Excerpted from Baise Education Network) X6m; @9 p6 u8 y' H
One of the most important and difficult contents in Algebra, a textbook for junior middle school students, is the nature and images of functions, especially quadratic functions, which have many properties and involve a wide range, and many topics are often combined with geometric figures, making students feel helpless, at a loss and full of difficulties. In fact, as long as we understand the connotation of quadratic function, it is not difficult to understand many skills that students need to master. The coordinate system we have learned is to establish a close relationship between geometric objects and numbers, geometric relations and functions, so that the study of spatial form can be attributed to a relatively mature and easy-to-control quantitative relationship. That is to say, as long as you master the connection point between algebraic relationship and geometric relationship, you can easily master the problem-solving methods, methods and answers of this kind of problems. So what is the connection point between them? Let's analyze it. Geometric conditions can get the line segment relationship of graphics, line segment relationship can get the coordinate line segment relationship, coordinate line segment relationship can get the coordinate relationship, and coordinate relationship can get the analytical formula of the image where the coordinates are located, and vice versa. It can be seen that these two processes must involve coordinate line segments, so geometry and algebra are linked by coordinate line segments, so all our analysis must be carried out around coordinate line segments. Let's look at a few examples.
: p, k0 Z. v5 U. n7 [+ [8 N 1 N cases 1, known parabola y = ax2+bx+c (a > The vertex of 0) is c(0, 1), and the straight line L: Y =-AX+3 intersects with this parabola at points P and Q, and intersects with the X and Y axes at points M and N respectively. If the length ratio of MP to PN is 3: 1, try to find the functional relationship of parabola. ! ^9 Y: f- m) Z！ x2 Z" U
Conditional preparation: There are two situations in this question, that is, point P and point Q are located at two points respectively, and the situation analysis is shown in the figure. Firstly, it is easy to calculate B = 0 and C = 1 from the vertex coordinates (0,1), that is, the analytical formula is y=ax2+ 1.
)o9 ~ ' H+w v @； } & amp~
! k； L & amp_- j* y 1 Q coordinate line segment: the point passing through P is PA⊥X axis, and the vertical foot is A. This auxiliary line we made is to appear the coordinate line segment I mentioned above, namely OA, OM, ON, PA, etc.
4m % F； s: u+ m！ E( i Geometric Relation Transformation: How to transform the condition MP: PN = 3: 1 into those coordinate line segments? We know that parallel lines have parallel line segments, so we can get PA: ON = PM: nm = 3: 4, because ON = 3, so PA = 9/4, and the result of PA is the value of the ordinate Y of point P, and then bring Y = 9/4 into two analytical expressions to get 9/4 = AX2+ 1 and 9/4 = respectively. 8 b2 _9 F* Q- P* d
Example 2: As shown in the figure, it is known that the vertex coordinate of the quadratic function image is C( 1 0), and the straight line Y = X+M intersects with the quadratic function image at point A and point B, where the coordinate of point A is (3,4) and point B is on the Y axis.
r & ampX2 ~ % v； B3 A & ampY( 1) Find the relationship between the value of m and this quadratic function. ，p# t 1 w3 K，x* @
(2)P is a moving point on the line segment AB (which does not coincide with A and B). The vertical line with P as the X axis intersects the image of this quadratic function at point E. Let the length of line segment PE be H and the abscissa of point P be X. Find the functional relationship between H and X, and write the value range of X ...+b+ W 1 g/m2 & ampX.
(3)D is the intersection of straight line AB and the symmetry axis of this quadratic function image. Is there a point p on the AB line, which makes DCEP a parallelogram?
8 j！ C & ampH2 R+ {* H condition: In the first question, it is easy to find out that the secondary resolution function and the primary resolution function are y = x2-2x+ 1 and y = x+ 1 respectively, and the coordinates of point D can be obtained as (1, 2).
# {! i8 y8 B3 X！ V'N+N coordinate line segment: passing through point P makes the image of X and quadratic function intersect at point E and extend to point F intersect with X axis, and PF and EF are vertical line segments of point P and point E.: o8i.a3x * v # l/kz5n1p.
- O. }！ q“h”K
Transformation of geometric relations: (1) For line segment H, it is obviously the difference of vertical line segments. If there is H = YP- leaf, then H = (x+1)-(x2-2x+1) → H =-x2+3x; (2) There are many conclusions of parallelogram, and it is necessary to find the one related to the coordinate line segment from many conclusions. This conclusion is DC = PE, which is the ordinate of point D and H obtained from the first question, that is, H = 2 →-X2+3x = 2, and the purpose of solving the problem can be achieved after solving it.
(I, g7 j* [6 k] As can be seen from the above example, the difficulty in solving quadratic function lies in the transformation of geometric relations, which transforms a geometric condition into an algebraic condition or a mathematical formula. Then these transformations are realized by the coordinate line segments of some points, so as long as these factors can be considered in the analysis process, the purpose of solving the problem can be achieved. Next, let's look at an example:
6 g2 a( v9 F$ O3 B4 l6 {5 n) p Case 3. It is known that the vertex coordinate of parabola Y = AX2+BX+C is (2,4). :u5 W. J% w9 g
(1) Try to represent b and c with an algebraic expression containing a; ; y * H2 ~ 6v 0n " p A & amp； V
(2) If the intersection of the straight line y = kx+4 (k ≠ 0) with the Y axis and parabola is D, E, F, S △ ODE: S △ OEF = 1: 3 in turn, where O is the coordinate origin, try to represent K with an algebraic expression containing A. ..
! g！ L9 q6 c " Z； {Conditional preparation: B =-4a, C = 4a+4 can be obtained from the abscissa and ordinate formulas of parabola vertices, then the parabola can be obtained from Y = AX2-4x+4a+4. Because I don't know the situation of A, I will analyze it according to the diagram first, and so on in other situations.
(American q5 o L2 G, B% q)
7h/a . p(H6 %. G-]5f & amp； E, | coordinate line segment: the vertical line segments EM and FN passing through the two intersection points e and f respectively are the y axis, and the vertical feet m and n N..EM and FN are the key horizontal line segments of this problem.
(m！ h； I/ Y/ G) @3 H geometric condition transformation: the geometric condition is S △ODE: S △ OEF = 1: 3, and the area involves the base and height of the triangle. The base of △ODE is DO, which is obviously the vertical segment of point D, and the height is the horizontal segment of point EM and E; But what if the height and bottom of △OEF have nothing to do with the coordinate line segment? Because S△ODF = s △ ode+s △ OEF, so s △ ode: S△ODF = 1: 4, and s △ ODF can be connected by coordinate line segments, the base of which is DO and the height of which is f point, so we can further get the coordinate line segment FN of EM: FN = 1: 4.
3[& amp； t； P% y0 y: W+ c There are a series of inevitable connections between the conditions and objectives of a topic, and these connections are the bridges from conditions to objectives. Which connections are used to solve problems need to be determined according to the mathematical principles followed by these connections. The essence of solving problems is to analyze which mathematical principle these connections conform to, and this connection is very hidden and can only be revealed by careful analysis.