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Published byEugene Colie Modified over 2 years ago

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2.1 Quadratic Functions Use the graphing calculator and graphy = x 2 y = 2x 2 y = The quadratic function f(x) = a(x – h) 2 + k is said to be in standard form. Vertex (h,k) If a > 0, opens up. If a < 0, opens down.

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Ex. Write the quadratic function in standard form, find the vertex, identify the x-int.’s and sketch. f(x) = 2x 2 + 8x + 7 f(x) = 2x 2 + 8x + 7Factor out a 2 from the x’s f(x) = 2(x 2 + 4x ) + 7 Complete the square + 4 - 8 f(x) = 2(x + 2) 2 - 1 V(, ) Up or Down Take the original and set = 0 to find the x-int. 0 = 2x 2 + 8x + 7 Use quad. formula on calc. x = -1.2928, -2.7071Now sketch it. -2 -1

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(-2,-1)

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Ex.Sketch the graph of f(x) = -x 2 + 6x - 8 f(x) = -x 2 + 6x - 8Factor -1 out of the x terms f(x) = -(x 2 – 6x ) - 8 Complete the square + 9 f(x) = -(x – 3) 2 + 1V(3, 1)Down

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To find a vertex of the quadratic f(x) = ax 2 + bx + c, (without putting the equation into standard form), evaluate by letting Ex.P =.0014x 2 -.1529x + 5.855 y = 1.68

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Find the the equations of a parabola that cups up and down given the x-intercepts (1, 0) and (-3, 0). First, write the x-intercepts as factors. y = (x - 1)(x + 3) The parabola that cups up is y = x 2 + 2x - 3 The parabola that cups down requires a negative in front of the quadratic. y = -(x 2 + 2x - 3) or y = -x 2 - 2x + 3

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Given the vertex and point on a given parabola, find the standard equation of the parabola. V(4, -1); point (2, 3) Fill in the points in the standard parabola form y = a(x - h) 2 + k and solve for a. 3 = a(2 - 4) 2 + (-1) 3 = 4a -1 4 = 4a 1 = a The answer is y = 1(x - 4) 2 - 1

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