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Published byEugene Colie Modified over 3 years ago

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2.1 Quadratic Functions Use the graphing calculator and graph y = x2 y = 2x2 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) = 2x2 + 8x + 7 f(x) = 2x2 + 8x + 7 Factor out a 2 from the x’s f(x) = 2(x2 + 4x ) + 7 + 4 - 8 Complete the square f(x) = 2(x + 2)2 - 1 V( , ) Up or Down Take the original and set = 0 to find the x-int. 0 = 2x2 + 8x + 7 Use quad. formula on calc. x = , Now sketch it.

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(-2,-1)

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**Ex. Sketch the graph of f(x) = -x2 + 6x - 8**

Factor -1 out of the x terms f(x) = -(x2 – 6x ) - 8 + 9 + 9 Complete the square f(x) = -(x – 3)2 + 1 V(3, 1) Down

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**To find a vertex of the quadratic f(x) = ax2 + bx + c,**

(without putting the equation into standard form), evaluate by letting Ex. P = .0014x x 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 = x2 + 2x - 3 The parabola that cups down requires a negative in front of the quadratic. y = -(x2 + 2x - 3) or y = -x2 - 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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