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The function y = 2x & # 178; - BX + C passes through (1,2), (0,4) point 1. Find the analytic expression of quadratic function 2. When the independent variable x takes a value in what range, y follows X And increase 3, the quadratic function has a maximum or a minimum? If so, when x takes what value, the function gets the maximum or the minimum? And find the maximum and minimum
1. If the function y = 2x & # 178; - BX + C passes through the point (1,2), (0,4), the two-point coordinates are substituted into the analytic formula of the function to get 2-B + C = 2 and the solution of C = 4 to get b = 4, so the analytic formula of the function is y = 2x & # 178; - 4x + 42
(1) Translate the image of the quadratic function y = 2x & # 178; up one unit along the y-axis, and find the analytic expression of the translation
The image of quadratic function y = 2x & # 178; is translated up one unit along the Y axis,
The analytical expression after translation is: y = 2x & # 178; + 1
It is known that the relationship between the independent variable X of quadratic function y = ax & # 178; + BX + C (a ≠ 0) and the function value y satisfies the following quantitative relationship:
x: - 4, y = 24; X = - 3, y = 15; X = - 2, y = 8; X = - 1, y = 3; X = 0, y = 0; X = 1, y = - 1; X = 2, y = 0; X = 3, y = 3; X = 4, y = 8; X = 5, y = 15; find (1) observe the data in the table, when x = 6, the value of Y is -;
(2) The coordinate of the intersection of the quadratic function and the x-axis is -;
(3) The value of the algebraic formula - B + √ B & # 178; - 4ac / 2A + - B - √ B & # 178; - 4ac / 2A + (a + B + C) (a-b + C) is
(4) If s and T are two unequal real numbers, when s ≤ x ≤ T, the quadratic function y = ax & # 178; + BX + C (a ≠ 0) has a minimum value of 0 and a maximum value of 24, then the analytic expression of the inverse proportional function passing through the point (s + 1, t + 1) is————
egdfgdgdrtutjfghfhyurtu
The partial corresponding values of independent variable x and function y of quadratic function y = ax ^ 2 + BX + C are shown in the table below
x -1 0 1 2 3
y -1 -7/4 -2 -7/4 （ ）
1, when x = 3, y = -——
2. When x = - -, y has the maximum value of -——
3 if points a (x1, Y1), B (X2, Y2) are two points on the image of the quadratic function, and - 1 < X1 < 0,1 < x2 < 2, try to compare the values of the two functions: Y1_____ y2
(1) According to the table, the symmetry axis of the quadratic function is: x = 1 (because the Y values corresponding to x = 0 and x = 2 are equal), then we can know that when x = 3, y = - 1; (2) according to (1), on the symmetry axis X = 1, that is, at the vertex (1, - 2) of the function image, we can get the minimum value of - 2; (3) geometric method: we can draw this quadratic function
It is known that the partial corresponding values of the independent variable x and the function value Y in the quadratic function y = ax ^ 2 + BX + C (a ≠ 0) are as follows
x.-3 -2 -1 0 1
y.-6 0 4 6 6
(1) Find the analytic expression of quadratic function and write out the vertex coordinates (detailed process);
(2) The parabola image is drawn in the linear coordinate system;
(3) If the abscissa of two points a (x1, Y1) and B (X2, Y2) on the parabola meets x1 ＜ x2 ＜ - 1, try to compare the size of Y1 and Y2
I just started teaching quadratic function last semester, and I learned the general form matching method of quadratic function. How can I draw this graph? I can't even find the vertex, but the teacher said that there was no wrong print on the test paper, and that's how the table is
Like you, I just learned quadratic function in the first semester of junior three, but I can tell you (1) with 9a-3b + C = 0, C = 6, 4A + 2B + C = 0A = - 1, B = 1, C = 6y = - x2 + X + 6h = - B / 2A = 1 / 2, k = 4A / 4ac-b2 = 4 / 25y = - (x-1 / 2) 2 + 4 / 25 vertex coordinates (1 / 2, 4 / 25) (2) according to the first four points
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