On quadratic function: using area to find analytic formula, using the relationship between root and coefficient to find analytic formula Example 1: given that the image of quadratic function y = ax ^ 2-4ax + B passes through a (1,0), B (x2,0), intersects with the positive half axis of Y axis and C point, and s Δ ABC = 2. Find the analytic expression of quadratic function Example 2: given that the parabola y = ax ^ 2-2ax-8a + 5 passes through the point P (- 2,5), intersects with the x-axis and a (x1,0), B (x2,0), X1 < X2, s △ PAB = 10, find the analytical formula of the parabola

1. First of all, we use Weida's theorem. Because X1 and X2 are on the X axis, they are two solutions of quadratic function, and we get X1 + x2 = - B '\ a = 4A \ \ a = 4
Substituting x = 0 into the equation, we can get y = B, C (0, b) and the ordinate of C point is the height of triangle. Because X1 = 1, we can get x2 = 4-x1 = 3, and the bottom edge of B (3,0) triangle is x2-x1 = 3-1 = 2
Because the area of triangle is (x2-x1). B = 2, we can get b = 2
Then we use the Veda theorem. X1. X2 = C / a '= B / a = 3 and substitute B = 2 to get a = 2 / 3
Finally, a = 2, B = 2 can be substituted into the equation to solve the function
2. We also use Weida's theorem X1 + x2 = - B / a '= 2A / a = 2, and because both X1 and X2 are on the x-axis, the vertical and horizontal coordinates of point P are the height of the triangle, s △ PAB = (x2-x1). 5 / 2 = 10, we can get x2-x1 = 4, then we can get x2 = 3 and X1 = - 1 by combining X1 + x2 = 2 and x2-x1 = 4, finally we use Weida's theorem x1. X2 = C / a' = - 8A + 5 / a = - 3, a = 1, and substitute y = x ^ 2-2x-3 in the function

RELATED INFORMATIONS