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2.1 Quadratic Functions Completing the square Write Quadratic in Vertex form Definition of a Polynomial Function F(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + .. + a 1 x 1 + a0 an , a n-1,… are numbers In “x n” n is the exponents going down in degree. This would be a polynomial function of x with degree n. Linear functions are first degree Quadratic functions are second degree f(x) = ax2 + bx + c Would be a graph of a Parabola A parabola is Axis of symmetry symmetric; “x = h” Where h is from the vertex (h, k) If a > 0, then the parabola opens up If a < 0, then the parabola opens downward What equation do you think of when you hear Quadratic ? This equation? Do you remember how to use it? Standard form of a Quadratic equation f(x) = a(x – h) 2 + k How would you graph the equation f(x) = 2(x + 3) 2 + 1 The vertex has moved off the origin 3 units to the left and 1 unit up. Since a = 2,the parabola opens up and gets skinny. The vertex is at (-3, 1) f(x) = 2(x + 3) 2 + 1 Finding the vertex when it is not in Standard form f(x) = x 2 + 12x – 9 We need to complete the square to find the standard form Take half of b squared and add and subtract to the function. f(x) = x 2 + 12x + (6)2+ ( -36) – 9 Factor the first three terms, then add the last terms f(x) = (x + 6) 2 – 45 vertex ( - 6, -45) Find the standard form of f(x) = x 2 + 10x + 8 Write the equation with the vertex ( - 4, 8) and the point (1, 4) Start with the standard equation f(x) = a(x – h) 2 + k from the vertex h = - 4, k = 8 from the point x = 1, f(x) = 4 4 = a(1 – (- 4)) 2 + 8 4 = a(5)2 + 8 4 = 25a + 8 -4 = 25a -4/25 = a Rewrite with a, h and k f(x) = -4/25 (x +4)2 + 8 Find the vertex from f(x) = ax2 + bx + c To find “h” we use “b” and “a” To find “k” we place h back in the equation. Lets find the vertex of the equation f(x) = 4x 2 + 3x – 8 Now let h = -⅜ k = 4(- ⅜)2 + 3(⅜) – 8 Vertex is ( - ⅜, - 137/16) k = -137/16 Homework Page 116 – 117 #3, 11, 17, 19, 25, 31, 45, 59, 63 Homework Page 116 #22, 33, 51, 64, 73, 77, 80, 83, 89, 91, 104