# 2.5 Graphing Quadratic Functions. There are two main methods for graphing a quadratic I. From Standard Form f (x) = ax 2 + bx + c Axis of Symmetry: Vertex:

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2.5 Graphing Quadratic Functions

There are two main methods for graphing a quadratic I. From Standard Form f (x) = ax 2 + bx + c Axis of Symmetry: Vertex: y-int: (when x = 0) x-int: (when y = 0) set = 0 & solve (aka parabola) (s 1, 0) (s 2, 0)

Ex 1) f (x) = 3x 2 – 9x –12 a) Axis of Symmetry: b) Vertex: c) y-int: (0, ??) (0, –12)

Ex 1) f (x) = 3x 2 – 9x –12 cont… d) x-int: (??, 0) e) Graph (4, 0) and (–1, 0) 0 = 3x 2 – 9x –12 0 = 3(x 2 – 3x –4) 0 = 3(x – 4)(x + 1) x = 4, –1 (0, –12) from prev:

Ex 2) Finding zeros using calculator f (x) = 4x 2 – 13x + 5 y = type in 4x 2 – 13x + 5 2 nd CALC push graph arrow over to right of it & hit ENTER ENTER again do same for other int 2: zero put cursor on left side & hit ENTER 0.446 & 2.804

II. From Vertex Form Change standard to vertex form by completing the square and then use transformations of x 2 to graph Ex 3) Graph f (x) = 2x 2 + 12x + 17 f (x) = 2(x 2 + 6x ) + 17+ 9 f (x) = 2(x + 3) 2 – 1 The graph of x 2 has moved 3 left, 1 down and gotten skinnier by the number 2 – 18

Ex 4) Do it the other way! Write equation from a picture y = ____(x – ____ ) 2 + ____ vertex: (2, 4) y = –2(x – 2) 2 + 4 skinny by 2  a = –2 upside down  a is (–) How??? To find ‘a’: plug in a point (1, 2): y = a(x – 2) 2 + 4 2 = a(1 – 2) 2 + 4 –2 = a

Homework #205 Pg 91 #1–29 odd, 30–35 all, 36, 38, 46

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