Chapter II Quadratic Function (Volume II)

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1 1, upward straight line X =- 1 (- 1, -5) 12, Y = (X-2) 2- 1 etc. 13,18.-116, y = y = x 217, (2)(3)(4) 18, all curves are symmetrical.

2 1, image omitted, x= 1, y= 122. (1) From the image, we know that the equation x 2-2x-3 = 0, and we get the solution x 1 = 3, x2 =- 1 (. = 1 or x>3, and the function value is greater than 0 (3)- 1

23.Y =-2x 2+4x-5224, ∵PA⊥x axis, and the ordinate of point ap= 1∴p is 1. When y= 1, 3/4x 2-3/2x+. ∵ Parabolic symmetry axis is x= 1, the right side of symmetry axis is point p, ∴x= 1+ root number 2, and the area of rectangular PAOB is (1+ root number 2) square unit 25. ( 1) A =-。 Therefore, the column length MN is 10-4.5 = 5.5m. (3) Let DE be the width of the isolation belt and EG be the sum of the widths of three vehicles, then the coordinate of G point is (7,0) (7 = 2 ÷ 2+2× 3), YH = 3/ When passing through G point, it is a gh vertical AB parabola at H point. 26. The expression of quadratic function that meets the requirements is: y =1/3(x- 1)2-1,y = root number 3 (x-1) 2-root number 3, y =-.