = 0 Put the equation in standard form. 6x + 18x 2 = 0 6x(3x + 1) = 0Factor. 6x = 0 3x + 1 = 0Set each factor equal to zero. x = 0 3x =  1

Solving Quadratic Equations Your Turn: Solve: 9x 2 = 81 9x 2 – 81 = 0Rewrite the equation in standard form. 9(x 2 – 9) = 0Factor out the common factor, 9. 9(x – 3)(x + 3) = 0Factor the quadratic equation. x = 3Solve each equation. x =  3 9 = 0 x – 3 = 0 x + 3 = 0Set each expression equal to 0.

Solving Quadratic Equations Example: Solve: 3x 3 + x 2 = 14x 3x 3 + x 2 – 14x = 0Rewrite the equation in standard form. x(3x 2 + x – 14) = 0Factor out the common factor, x. x(x – 2)(3x + 7) = 0Factor the quadratic equation. x = 0 Solve each equation. x = 2 x = 0 x – 2 = 0 3x + 7 = 0Set each expression equal to 0. The solution set is

Quadratic Factorization Warning!! Problem... many (most) quadratic functions are NOT easily factored!! Example: Use Completing the square or Quadratic Formula – can solve all Quadratic equations. Factoring can only be used if the quadratic equation can be factored to integer factors. All quadratic equations can be solved.

Solving Quadratic Equations by Completing the Square

Square Root Property If b is a real number and a 2 = b, then

Square Root Property Solving Quadratic Equations Using the Square Root Property Step 1: Isolate the expression containing the square term. Step 2: Use the Square Root Method. Don’t forget the  symbol. Step 3: Isolate the variable, if necessary. Step 4: Check. Verify your solutions.

Square Root Property Solve x 2 = 49 Solve (y – 3) 2 = 4 Solve 2x 2 = 4 x 2 = 2 y = 3  2 y = 1 or 5 Example:

Square Root Property Solve x 2 + 4 = 0 x 2 =  4 There is no real solution because the square root of  4 is not a real number. Example:

Square Root Property Solve (x + 2) 2 = 25 x =  2 ± 5 x =  2 + 5 or x =  2 – 5 x = 3 or x =  7 Example:

Square Root Property Solve (3x – 17) 2 = 28 3x – 17 = Your Turn:

Completing the Square We now look at a method for solving quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section. Example:

Completing the Square Solving a Quadratic Equation in x by Completing the Square 1)If the coefficient of x 2 is 1, go to Step 2. Otherwise, divide both sides of the equation by the coefficient of x 2. 2)Isolate all variable terms on one side of the equation. 3)Complete the square for the resulting binomial by adding the square of half of the coefficient of the x to both sides of the equation. 4)Factor the resulting perfect square trinomial and write it as the square of a binomial. 5)Use the square root property to solve for x.

Solving Quadratics by Completing the Square Solve by completing the square. y 2 + 6y =  8 y 2 + 6y + 9 =  8 + 9 (y + 3) 2 = 1 y =  3 ± 1 y =  4 or  2 y + 3 =± = ± 1 Example:

Solving Quadratics by Completing the Square Solve by completing the square. y 2 + y – 7 = 0 y 2 + y = 7 y 2 + y + ¼ = 7 + ¼ (y + ½) 2 = Example:

Solving Quadratics by Completing the Square Solve by completing the square. 2x 2 + 14x – 1 = 0 2x 2 + 14x = 1 x 2 + 7x = ½ x 2 + 7x + = ½ + = (x + ) 2 = Your turn:

π Quadratic Formula To derive, use completing the square with the standard form ax 2 + bx + c = 0. Once this is done, we can use the formula for any quadratic function.

The Quadratic Formula Quadratic Formula The solution(s) to the quadratic equation ax 2 + bx + c = 0, a  0, are given by the quadratic formula Solving a Quadratic Equation Using the Quadratic Formula Step 1: Write the equation in standard form ax 2 + bx + c = 0 and identify the values of a, b, and c. Step 2: Substitute the values of a, b, and c into the quadratic formula. Step 3: Simplify the expression found in Step 2. Step 4: Check. Verify your solutions.

Solve 11n 2 – 9n = 1 by the quadratic formula. 11n 2 – 9n – 1 = 0, so a = 11, b = -9, c = -1 The Quadratic Formula Your Turn:

The Quadratic Formula Solve x(x + 6) =  30 by the quadratic formula. x 2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution. Example:

The Discriminant The expression under the radical sign in the formula (b 2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

The Discriminant π

Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x 2 = 0 a = 12, b = –4, and c = 5 b 2 – 4ac = (–4) 2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions. Example:

The Discriminant and the Quadratic Function

Quadratic Function Transformations 3 Types of Transformations  Translations Horizontal Vertical  Dilations  Reflections

Quadratic Function Transformations Translations – Horizontal & Vertical

Quadratic Function Transformations Transformations in the horizontal direction are done to the input of a function (x). Transformations in the vertical direction are done to the output of a function (f(x)). Parent Quadratic Function f(x) = x 2. f(x) = x 2 Output Input

If k is positive, the graph of f(x) = x 2 + k is the graph of y = x 2 shifted upward k units. If k is negative, the graph of f(x) = x 2 + k is the graph of y = x 2 shifted downward |k| units. The vertex is (0, k), and the axis of symmetry is the y-axis. Graphing Quadratic Functions Graphing the Parabola Defined by f(x) = x 2 + k

x y f(x) = x 2 g(x) = x 2 + 3 h(x) = x 2 – 3 Example:

If h is positive, the graph of f(x) = (x – h) 2 is the graph of y = x 2 shifted to the right h units. If h is negative, the graph of f(x) = (x – h) 2 is the graph of y = x 2 shifted to the left |h| units. The vertex is (h, 0), and the axis of symmetry is the vertical line x = h. Graphing Quadratic Functions Graphing the Parabola Defined by f(x) = (x – h) 2

x y f(x) = x 2 g(x) = (x – 3) 2 h(x) = (x + 3) 2 Example:

Graph f(x) = x 2 Graph g(x) = (x – 2) 2 + 4 Example: Continued Graphing Quadratic Functions

f(x) = x 2 g(x) = (x – 2) 2 + 4 x y Example continued:

Quadratic Function Transformations Dilations – Change the shape of curve (narrower or wider)

If |a | > 1, the graph of the parabola is narrower than the graph of f(x) = x 2. If |a | < 1, the graph of the parabola is wider than the graph of f(x) = x 2. Graphing Quadratic Functions Graphing the Parabola Defined by f(x) = ax 2

x y f(x) = x 2 g(x) = 3x 2 h(x) = (1/3)x 2 Example continued: Comparing Quadratic Functions

If a is positive, the parabola opens upward, and if a is negative, the parabola opens downward. Graphing Quadratic Functions Graphing the Parabola Defined by f(x) = ±ax 2

Quadratic Functions of the Form f(x) = ax 2 Example: g(x) = – x 2 f(x) = x 2 x y 22 44 66 88 246 8 44 66 88 2 4 6 8 Notice that the graph of g(x) is a reflection of the graph of f(x) over the x-axis.

Summary of Transformations f(x) = ± a(x – h) 2 + k Reflection about the x-axis. + opens upward - opens downward Dilation a>1 narrower 0<a<1 wider Horizontal Translation Opposite direction of sign + to the left - to the right Vertical Translation Same direction of sign + up - down

5 y x 5-5-5 Example: Compare the graphs of, and

Example: Graph f (x) = (x – 3) 2 + 2 and find the vertex and axis. f (x) = (x – 3) 2 + 2 is the same shape as the graph of g (x) = (x – 3) 2 shifted upwards two units. g (x) = (x – 3) 2 is the same shape as y = x 2 shifted to the right three units. f (x) = (x – 3) 2 + 2 g (x) = (x – 3) 2 y = x 2 - 4- 4 x y 4 4 vertex (3, 2)

Vertex (or Turning Point ) Form

Vertex Form Combining the transformations produces a general expression of the form y = a(x - h) 2 + k (Vertex Form). The graph is a parabola  With vertex (h, k)  And line of symmetry x = h.

From the graph of f(x) = x 2 the graph of g(x) = (x – 2) 2 + 4 is obtained. 1)What is the vertex of g(x)? 2)What is the axis of symmetry of g(x)? Example: (2, 4) x = 2 Vertex Form

Vertex Form – Order of Transformations When graphing a quadratic in vertex form, apply each transformation in the following order. 1)Reflection

Vertex Form – Order Transformations Continued: 2)Dilation – narrower or wider

Vertex Form – Order Transformations Continued: 3)Horizontal Translation

Vertex Form – Order Transformations Continued: 4)Vertical Translations

Without graphing, how does the graph of g(x) = – 4(x + 9) 2 – 1 compare with the graph of f(x) = x 2 ? Reflect about the x-axis so that it opens down and then the parabola becomes narrower by a factor of 4. Translate 9 units to the left and then translate one unit down. The vertex will be at ( – 9, – 1) and the axis of symmetry is the vertical line x = – 9. Example: Quadratic Function Transformations

Graphing Using Transformations: The Pattern of “a” The shape of a parabola follows a specific pattern determined by the value of “a” in the equation of the quadratic, y = ax 2 + bx + c or y = a(x – h) 2 + k. Starting at the vertex, the pattern of “a”, along with the axis of symmetry, can be used to find points on the curve of a quadratic equation. The pattern of “a” starts at the vertex, then follows the sequence; over 1 – up/down “a”, over 1 – up/down 3“a”, over 1 - up/down 5“a”, …

CHANGECHANGE Standard Form Vertex Form

Axis of Symmetry Use the pattern of “a” to find additional points.

Standard Form Vertex Form CHANGECHANGE

Axis of Symmetry

Use the pattern of “a” to find additional points.

Method 1  Vertex form f(x)=a(x-h) 2 +k 1.Calculate the vertex (h,k). 2.Calculate y-intercept (calculate f(0)). 3.Calculate x-intercept (set f(x) =0 and solve a(x-h) 2 +k=0). 4.Use the pattern of “a” or substitute values of x into f(x) to calculate additional points. 5.Use the axis of symmetry, x=h, to calculate points. 6.Sketch the parabola passing through the calculated points. Sketching the Graph of a Quadratic

Standard Form

Properties of Quadratic Function y-intercept x-intercept

The Vertex of a Parabola Any quadratic function f(x) = ax 2 + bx + c, a  0, will have vertex The x-intercepts, if there are any, are found by solving the quadratic equation f(x) = ax 2 + bx + c = 0.

Graphing Using Properties Graphing a Quadratic Function Using Its Properties Step 1: Determine whether the parabola opens up or down (a positive – up, a negative – down). Step 2: Determine the vertex and axis of symmetry (axis of symmetry x = -b/2a, vertex = (-b/2a, f(-b/2a)). Step 3: Determine the y-intercept, f(0). Step 4: Determine the discriminant, b 2 – 4ac. If b 2 – 4ac > 0, then the parabola has two x-intercepts, which are found by solving f(x) = 0. If b 2 – 4ac = 0, the vertex is the x-intercept. If b 2 – 4ac < 0, there are no x-intercepts. Step 5: Plot the points. Use the axis of symmetry to find an additional point. Draw the graph of the quadratic function.

Graphs of y = ax 2 will have similar form and the value of the coefficient ‘a ’ determines the graph’s shape. x y -3 -2 -1 1 2 3 4321 y = x 2 y = 2x 2 y = 1 / 2 x 2 a > 0 opening up

Consider f (x ) = ax 2 +bx + c In a general sense the linear term bx acts to shift the plot of f (x ) from side to side and the constant term c acts to shift the plot up or down. x y c a > 0 c a < 0 y-intercept x-intercept Notice that c is the y -intercept where x = 0 and f (0) = c Note also that the x -intercepts (if they exist) are obtained by solving: y = ax 2 +bx + c = 0

It turns out that the details of a quadratic function can be found by considering its coefficients a, b and c as follows: (1) Opening up (a > 0), down (a < 0) (2) y –intercept: c (3) x -intercepts from solution of y = ax 2 + bx + c = 0 (4) vertex = You solve by factoring or the quadratic formula Consider f (x ) = ax 2 +bx + c

Graphing Using Standard Form Example: Graph f(x) = –2x 2 – 8x + 4. Continued. f(x) = –2x 2 – 8x + 4 abc The x-coordinate of the vertex is The y-coordinate of the vertex is

Graphing Using Standard Form Example continued: The x-intercepts occur where f(x) = 0. –2x 2 – 8x + 4 = 0 Use the quadratic formula to determine the x-intercepts. x   4.4 x  0.4 x y 22 44 6 6 8 8 24 6 8 88  12  16 4 8 12 16 f(x) = –2x 2 – 8x + 4 The y-intercept is f(0) = –2(0) 2 – 8(0) + 4 = 4 The vertex is (–2, 12). y-intercept (0, 4) x-intercepts vertex (– 2, 12) The axis of symmetry is x = – 2.

a = 1, b = -1 and c = -2 (1) opens upwards since a > 0 (2) y –intercept: -2 (3) x -intercepts from x 2 - x - 2 = 0 or (x -2)(x +1) = 0 x = 2 or x = -1 (4) vertex: Plug in x = to find the y-value Your Turn: f (x ) = x 2 - x - 2

x y -2 -1 0 1 2 -2-3 (-1, 0) (2, 0) y = x 2 - x - 2

Sketching the Graph of a Quadratic Method 2  Standard form f(x)=ax 2 +bx+c 1.Calculate x-intercepts (set f(x) =o and solve ax 2 +bx+c=0). 2.Calculate y-intercept (0,c). 3.Calculate vertex (-b/2a, f(-b/2a)). 4.Substitute values of x into f(x) to calculate additional points. 5.Use the axis of symmetry, x=-b/2a, to calculate points. 6.Sketch the parabola passing through the calculated points.

Finding the Quadratic Equation from a Graph

Your Turn: Find the equation of the parabola that has its vertex at (2,3) and passes through the point (0,2) Vertex: (2,3) Point: (0, 2)

Finding the Quadratic Equation from a Graph Find the equation of the parabola that passes through the points (-2,15), (1,12) and (4,-9).

Your Turn: Find the equation of the parabola that passes through the points (-2,19), (1,4) and (3,14). Find the equation of the parabola that passes through the points (-3,-12), (2,8) and (0,-6).

Maximum and Minimum Values The graph of a quadratic function has a vertex at The vertex will be the highest point on the graph if a < 0 and will be the maximum value of f. The vertex will be the lowest point on the graph if a > 0 and will be the minimum value of f. Maximum Minimum Opens up a > 0 Opens down a < 0

Maximum and Minimum Values Example: Determine whether the quadratic function f(x) = –3x 2 + 12x – 1 has a maximum or minimum value. Find the value. Because a < 0, the graph will open down and will have a maximum. f(x) = –3x 2 + 12x – 1 abc The maximum of f is 11 and occurs at x = 2.

Maximum and Minimum Values Example Give the coordinates of the maximum or minimum value for each function. (a)(b) The vertex of the graph is (–1,–18). Since a > 0, the minimum point is (–1,–18), and the minimum value is –18. The vertex of the graph is (–3,1). Since a < 0, the maximum point is (–3,1), and the maximum value is 1.

Applications and Models Sometimes when a quadratic function f is used in applications, the vertex provides important information. The reason is that the y-coordinate of the vertex is the minimum value of f(x) when its graph opens upward and is the maximum value of f(x) when its graph opens downward.

Example 1 A rancher is fencing a rectangular area for cattle using the straight portion of a river as one side of the rectangle. If the rancher has 2400 feet of fence, find the dimensions of the rectangle that give the maximum area for the cattle. Solution Let W be the width and L be the length of the rectangle. Because the 2400-foot fence does not go along the river, it follows that W + L + W = 2400 or L = 2400 – 2W

Solution Solution continued Area of the rectangle equals length times width. This is a parabola that opens downward, and by the vertex formula, the maximum area occurs when

Solution Solution continued The corresponding length is L = 2400 – 2W = 2400 – 2(600) = 1200 feet. The dimensions are 600 feet by 1200 feet.

Applications and Models Another application of quadratic functions occurs in projectile motion, such as when a baseball is hit up in the air. If air resistance is ignored, then the formula s(t) = –16t 2 + v 0 t + h 0 calculates the height s of the object above the ground in feet after t seconds. In this formula h 0 represents the initial height of the object in feet and v 0 represents its initial vertical velocity in feet per second. If the initial velocity is upward, then v 0 > 0 and if the initial velocity is downward, then v 0 < 0.

Example 2 A baseball is hit straight up with an initial velocity of v 0 = 80 feet per second (or about 55 miles per hour) and leaves the bat with an initial height of h 0 = 3 feet, a) Write a formula s(t) that models the height of the baseball after t seconds. b) How high is the baseball after 2 seconds? c) Find the maximum height of the baseball. Support your answer graphically.

Solution a) b) Baseball is 99 feet high after 2 seconds. c) Because a is negative, the vertex is the highest point on the graph, with a t-coordinate of

Example 3 If a metal ball is dropped 100 feet from a water tower, its height h in feet above the ground after t seconds is given by h(t) = 100 – 16t 2. Determine how long it takes the ball to hit the ground. Solution The ball strikes the ground when the equation 100 – 16t 2 = 0 is satisfied.

Example 3 Solution continued The ball strikes the ground after 10/4, or 2.5, seconds.

Example 4 A box is being constructed by cutting 2-inch squares from the corners of a rectangular piece of cardboard that is 6 inches longer than it is wide. If the box is to have a volume of 224 cubic inches, find the dimensions of the piece of cardboard.

Example 4 Solution Step 1:Let x be the width and x + 6 be the length. Step 2:Draw a picture.

Solution continued Since the height times the width times the length must equal the volume, or 224 cubic inches, the following can be written 2(x – 4)(x + 2) = 224 Step 3: Write the quadratic equation in the form ax 2 + bx + c = 0 and factor.

Solution continued The dimensions can not be negative, so the width is 12 inches and the length is 6 inches more, or 18 inches. Step 4:After the 2-inch-square corners are cut out, the dimensions of the bottom of the box are 12 – 4 = 8 inches by 18 – 4 = 14 inches. The volume of the box is then 2814 = 224 cubic inches, which checks.

Your Turn: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is: The path is a parabola opening downward. The maximum height occurs at the vertex. At the vertex, So, the vertex is (9, 15). The maximum height of the ball is 15 feet.

Your Turn: A fence is to be built to form a rectangular corral along the side of a barn 65 feet long. If 120 feet of fencing are available, what are the dimensions of the corral of maximum area? barn corral x x 120 – 2x Let x represent the width of the corral and 120 – 2x the length. Area = A(x) = (120 – 2x) x = –2x 2 + 120 x The graph is a parabola and opens downward. The maximum occurs at the vertex where a = –2 and b = 120 120 – 2x = 120 – 2(30) = 60 The maximum area occurs when the width is 30 feet and the length is 60 feet.

Assignment Pg. 99 – 102: Voc. Check #1 – 5 all, Exercises #1 – 47 odd, 55 – 59 odd.

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