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Warm-Up Find the x and y intercepts: 1. f(x) = (x-4)2-1 2. f(x) = -2(x-3)(x+4) 3. f(x) = x2-2x -15 Homework: 3.1 A Worksheet due 10/18 3.1B Worksheet due 10/19 Lesson 3.1 Quadratic Functions– The Three Forms 1. Standard or General Form y = ax2+ bx + c y-intercept This form tells me the __________________. crosses the y-axis That is where the graph ________________. (0,c) The y-intercept is _____________________. 2. Factored Form y = a( x – r1) (x-r2) roots, zeros, or x-intercepts This form tells me the __________________. crosses the x-axis That is where the graph ________________. ( x – r1) (x-r2) = 0 To find the roots, zeros or x-intercepts Set _______________. 3. Vertex Form y = a( x –h)2+k vertex This form tells me the __________________. or maximum of the parabola That is where the graphMinimum ________________. (h,k) The vertex is _____________________. How to convert from different forms? Given: General Form Factored Form Vertex Form Given: Factored Form General Form Vertex Form Given: Vertex Form General Form Factored Form Ex. 1: y = x2-6x + 5 General Form What form? Factored Form Change from general to ____________ by Factoring using either _______________ or ___________________. Quadratic Formula X2 – 6x + 5 (x-5)(x-1) = 0 X = 5, x = 1 Ex. 2: y = x2-6x + 5 Vertex form Change from general form to __________ Completing the square by using _________________________. y = x2-6x + 5 Complete the square 9 9 (x2-6x + _____) = -5 + _____ (x-3)2 -4 = y Now Graph: (0,5) Y-intercept:_______ (3,-4) Vertex :_______ (1,0) X-intercepts: (5,0) ______, ______ Ex. 3: Using f(x) = -3x2 +6x - 13 find the vertex when given the standard form? Use the formula to find x : b x 2a In standard form a = -3, b = 6, c= -13 6 6 x 1 2(3) 6 Substitute in x to find y. x = 1 , then y = -3(1)2 +6(1) -13 = -10 Vertex = (1, -10) 3b. You try: y = (x+4)2-13 HINT: If you cannot factor it, you must use the Quadratic Formula!! Answers: GF: x2+8x+3 FF: y = (x+.39) (x+7.61) Ex. 4: Minimum and Maximum Quadratic Functions Consider the quadratic function: f(x) = -3x2 + 6x – 13 a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the max. or min. value and determine where it occurs c. Identify the function’s domain and range. Ex. 4 Continued f(x) = -3x2 + 6x – 13 Begin by identifying a, b, and c. a. Because a < 0, the function has a max. value. If a>0 then the function would have a min. b. The max. occurs at x = -b/2a = - 6 /2(-3) = -6/-6 = 1. The maximum value occurs at x = 1 and the maximum value of f(x) = -3x2 + 6x – 13 f(1) = -3*12 + 6*1 – 13 = -3 + 6 – 13=-10 Plug in one for x into original function. We see that the max is -10 at x = 1. c. Domain is (-∞,∞) Range (-∞,-10]. 4b. YOU TRY!! Repeat parts a through c using the function: f(x) = 4x2 – 16x + 1000. Answer: a. Min b. Min is 984 at x = 2 c. Domain is (-∞,∞) Range is [984,∞) Summary: Describe how to find a parabola’s vertex if its equation is expressed in vertex form.