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標題:
Quadratic Function
發問:
最佳解答:
y = 3x^2/4 - 5x/11 + 4/9 = (3/4)(x^2 - 20x/33) + 4/9 = (3/4)(x^2 - 20x/33 + 100/121) + 4/9 - 75/121 = (3/4)(x - 10/11)^2 - 191/1089 so, the vertex is (10/11, -191/1089) 2015-03-03 09:01:24 補充： Sorry, correction from the third line, y = (3/4)(x^2 - 20/33 + 100/1089) + 4/9 - 25/363 = (3/4)(x - 10/33)^2 + 409/1089 so, the co-ordinates of the vertex is (10/33, 409/1089).
其他解答:
Vertex = (-b/2a , (4ac-b^2) /4a) You may use it to ko mcq. X coordinates = -(-5/11) / (2x3/4) = 10/33 Y coordinates = [4(3/4)(4/9)-(-5/11)^2] / [4(3/4)] = 409/1089
Quadratic Function
發問:
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For the function y = 3x^2/4 - 5x/11 + 4/9, find the co - ordinates of the vertex. PLEASE SHOW STEPS, THANKS.最佳解答:
y = 3x^2/4 - 5x/11 + 4/9 = (3/4)(x^2 - 20x/33) + 4/9 = (3/4)(x^2 - 20x/33 + 100/121) + 4/9 - 75/121 = (3/4)(x - 10/11)^2 - 191/1089 so, the vertex is (10/11, -191/1089) 2015-03-03 09:01:24 補充： Sorry, correction from the third line, y = (3/4)(x^2 - 20/33 + 100/1089) + 4/9 - 25/363 = (3/4)(x - 10/33)^2 + 409/1089 so, the co-ordinates of the vertex is (10/33, 409/1089).
其他解答:
Vertex = (-b/2a , (4ac-b^2) /4a) You may use it to ko mcq. X coordinates = -(-5/11) / (2x3/4) = 10/33 Y coordinates = [4(3/4)(4/9)-(-5/11)^2] / [4(3/4)] = 409/1089
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