Quadratic Functions Vertex & Axis of Symmetry Dr. Fowler  AFM  Unit 2-3 Quadratic Functions Vertex & Axis of Symmetry Transformations Maximizing Revenue

Graphing Quadratics – Introduction and Important Terms: https://www.youtube.com/watch?v=vpTvNFXvj38

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Without graphing, locate the vertex and axis of symmetry of the parabola defined by . Does it open up or down?

Since a = 2 > 0 the parabola opens up and therefore will have no x-intercepts.

The domain of f is the set of all real numbers.

GRAPHING Standard form: y = ax² + bx + c Vertex form: y = a(x-h)² + k

From Standard to Vertex Form • What's the pattern? + x 6 x2 6x 36 (x + 6)2 x2 + 12x + 36 • How about these? x2 + 4x ______ (x _____ )2 + 4 + 2 x2 + 10x ______ (x _____ )2 + 25 + 5 x2 – 14x ______ (x _____ )2 + 49 – 7

From Standard to Vertex Form • Converting from standard form to vertex form can be easy… x2 + 6x + 9 (x + 3)2 x2 – 2x + 1 = (x – 1)2 x2 + 8x + 16 = (x + 4)2 x2 + 20x + 100 = (x + 10)2 … but we're not always so lucky

From Standard to Vertex Form • The following equation requires a bit of work to get it into vertex form. y = x2 + 8x + 10 y = (x2 + 8x ) + 10 + 16 – 16 16 is added to complete the square. 16 is sub-tracted to maintain the balance of the equation. y = (x + 4)2 – 6 The vertex of this parabola is located at ( –4, –6 ).

From Standard to Vertex Form • Lets do another. This time the x2 term is negative. y = –x2 + 12x – 5 Un-distribute a negative so that when can complete the square y = (–x2 + 12x ) – 5 y = –(x2 – 12x ) – 5 y = –(x2 – 12x ) – 5 + 36 + 36 The 36 in parentheses becomes negative so we must add 36 to keep the equation balanced. y = – (x – 6)2 + 31 The vertex of this parabola is located at ( 6, 31 ).

From Standard to Vertex Form • The vertex is important, but it's not the only important point on a parabola y-intercept at (0, 10) x-intercepts at (1,0) and (5, 0) Vertex at (3, -8)

Standard Form to Vertex Form Write the equation in vertex form: y = 3x2 – 6x + 8

Standard Form to Vertex Form Write the equation in vertex form: y = 3x2 – 6x + 8 y = 3(x2 – 2x ) + 8 factor out “a”

Standard Form to Vertex Form Write the equation in vertex form: y = 3x2 – 6x + 8 y = 3(x2 – 2x ) + 8 factor out “a” y = 3(x2 – 2x + 1) + 8 – 3 complete the square

Standard Form to Vertex Form Write the equation in vertex form: y = 3x2 – 6x + 8 y = 3(x2 – 2x ) + 8 factor out “a” y = 3(x2 – 2x + 1) + 8 – 3 complete the square y = 3(x – 1)2 + 5 factor and combine

Example 1: Graph y = (x + 2)2 + 1 Step 1 Plot the vertex (-2 , 1) Step 2 Draw the axis of symmetry, x = -2. Step 3 Find and plot two points on one side, such as (-1, 2) and (0 , 5). Step 4 Use symmetry to complete the graph, or find two points on the left side of the vertex.

Video Angry Birds Parabalos https://www.youtube.com/watch?v=X-mXg9MTTW0

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Since a = 2 > 0 the parabola opens up.

The domain of f is the set of all real numbers.

Since a is negative, the parabola opens down.

The domain of f is the set of all real numbers.

Since a is negative, the graph of f opens down so the function will have a maximum value.

Excellent Job !!! Well Done