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Quadratic Functions Review / Warm up

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f(x) = ax^2 + bx + c. In this form when: a>0 graph opens up a<0 graph opens down b2 - 4ac > 0 Graph has 2 x-intercepts b2 - 4ac = 0 Graph has 1 x-intercept b2 - 4ac < 0 Graph has no x-intercepts Vertex Point at x = b/2a to find y substitute above value for x x-intercepts - factor or use quadratic formula y-intercept - “c” value in ax^2 + bx + c

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Sample Problems 1) f(x) = x^2 + 2x – 3 Find the vertex point Find the axis of symmetry Find the x and y intercepts Graph the function

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Solution: a > 0, graph opens up The vertex point is at: x = -b/2a = -2/2 = -1 y = (-1)2 + 2(-1) - 3 = -4 Vertex point is at (-1, -4)

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Solution: Axis of symmetry: x = -1 x-intercepts found by factoring (x + 3)(x - 1) = 0 They are at x = -3, x = 1 y-intercept at (0, c) at (0, -3)

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The graph is as follows:

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Alternate method! A second form of a quadratic can be written in the following form: f(x) = a(x - h)2 + k Where (h, k) is the vertex point

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