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Chinese Normal University version of mathematics ninth grade book plan: practice and exploration exercises (with answers) data download Huashida version of mathematics ninth grade book plan: practice and exploration exercises ( There are answers) China Normal University version of the ninth grade book practice and exploration exercises 1. Selective question 1, the known parabola y = x2-x - an intersection with the x axis is (m, 0), then algebraic m2-m + The value of 2014 is ( ) A. 201B. 201C. 201D. 2012, the quadratic function y=ax2+bx+c (a鈮?, a, b, c is a constant) image, as shown, ax2+bx+c=m has a real root condition () A. m鈮?2B. m鈮?C. m鈮?D. m>4 The corresponding value of the y and x parts of the known quadratic function y=ax2+bx+c is as follows: The correct one in the following judgment is ( ) A. Parabolic opening up B. The parabola intersects the y axis on the negative half axis C. When x=3, y<0D. The equation ax2+bx+c=0 has two equal real root quadratic functions y=ax2+bx+c(a鈮?) as shown in the figure, then the function value y>0, the value range of x Yes ( ) A. x<-1B. x>3C. -13 quadratic function y=x2+bx is shown in the figure, the axis of symmetry is a straight line x=1, if the quadratic equation x with respect to x is x2+bx-t=0 (t is a real number) If there is a solution in the range of -10) between the speed x (m/s) at the start of braking, and if the braking distance of the car is 5m, the brake starts. The speed is ( ) A. 40m/sB. 20m/sC. 10m/sD. 5m/s, the image of the quadratic function y=ax2+bx+c(a鈮?) intersects the x-axis at points A and B, and the y-axis intersects at point C, point B coordinates (-1, 0), the following four conclusions: 1OA=3; 2a+b+c<0; 3ac>0; 4b2-4ac>0. The correct conclusion is ( ) A. 14B. 13C. 24D. 12, the symmetry axis of the parabola y=ax2+bx+c(a>0) is a straight line passing the point (1,0) and parallel to the y axis. If the point P(4,0) is on the parabola, then The value of 4a-2b+c is . As shown in the figure, the image of the quadratic function y=ax2+bx+3 passes through the point A(-1,0), B(3,0), then the root of the quadratic equation ax2+bx=0 is . The abscissa of the intersection A of the parabola y=x2+1 and the hyperbola y= is 1, and the solution set for the inequality of x-+x2+1<0 is . As shown in the figure, the parabola y=ax2+bx intersects the straight line y=kx at two points O(0,0) and A(3,2), then the solution set of the inequality ax2+bx0. Solution: Let y=x2-3x+2, draw y=x2-3x+2 as shown in the figure. It can be seen from the image that when x<1 or x>2, y>0. So the solution set of the quadratic inequality x2-3x+2>0 is x<1 or x>2. Fill in the blank: (1) The solution set of x2-3x+2<0 is 10 is x<-1 or x>1; a similar method is used to solve the one-two Sub inequality -x2-5x+6>0. As shown in the figure, the volleyball player stands at the point O to practice the serve, and the ball is sent from the point A at the top of the point O 2 meters to see the ball as a point, the height of its operation y (meter) and the horizontal distance of the operation x (m) Satisfying the relationship y=a(x-6)2+h, the horizontal distance between the net and the point O is 9 meters, the height is m, and the horizontal distance of the boundary of the stadium is 18 meters. (1) When h =, find the functional relationship between y and x. (2) When h=, can the ball cross the net? Will the ball go out of bounds? Please explain why. (3) If the ball must be able to cross the net, there is no border. What is the range of values 鈥嬧€媜f h? As shown in the figure, the parabola y=ax2+bx+c(a鈮?) intersects the x-axis at points A(-1,0), B(3,0), and the y-axis intersects at point C(0,- 3). (1) Find the analytical expression of the parabola and the vertex M coordinate; (2) Find the ratio of the 螖BCM area to the 螖ABC area; (3) If P is a moving point on the x-axis, pass P as the ray PQ鈭C parabola At point Q, with the movement of point P, is there such a point Q on the parabola, so that the quadrilateral with A, P, Q, and C as the vertices is a parallelogram? If it exists, request the coordinates of Q point; if it does not exist, please explain the reason. X1=0, x2=200 is x<-1 or x>1; let y=-x2-5x+6, solution -x2-5x+6 =0, x1=-6, x2=1, so the solution set of the quadratic inequality-x2-5x+6>0 is -6 , so the ball can pass the net; when y = 0, (x-6) 2+ = 0, the solution is: x1 = 6 + > 18, x2 = 6 - (toss) is out of bounds; (3) when When the ball just passes the point (18,0), the parabola y=a(x-6)2+h is also past the point (0,2), and is substituted into the analytical formula:, the solution is obtained. At this time, the quadratic function analytical expression is: y=(x-6)2+, if the ball does not exit the boundary h鈮? when the ball can just pass the net, the function resolves (9,), and the parabola y=a(x-6)2+h Also past the point (0,2), substituted into the analytical formula:, the solution, at this time the ball has to pass the network h 鈮? so if the ball must be able to cross the net, and no boundaries, the value range of h is: h 鈮?. Solution: (1) Let the parabolic equation be y=a(x+1)(x-3), 鈭?parabola over point (0,3), 鈭?3=a(0+1)(0-3), 鈭碼=1, 鈭?parabola analytical expression is y=(x+1)(x-3)=x2-2x-3, 鈭祔=x2-2x-3=(x-1)2-4, 鈭碝( 1,-4). (2) As shown in Figure 1, connect BC, BM, CM, MD 鈯 axis to D, 鈭礢鈻矪CM=S trapezoid OCMD+S鈻矪MD-S鈻矪OC=(3+4)1+2-4- 33=+-=3S鈻矨BC=ABOC=43=6, 鈭碨鈻矪CM: S鈻矨BC=3:6=1:2. (3) Exist, for the following reasons: 1 as shown in Fig. 2, when Q is below the x-axis, QE鈯 axis is E, 鈭?quadrilateral ACQP is parallelogram, 鈭碢Q is parallel and equal to AC, 鈭粹柍PEQ鈮屸柍AOC, 鈭碋Q=OC=3, 鈭?3=x2-2x-3, solve for x=2 or x=0 (coincident with point C, round off), 鈭碤(2,-3). 2, as shown in Fig. 3, when Q is above the x-axis, QF 鈯?x-axis is F, 鈭?quadrilateral ACPQ is parallelogram, 鈭碤P is parallel and equal to AC, 鈭?鈻?PFQ 鈮?鈻?AOC, 鈭?FQ = OC = 3, 鈭?3=x2-2x-3,.


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