If the coordinates of three vertices of triangle ABC are a (- 3, - 1), B (2, - 1), C (1,3), then the area of triangle ABC is

If the coordinates of three vertices of triangle ABC are a (- 3, - 1), B (2, - 1), C (1,3), then the area of triangle ABC is

= (1/2)|-3( -1 - 3) + 2(3 + 1) + 1(-1 + 1)|
= 10
Or AB = 5 (same ordinate)
Height on AB = 4 (difference between ordinate C and ordinate AB)
S = 5*4/2 = 10

It is known that the image vertex of quadratic function is (- 1,2), and the analytic expression of this function is obtained through (1, - 3)

Vertex (- 1,2)
y -2 = a（x+1)^2
After (1, - 3)
-3-2 = a(1+1)^2
a = -5/4
y-2 = -5/4(x+1)^2
y = -5/4x^2-5/2x+3/4

According to the following conditions, the relation of quadratic function (1) is obtained: the image passes through a (0,3) B (2,0) C (4,3) (2): the vertex is (2,3), and it passes through a point（
According to the following conditions, find the relationship of quadratic function
(1) : image passing a (0,3) B (2,0) C (4,3)
(2) : vertex is (2,3), and passing through point (1,1)
(3) : the point of intersection with x-axis is (1,0) and (5,0), passing through (- 1,4)
(4) The vertex is (- 2,3)
To process! To process! To process! Please! Everyone! Please!

The beauty of Mathematics
Let the quadratic function be y = ax ^ 2 + BX + C, the intersection of the function and the Y axis be (0, c), and the axis of symmetry be x = - B / (2a)
Vertex coordinates (- B / (2a), (4ac-b ^ 2) / (4a))
(1) The intersection of function and Y axis is a, that is, C = 3, B on function, then: 4A + 2B + 3 = 0 (1)
And point C is on the function, so 16A + 4B + 3 = 3 (2)
Simultaneous (1) (2) can be solved: a = 3 / 4, B = - 3, so the function: y = (3 / 4) x ^ 2-3x + 3
(2) - B / 2A = 2, (4ac-b ^ 2) / (4a) = 3, that is, 4ac-16a ^ 2 = 12a, that is, a (4a + 3-C) = 0, so C = 4A + 3, or a = 0 (rounding off)
Then the function passes through point (1,1), so a + B + C = 1, so we can get 5A + B = - 2, combining B = - 4a to get a = - 2, then C = - 5, B = 8
So the function: y = - 2x ^ 2 + 8x-5
(3) Weida theorem: the two intersection points X1 and X2 of function and X axis satisfy: X1 + x2 = - B / A, X1 * x2 = C / A
That is: - B / a = 6, C / a = 5, that is, B = - 6a, C = 5a, and point (- 1,4) is on the function, then: A-B + C = 4, that is, a = 1 / 3
B = - 2, C = 5 / 3, so the function: y = (1 / 3) x ^ 2-2x + 5 / 3
(4) If - B / 2A = - 2, then B = 4A; (4ac-b ^ 2) / (4a) = 3, then 4ac-b ^ 2 = 12a (1)
It intersects with X axis at two points AB, and ab = 6, so sqrt (b ^ 2-4ac) = 6a, that is, B ^ 2-4ac = 36a ^ 2, so 36a ^ 2 = - 12a
That is (a ≠ 0): a = - 1 / 3, then: B = - 4 / 3, substituting (1) to get: C = 5 / 3, so the function: y = (- 1 / 3) x ^ 2-4x / 3 + 5 / 3

According to the following conditions, the relationship of quadratic function is obtained: 1) the vertex coordinates of the image are (- 3, - 2), and pass through the points (1,2);
2) The image and X-axis intersect at points m (- 5,0) and n (1,0), and the ordinate of vertex coordinate is 3

(1) Let y = a (x + 3) ² - 2
Because after (1,2), then 2 = 16a-2, then a = 1 / 4
y=1/4(x+3)²-2
(2) Let y = a (x + 5) (x-1) = a (X & # 178; + 4-5)
Then the vertex is (- 2,3)
Then 3 = a × 3 × (- 2-1)
a=-1/3
y=-1/3(x+5)(x-1)