Quadratic function y = - 1 / 2x ²+ The focus of X + 4 image and X axis are a, B, C and D respectively from right to left 1. Find the quadrilateral ABCD area 2. Find a point d# on the parabola in the first quadrant, which is the largest quadrilateral abcd# area

Quadratic function y = - 1 / 2x ²+ The focus of X + 4 image and X axis are a, B, C and D respectively from right to left 1. Find the quadrilateral ABCD area 2. Find a point d# on the parabola in the first quadrant, which is the largest quadrilateral abcd# area

1、y = (-1/2)x ² + X + 4Y = (- 1 / 2) (x-4) (x + 2) when y = 0, x = 4 or - 2A: (4,0), B: (- 2,0) when x = 0, y = 4C: (0,4) y = (- 1 / 2) (x-1) ² + 9 / 2D: (1,9 / 2) using the segmentation method, it is divided into three regions. The vertices of the first region are a, D and (1,0)

Known quadratic function y = x ²- (m ²+ 5)x+2m ²+ 6. Does the image of the function have two intersections with the x-axis? If there are two intersections Find the coordinates of one of the points. If there are no two intersections, please explain the reason. I'm Xiaobai,

Discriminant Δ= (m^2+5)^2-4(2m^2+6)
=m^4+10m^2+25-8m^2-24
=m^4+2m^2+1
=(m^2+1)^2
Because m ^ 2 ≥ 0, m ^ 2 + 1 ≥ 1 > 0
therefore Δ> 0, so. The image of the function must have two intersections with the X axis
x=[(m^2+5)±(m^2+1)]/2
x1=m^2+3,x2=2
The coordinates of the two intersections are (m ^ 2 + 3,0), (2,0)

Known quadratic function y = - x ²+ 2m image passes through the point (- 4, m) to find the intersection of the quadratic function image with the x-axis and y-axis

Known quadratic function y = - x ²+ 2m image passes through the point (- 4, m), and find the intersection of the quadratic function image with the x-axis and y-axis, x = - 4; y=-16+2m=m; ∴m=16; ∴y=-x ²+ 32; ∴x=0; y=32; Y-axis intersection (0,32) y = 0; x=±4√2; The intersection of X axis is (4 √ 2,0) or (- 4 √ 2,0). Hello, I'm glad to answer for you

When the function y = MX squared- When the image of (3m + 1) x + 2m + 1 (M is a constant) has only two intersections on the coordinate axis, find the value of M

There are and only two intersections
Then the discriminant is greater than 0 and X ² Coefficient not equal to 0
[-(3m+1)] ²- 4m(2m+1)>0
9m ²+ 6m+1-8m ²- 4m>0
(m+1) ²> 0
So m ≠ - 1 and m ≠ 0

Given that the graph of the quadratic function passes through (2, - 3), the axis of symmetry x = 1, and the distance between the intersection of the parabola and the X axis is 4, the analytical formula of the quadratic function is obtained

∵ the distance between the two intersection points of parabola and X axis is 4, and x = 1 is the axis of symmetry
The coordinates of two intersections of parabola and X axis are (- 1, 0), (3, 0)
Let the analytical formula of parabola y = a (x + 1) (x-3)
Another ∵ parabola passes through (2, - 3) points
∴-3=a（2+1）（2-3）
The solution is a = 1
The analytical formula of quadratic function is
y=（x+1）（x-3）=x2-2x-3．

Given that the image of the quadratic function passes through (2, - 3), then the axis of symmetry x = 1, and the distance between the intersection of the parabola and the X axis is 4, find the analytical formula of the quadratic function?

Let y = ax ^ 2 + BX + C, the axis of symmetry is x = 1, so - B / (2a) = 1, that is, B = - 2A. ① the distance between the intersection of the parabola and the x-axis is 4, and the axis of symmetry is x = 1. The function passes through two points (3,0), (1,0), and substituting (2, - 3), (3,0), (1,0) into the algebraic formula, - 3 = 4A + 2B + C... ②, 0 = 9A + 3B + C... ③, 0 = A-B + C... ④