 # The image of a quadratic function intersects the x-axis at points (- 1,0), (4,0) and its shape is the same as that of y = - x2. Then the analytical formula of this quadratic function is __

## The image of a quadratic function intersects the x-axis at points (- 1,0), (4,0) and its shape is the same as that of y = - x2. Then the analytical formula of this quadratic function is __

∵ the function image intersects the x-axis at points (- 1,0), (4,0),
Let the analytical formula be y = a (x + 1) (x-4),
And because the shape of the image is the same as that of y = - X2,
So a = - 1 or 1,
Therefore, the analytical formula is y = ± (x + 1) (x-4),
After finishing, y = - x2 + 3x + 4 or y = x2-3x-4
So the answer is: y = - x2 + 3x + 4 or y = x2-3x-4

### It is known that the image with quadratic function y = x ^ - 8x = 15 intersects with the X axis at two points a and B, and C moves on the parabola, which can make the area of triangle ABC equal to point C of 1

Y = x ^ - 8x + 15 = (x-4) ^ 2-1 [vertex type, vertex (4, - 1)] = (x-3) (X-5) [intersection type, intersection points with X axis are (3,0), (5,0)], i.e. point a coordinate (3,0), point B coordinate (5,0) AB = XB XA = 5-3 = 2S △ ABC = 1 / 2 * AB * |yc| = 1yc = ±| 2 / ab| = ± 1yc = + 1, 1 = (x-4) ^ 2-1, X

### It is known that the image of parabola y = x2-2x-3 intersects the x-axis at two points a and B. there is a point C on the parabola above the x-axis, so that the area of △ ABC is 10, and the coordinate of point C is __

From x2-2x-3 = 0, X1 = 3, X2 = - 1,
So AB distance is 4,
To make the area of △ ABC 10, the ordinate of C should be 5,
Put y = 5 into the function y = x2-2x-3 to get x2-2x-3 = 5,
The solution is X1 = 4, X2 = - 2
Therefore, the coordinates of point C are (4, 5) or (- 2, 5)

### How to draw the quadratic function image of y = - x square

It is a parabola with an opening downward whose axis of symmetry is y = 0. You can take points in turn and follow the point drawing method. If you can draw the quadratic function image of y = x square and y = - x square, it is the image symmetrical about the X axis

### How to draw the opening size of parabola in quadratic function image For example, how to determine the opening size when y = 2 / 1X ^ + 5x + 1

In terms of image, you can only trace points or draw with a function computer
Numerically speaking, this is not a fixed value
Parabola is infinitely extended (within the range of real numbers), and its opening is infinitely extended, in other words, infinitely expanded
If you want to know the approximate size of an opening, that is, the appearance of a graph, you can judge by the coefficient of x ^ the closer the coefficient is to 0, the larger the opening is

### It is known that the image of the function y = - x + 1 intersects with the x-axis and y-axis at points c and B respectively, and the hyperbola y = K X intersects at points a and D. if AB + CD = BC, the value of K is _

It is known that the image of the function y = - x + 1 intersects with the x-axis and y-axis at points c and B respectively, then the coordinates of B and C are (0,1), (1,0), then ob = 1, OC = 1, BC = 2, set the coordinates of point a as (- m, n), make AE ⊥ X-axis at point e through a, then △ CBO