Quadratic function is one of the main contents of junior high school mathematics, and it is also an important link to contact senior high school mathematics. It is a difficult point in algebra "function and its image" in junior high school, and it is also the key to finding quadratic resolution function. To find the analytic expression of quadratic function, the form of quadratic resolution function should be properly selected, which is simple and easy to solve, otherwise the solution will be complicated. When solving problems, we should flexibly choose the form of quadratic resolution function according to the characteristics of the topic and solve it by undetermined coefficient method. Here is an example.
Thought 1. Given that the image passes through three points, the analytical formula of quadratic function can be easily obtained in general.
For example 1, it is known that the image of quadratic function passes through (-1, -9), (1, -3), (3, -5), and the analytic expression of this quadratic function is found.
Solution: Let the analytic expression of this quadratic function be derived from the meaning of the problem:
Get a solution
∴ The analytic formula of quadratic function is
Idea 2: Given the vertex coordinates, symmetry axis, maximum value or minimum value, it is generally convenient to find the quadratic resolution function with its vertices.
Example 2: Given the vertex of a parabola (-1, -2) and the passing image (1, 10), find the analytical formula.
Solution: Set a parabola, which is derived from the meaning of the question:
Parabolic intersection (1, 10)
In other words, the analytical formula is
Idea 3. It is known that the coordinates of the two intersection points of the image and the axis can be used in the form, where is the abscissa of the intersection point of the parabola and the axis and the two roots of the unary quadratic equation.
Example 3: Given that the intersection points of the image and the axis of a quadratic function are (-5,0) and (2,0), if the image crosses (3,4), find the analytical formula.
Solution: Let the required analytical formula be
Image channels (3, -4)
∴ ∴
Namely:
Then the analytical formula is.
Idea 4: Knowing the distance between the two intersections of the image and the axis, we can find an analytical formula in the form, where the distance between the two intersections is the abscissa of an intersection with the axis.
Example 4: The distance between the image of the quadratic function and the two intersections of the axis is 2. After passing through two points (2, 1) and (-1, -8), the analytic expression of this quadratic function is found.
Solution: The quadratic resolution function is known.
∴
It is also known that:
Solution: or
∴ Find the second analytic function as follows:
Idea 5: Find the analytical formula from the translation of the known image, and generally write it in the form of the analytical formula of the known image. If the image moves one unit to the left (right), the value in brackets will increase (decrease) by one unit; If the image is translated up (down) by one unit, the value of will increase (decrease) by one unit, that is, the left will increase and the right will decrease, and the shape and size of the translated parabola will remain unchanged.
Example 5. Translate the image of quadratic function by one unit to the right, and then by one unit to the upward, and find the analytical expression of quadratic function.
Solution:
Pan 2 units to the right:
Namely:
Pan up by 3 units:
Namely:
∴ Find the quadratic resolution function.
Idea 6: a quadratic function is known, and its image is required to be symmetrical (or folded along the axis); The resolution function of the image of a straight line that is symmetrical and passes through its vertex and is parallel to the axis (it can also be said that the parabola image rotates around the vertex180) is transformed into the form of the original function.
(1) Vertices of two images about axial symmetry are about axial symmetry, and the opening directions of the two images are opposite, that is, opposite numbers.
(2) Vertices of two images that are symmetrical are symmetrical, and the shapes and sizes of the two images are the same, that is, the same.
(3) The vertex coordinates of the images of two functions that pass through their vertices and are symmetrical on a straight line parallel to the axis are unchanged, and the opening directions are opposite, that is, mutually opposite numbers.
Example 6; Given a quadratic function, find the analytical formula of the quadratic function that satisfies the following conditions: (1) The image is symmetric about the axis; (2) image symmetry; (3) The image is symmetrical about a straight line passing through its vertex and parallel to the axis.
Solution: It can be transformed into, according to the symmetry formula, ① the image analytical formula about the axisymmetric image is:
(2) The image analytical formula for axisymmetric images is:
, namely:;
(3) The analytical expression of an image that is symmetrical with a straight line passing through its vertex and parallel to the axis is, that is.
Idea 7. The Solution of Quadratic resolution function Combining Numbers and Shapes. In this case, algebra and geometry are integrated, and the algebraic problem is transformed into a geometric problem to solve, so long as the relevant geometric knowledge is fully utilized, the goal can be achieved.
Example 7: Let the image of a quadratic function intersect with the axis at two points and intersect with the axis at two points. If yes, find the analytic expression of this quadratic function.
Solution: In,
∵
∴
It's also VIII
∴rtδobc∽rtδabc,rtδoac∽rtδabc，
rtδOAC∽rtδOBC
∴
∴
Let the analytical expression of parabola be
In other words, you must:
So the parabola is:
Idea 8: Synthesize the solution of quadratic decomposition function. Most of the comprehensive questions designed with quadratic function as the background are the finale of the senior high school entrance examination, which is used to open the score. Generally, it is centered on quadratic function and organically integrated with algebra, geometry and trigonometry. This type of question combines junior high school algebra, geometry, trigonometry and other knowledge, and communicates the vertical and horizontal relations among many knowledge points. When solving problems, we should establish equivalence relations according to the related properties of geometric figures, find out the function relations or geometric images in the function images, and solve geometric problems related to functions by using the number-shape synthesis method. As long as the problem of each knowledge point is solved, it can be solved.
Example 8: As shown in the figure, EB is the diameter of semicircle O, and EB=6. Take point P on the extension line of BE, so that EP=EB, A is the last moving point of EP (point A and point E do not coincide), the tangent point of A is ⊙O, the tangent point of D is DF⊥AB, the vertical foot of D is F, and the vertical line BH of B is AD.
Solution: link BD
∫EB is the diameter of the semicircle O.
∴∠EDB=90
∴
∵BH⊥DA
∴∠BHD=90
∵AH is the tangent of⊙ o.
∴∠BDH=∠DEB
∴rtδbdh∽rtδbed
∴
∴
∴
According to the condition, when it coincides with a and p, there is
∴
∑δPDE∑δPBD
∴
∴
The range of independent variables is.