Practice questions on all things quadratic functions in junior high school math

13.Solution: Let y=a(x-1)2-4, and substitute b(3,0) to get a=1

Therefore y=(x-1)2-4 or y=x2-2x-3 14Solution: by the meaning of the topic.

1）y=（x-50）w

x-50）（-2x+240）

2x*2+340x-12000；

2) y=-2x*2+340x-12000=-2(x-85)*2+2450, when x=85, the value of y is maximum, y is maximum=2450 or a=-2, when x=-3402 (-2)=85, the value of y is maximum, and y is maximum=2450

1, according to the meaning of the question set y=k(x-1) -4, substituting (3,0) to solve k=1, so the function interpretation formula is y=(x-1), substituting (a,0) solution to obtain a=-1, or a=3, so the function can pass through the origin by translating one unit to the right, after translation, the interpretation formula is y=(x-2) -4, substituting (b,0) to solve b=0, or b=4, so the coordinates of the other intersection point are (4,0).

You can type out the "image of a quadratic function" on the Internet to understand its properties, or draw some diagrams by yourself while listening to the teacher's lectures, and determine the properties of the image according to the general formula y=ax2+bx+c and the vertex formula y=a(x-h)2+k.

Brothers, they are all junior high three parties, and no one has died in life since ancient times.

(1) I can't see the figure--- but I guess the figure must tell you a few data, convert it into the corresponding coordinates, and substitute it into the analytical formula of the function.

2) The profit in this question should not be represented by y, but can be represented by w, then w= bring the above function analytic formula into a functional relationship between profit and month x.

3) Using the maximum and minimum value of the quadratic function, you can find the maximum profit.

There are also tons of free workbooks (with answers) at your disposal.

(3) clearly pointed out that before the "May One", of course, there are only 1,2,3,4 these 4 months, because X is a sales month, of course, only need to consider X<5.

The image of y=kx(x-square)-6x+3 has an intersection point with the x-axis, i.e., the equation kx 2-6x+3=0 has a solution.

So, discriminant 36-12k>=0

Get: k<=3

is a quadratic function again, then k is not 0

So, k is in the range of k<=3 and k is not 0

If there is an intersection with the x-axis, then its discriminant formula =b -4ac 0 is sufficient.

i.e. 36-12k 0

Solution k 3 (k≠0).

1,k>0

Discriminant = 36-12k is greater than or equal to 0

The solution is less than or equal to 3

Therefore, 0 satisfies the topic.

0 satisfies the topic.

In summary, k is less than or equal to 3

Analysis: (1) Find the coordinates of b, substitute the coordinates of b and d into the analytic formula of the quadratic function to obtain the system of equations, and find the solution of the system of equations;

2) Find the coordinates of the intersection of the straight line and the quadratic function, find the coordinates of C, find the coordinates of E, pass C as the cn x axis to n, and find the area of the trapezoidal boec, bod, and cne respectively according to the image, and then find the answer;

Solution: (1) y=ax 2+bx+c

a(x^2+b/ax+c/a)

a(x 2+b ax+b 2 4a 2-b 2 4a 2+c a) = a(x+b 2ax) 2+(4ac-b 2) 4a(2) vertex coordinates.

x=-b 2a y=(4ac-b 2) 4a axis of symmetry x=-b 2a

Opening: a>0, opening up:

a<0, the opening is down.

3）y=a(x-1)^2+h=a(x-1)(x-3)=a(x^2-2x-3)=ax^2-2ax-3a

ax^2-2ax+a+h

Know: a+h=-3a (pending coefficient).

b/2a=1

3=a(1)(-1)->a=-3(past (2,3)) y=-3x 2+6x+9

4) The real root of the quadratic function is the x-coordinate value of the intersection point of the function image and the x-axis when the y value of the quadratic function is =0.

The function image and the x-axis have several intersections and several real roots. The real root is relative to the imaginary root, that is, when the value of the discriminant of the quadratic function is less than 0, the function has no intersection point with the x-axis, and there is a pair of imaginary roots.