Test point scanning

The image of 1. quadratic function will be drawn by drawing.

2. The opening direction, symmetry axis and vertex position of parabola can be determined by image or matching method.

3. According to the coordinates of three points on the known image, the analytic expression of quadratic function will be obtained.

Master Jing Jiang.

1. quadratic function y=ax2, y=a(x-h)2, y=a(x-h)2+k, y=ax2+bx+c (a≠0, among others) have the same image shape, but different positions. Their vertex coordinates and symmetry axes are as follows:

The analytical formula y = ax2y = a (x-h) 2y = a (x-h) 2+k y = ax2+bx+c.

Vertex coordinates (0,0) (h,0) (h, k) ()

Symmetry axis x=0 x=h x=h x=

When h>0, the parabola y=ax2 is moved to the right by H units in parallel, and the image of y=a(x-h)2 can be obtained.

When h < 0, it is obtained by moving |h| units in parallel to the left.

When h>0, k>0, the parabola y=ax2 is moved to the right by H units in parallel, and then moved up by K units, the image of y=a(x-h)2+k can be obtained;

When h>0, k<0, the parabola y=ax2 is moved to the right by h units in parallel, and then moved down by | k units, and the image of y=a(x-h)2+k is obtained;

When h < 0, k >; 0, moving the parabola to the left by |h| units in parallel, and then moving it up by k units to obtain an image with y=a(x-h)2+k;

When h < 0, k<0, move the parabola to the left by |h| units in parallel, and then move it down by |k| units to obtain an image with y=a(x-h)2+k;

Therefore, it is very clear to study the image of parabola y=ax2+bx+c(a≠0) and change the general formula into the form of y=a(x-h)2+k through the formula, so as to determine its vertex coordinates, symmetry axis and approximate position of parabola, which provides convenience for drawing images.

2. the image of parabola y=ax2+bx+c(a≠0): when a >; 0, the opening is upward, and when

3. parabola y=ax2+bx+c(a≠0), if a >；; 0, when x≤, y decreases with the increase of x; When x≥, y increases with the increase of x. If a

4. The intersection of the image with the parabola y=ax2+bx+c and the coordinate axis:

(1) The image must intersect with the Y axis, and the coordinate of the intersection point is (0, c);

(2) when △ = B2-4ac >; 0, the image and the X axis intersect at two points A(x 1, 0) and B(x2, 0), where x 1, x2 is the unary quadratic equation ax2+bx+c=0.

(a≠0)。 The distance between these two points AB = | x2-x 1 | =.

When △ = 0, the image has only one intersection with the X axis;

When delta < 0. The image does not intersect with the x axis. When a >; 0, the image falls above the X axis, and when X is an arbitrary real number, there is y >；; 0; When a<0, the image falls below the X axis, and when X is an arbitrary real number, there is Y.

5. the maximum value of parabola y=ax2+bx+c: if a >；; 0(a & lt； 0), then the minimum (maximum) value of y = when x=.

The abscissa of the vertex is the value of the independent variable when the maximum value is obtained, and the ordinate of the vertex is the value of the maximum value.

6. Find the analytic expression of quadratic function by undetermined coefficient method.

(1) When the given condition is that the known image passes through three known points or three pairs of corresponding values of known x and y, the analytical formula can be set to the general form:

y=ax2+bx+c(a≠0)。

(2) When the given condition is the vertex coordinate or symmetry axis of the known image, the analytical formula can be set as the vertex: y = a (x-h) 2+k (a ≠ 0).

(3) When the given condition is that the coordinates of the two intersections of the image and the X axis are known, the analytical formula can be set as two formulas: y = a (X-X 1) (X-X2) (A ≠ 0).

7. The knowledge of quadratic function can be easily integrated with other knowledge, resulting in more complex synthesis problems. Therefore, the comprehensive question based on quadratic function knowledge is a hot topic in the senior high school entrance examination, which often appears in the form of big questions.

Typical examples of senior high school entrance examination

1. (Xicheng District, Beijing) The symmetry axis of parabola y=x2-2x+ 1 is ().

(a) row x= 1 (B) row x=- 1 (C) row x=2 (D) row x=-2.

Test Center: Symmetry axis of quadratic function Y = AX2+BX+C. 。

Comments: Because the symmetry axis equation of parabola y=ax2+bx+c is: y=-, substituting a= 1 and b=-2 into the known parabola to get x= 1, so option A is correct.

Another method: the parabola formula can be in the form of y=a(x-h)2+k, the symmetry axis is x=h, and the known parabola formula is y=(x- 1)2, so the symmetry axis is x= 1, so A should be chosen.

2. (Dongcheng District, Beijing) has an image of a quadratic function, and three students described some characteristics of it:

A: The symmetry axis is a straight line x = 4；;

B: the abscissa of the two intersections with the X axis is an integer;

C: The ordinate intersecting with the Y axis is also an integer, and the area of the triangle with these three intersections as its vertices is 3.

Please write a quadratic resolution function that satisfies all the above characteristics.

Test center: the solution of quadratic function y=ax2+bx+c

Note: let the analytical formula be y=a(x-x 1)(x-x2) and x 1 < x2, then the two intersections between the image and the y axis are a (x 1 0) and b (x2,0) respectively, and the coordinates of the intersections with the y axis are (x10).

The symmetry axis of parabola is the straight line x=4,

∴x2-4=4-x 1, that is, x 1+ x2=8 ①.

∵S△ABC=3,∴ (x2- x 1)？ |a x 1 x2|= 3，

Namely: x2- x 1= ②

① ② Two formulas are added and subtracted: x2=4+, x 1=4-

∵x 1, x2 is an integer, ax 1x2 is also an integer, ∴ax 1x2 is a divisor of 3, * * can be taken as: 1, 3.

When ax 1x2 = 1, x2=7, x 1= 1, and a =

When ax 1x2 = 3, x2=5, x 1=3 and a = 3.

So the analytical formula is: y = (x-7) (x- 1) or y = (x-5) (x-3).

That is, y= x2- x+ 1 or y=- x2+ x- 1 or y= x2- x+3 or y=- x2+ x-3.

Note: in this question, just fill in an analytical formula or guess and verify. For example, guess that the intersection with the X axis is A (5 5,0) and B (3 3,0). Then find out a from the conditions of the problem and see if c is an integer. If there is, the guess can be verified, just fill it in.

5. (Hebei Province) As shown in figure 13-28, if the image of quadratic function y=x2-4x+3 intersects with the X axis at points A and B, and intersects with the Y axis at point C, the area of △ABC is ().

a、6 B、4 C、3 D、 1

Test site: the image of quadratic function y=ax2+bx+c and the application of its properties.

Comments: From the function image, we can know that the coordinate of point C is (0,3), and then from x2-4x+3=0, we can get x1= kloc-0/,x2=3, so the distance between point A and point B is 2. Then the area of △ABC is 3, so C should be chosen.

Figure 13-28

6. Psychologists in Anhui Province have found that there is a functional relationship between students' ability to accept concepts y and the time to put forward concepts x (unit: minutes): Y =-0. 1x2+2.6x+43 (0 < x < 30). The greater the value of y, the stronger the acceptability.

In what range of (1)x, students' acceptance ability is gradually enhanced? In what range of X, students' acceptance is gradually decreasing?

(2) What is the acceptability of students when the score is 10?

(3) What scores do students accept the most?

Test site: properties of quadratic function y = AX2+BX+C

Comment: parabola y=-0. 1x2+2.6x+43 changed to vertex: y =-0.1(x-13) 2+59.9. According to the properties of parabola, it can be known that the opening is downward. When x≤ 13, y increases with the increase of x, and when 13, y decreases with the increase of x. The range of independent variables of this function is: 0≤x≤30, so the two ranges should be 0 ≤ x ≤13; 13≤x≤30. Substitute x= 10 to find the function value. From the vertex analytic formula, the acceptance ability is the strongest at 13 minutes. The problem solving process is as follows:

Solution: (1) y =-0.1x 2+2.6x+43 =-0.1(x-13) 2+59.9.

Therefore, when 0≤x≤ 13, students' acceptance ability is gradually enhanced.

When 13 < x ≤ 30, students' acceptance ability gradually decreases.

(2) When x= 10, y =-0.1(10-13) 2+59.9 = 59.

When the score is 10, the students' acceptance ability is 59.

(3) When x = 13, y takes the maximum value.

Therefore, in the score of 13, students' acceptance ability is the strongest.

9. A store in Hebei Province sells an aquatic product at a cost of 40 yuan per kilogram. According to market analysis, if it is sold in 50 yuan per kilogram, it can sell 500 kilograms a month; For every increase in the unit sales price of 1 yuan, the monthly sales volume will decrease by 10 kg. Please answer the following questions about the sale of this aquatic product:

(1) When the sales unit price is set to 55 yuan per kilogram, calculate the monthly sales volume and monthly sales profit;

(2) Let the sales unit price be X yuan per kilogram and the monthly sales profit be Y yuan, and find the functional relationship between Y and X (it is not necessary to write the value range of X);

(3) The store wants to make a monthly sales profit of 8,000 yuan when the monthly sales cost does not exceed 1 10,000 yuan. What should the sales unit price be?

Solution: (1) When the sales unit price is set to 55 yuan per kilogram, the monthly sales volume is: 500-(55–50) ×10 = 450 (kg), then the monthly sales profit is

: (55–40) × 450 = 6750 (yuan).

(2) When the sales unit price is X yuan per kilogram, the monthly sales volume is [500-(x–50) ×10] kilograms, and the sales profit per kilogram is (x–40) yuan, then the monthly sales profit is:

y =(x–40)[500-(x–50)× 10]=(x–40)( 1000– 10x)=– 10x 2+ 1400 x-

The resolution function of y and x is: y =–10x2+1400x–40000.

(3) Make the monthly sales profit reach 8000 yuan, that is, y=8000, ∴–10x2+1400x–40000 = 8000,

Namely: x2–140x+4800 = 0,

Solution: x 1=60, X2 = 80.

When the sales unit price is set at one kilogram of 60 yuan, the monthly sales volume is: 500-(60-50)× 10 = 400 (kg), and the monthly sales cost is:

40×400= 16000 (yuan);

When the sales unit price is set at one kilogram of 80 yuan, the monthly sales volume is: 500-(80-50)× 10 = 200 (kg), and the monthly sales unit price cost is:

40×200=8000 (yuan);

Since 8000 < 10000 < 16000, and the monthly sales cost cannot exceed 10000 yuan, the sales unit price should be set at 80 yuan per kilogram.

References:

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