It is known that the parabola and X-axis intersect at two points a-1,0, b1,0, and the analytic expression of the function is obtained through m0,1

It is known that the parabola and X-axis intersect at two points a-1,0, b1,0, and the analytic expression of the function is obtained through m0,1

Let the parabola be y = ax & # 178; + BX + C, (a ≠ 0)
Because the parabola passes through M (0,1)
So C = 1
So y = ax & # 178; + BX + 1
Because a (- 1,0), B (1,0)
So substituting: A-B + 1 = 0, a + B + 1 = 0,
So the solution is a = - 1, B = 0,
So the analytic expression of the function is y = - X & # 178; + 1

It is known that the quadratic function and X-axis intersect at a (- 2,0) B (4,0) and pass C () - 1,5. Find the analytical expression of the parabolic function

Let the analytic expression be
y=ax^2+bx+c
a. Two points B are the roots of the equation AX ^ 2 + BX + C = 0
According to Weida's theorem
-2+4=-b/a
-2×4=c/a
b=-2a
c=-8a
By substituting these two, together with the C coordinate, into the analytical expression, we get the following result:
5=a(-1)^2-2a(-1)+(-8a)
-5a=5
a=-1
b=2
c=8
The analytic formula is y = - x ^ 2 + 2x + 8