#### Transcript Lesson 1-2

Lesson 2.2, page 273 Quadratic Functions Objectives Recognize characteristics of parabolas. Graph parabolas. Determine a quadratic function’s minimum or maximum value. Solve problems involving a quadratic function’s minimum or maximum value. Vocabulary • Quadratic Function: a function that can be written in the form f(x) = ax2 + bx + c, a ≠ 0. • Standard form: f(x) = a(x - h)2 + k • Parabola: graph of quadratic function; Ushaped curve; symmetric • Symmetry: when folded, two sides match exactly f(x) = a(x - h)2 + k • Vertex: the point (h,k); highest or lowest point of parabola; on axis of symmetry • a: describes steepness and direction of parabola • Axis of symmetry: x = h; fold line that divides parabola into two matching halves • Minimum: if a > 0, parabola opens up; vertex is minimum point or lowest point of parabola; k is the minimum value • Maximum: If a < 0, parabola opens down; vertex the maximum point or lowest point of parabola; k is the maximum value Minimum (or maximum) function value for a quadratic occurs at the vertex. • Identify the vertex of each graph . Find the minimum or maximum value. • A. B. y y 10 10 x -10 -10 10 x -10 -10 10 Review Solving Quadratic Equations Example: Get in standard form. Factor. (3x + 7)(x – 3) = 0 3x+7=0 or x–3 = 0 Set each factor equal to zero. Solve. Check. 3x2 – 2x = 21 3x2 – 2x – 21 = 0 x = -7/3 or x = 3 If no “B” term, can use square root property to solve for zeros. For example, if x2 = 16, then x = ±4. • Solve f(x) = 1 – (x – 3)2. When factoring or square root property won’t work, you can always use the QUADRATIC FORMULA x • • • • b b 4 ac 2 2a Get in standard form: ax2 + bx + c = 0 Substitute for a, b, and c in formula. Solve. Check. This works every time!!! Try these. Solve using the Quadratic formula. • 3x2 - 2x - 1 = 0 x2 + 2x = -3 Graphing Quadratic Functions f(x) = a(x - h)2 + k 1. Determine whether the parabola opens up (a > 0) or down (a < 0). 2. Find the vertex, (h,k). 3. Find x-intercepts by solving f(x) = 0. 4. Find the y-intercept by computing f(0). 5. Plot the intercepts, vertex, and additional points. Connect with a smooth curve. Note: Use the axis of symmetry (x = h) to plot additional points. Check Point 1, page 276 • Graph f(x) = -(x – 1)2 + 4. y 10 x 10 -10 -10 Check Point 2, page 276 • Graph f(x) = (x – 2)2 + 1. y 10 x -10 -10 10 f(x) = ax2 + bx + c • If equation is not in standard form, you may have to complete the square to determine the point (h,k). 2 f ( x) 2 x 4 x 3 Graph of f ( x) 2 x 4 x 3 2 Finding the vertex when in ax2 + bx + c form. Vertex = b b , f 2a 2a Check Point 3, page 279 • Graph f(x) = -x2 + 4x +1. Identify domain & range. y 10 x -10 10 -10 Re-Cap Quadratic Equations & Functions ax2 + bx + c = 0 or f(x) = ax2 + bx + c, a 0 • “x” is squared. • Graph would be a curve. • Solutions are at the x-intercepts. The “zeros” are the solutions or roots of the equation. • To solve: factor & use zero product property, take square root of both sides (if no B term), or use other methods. See Example 4, page 279 Check Point 4, page 280: f(x) = 4x2 - 16x + 1000 a) Determine, without graphing, whether the function has a minimum value or a maximum value. b) Find the minimum or maximum value and determine where it occurs. c) Identify the function’s domain and range. Example 1- Application Problem A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball, in meters, t seconds after the hit is given by the quadratic function h ( t ) 4 . 9 t 34 . 3t 1 2 How long does it take for the baseball to reach its maximum height? What is the maximum height obtained by the baseball? Solution Vertex: x b 2a 34.3 2( 4.9) 3.5 seconds h 3.5 4.9 3.5 34.3(3.5) 1 2 61.025 m eters Example 2 - Application A farmer has 600 feet of fencing to enclose a rectangular field. If the field is next to a river, so no fencing is needed on one side, find the dimensions of the field that will maximize the area. Find the maximum area. Area = (600 2x)(x) A(x) = 2x2 + 600x Find Vertex – to find max. Solution x 600 2( 2) = 150 y = 2(150)2 + 600(150) = 45000 x x 600 – 2x Width (x) = 150 Length(600-2x) = 300 Max area = 45000