High School Math Quadratic Function Problems. Seek ideas. Is there an easy way?

Solution: The size of t is determined by the expression of f(x).

When t 0, we know that f(x)=x f(x+t)=(x+t) from f(x+t) 2f(x) can be obtained (x-t) 2t, which is the absolute value |x-t|≤√2 t

Because x [t,t+2].

0≤|x-t|≤2

2 t 2 gets t 2 when t + 2 0, i.e. t -2.

f(x)=-x f(x+t)=-(x+t) The same can be obtained|x-t|≥√2| t|

0≤|x-t|2 And t<-2 is obviously not valid.

At this point, t is empty.

When -2 t 0.

If x 0, f(x)=-x f(x+t)=(x+t), it is obvious that f(x+t) 2f(x) can be guaranteed

If x 0, f(x)=x f(x+t)=(x+t) to f(x+t) 2f(x) |x-t|≤√2| t|At this time|x-t|≤|t|(because x 0 and t is negative) hence |x-t|≤√2| t|It can also be established.

then 2 t 0 satisfies the requirement.

In summary, the value range of t is t 2 or -2 t 0

It is easy to know that this function is strictly monotonic.

And f(x+t)>=2f(x) is equivalent to f(x+t) f( 2*x), so the problem is equivalent to when x belongs to [t,t+2], x+t 2*x is constant, and x+t 2*x is transformed into ( 2+1)t x, so only ( 2+1)t t+2 is needed

Solution t 2

Solution: y=ax 2+bx passes the point (1,2) to get a+b=2, i.e., b=2-a joint equation: y=ax 2+bx and y=-x 2+2x get x1=(2-b) (a+1).

x1>0 again, so a -1

So s = [0,x1](ax 2+bx)-(x 2+2x)dx= [0,x1](a+1) x 2+(b-2)xdx

Substituting x1=(2-b) (a+1), b=2-a, we get s=-a 3 6(a+1) 2

Ream'=[-a^3/6(a+1)^2]'=0 to get a=-3 to get b=5

x1=a/(a+1)

Then s is equal to (a+1) x 2-ax from 0 to x1 is equal to -a 3 6(a+1) 2

x1>0

a<-1

DerivativeFind extremum.

This shouldn't be a sophomore question, and you can't finish it without calculus.

In high school, the problems of quadratic functions are generally divided into: quadratic functions with parameters in the first term, quadratic functions with parameters in other terms, problems with fixed axes and moving intervals, problems with fixed axes and moving intervals, problems with the maximum value of quadratic functions, and problems of the constant establishment of quadratic functions.

These questions are the most basic questions, for example, when solving the inequality of a quadratic function, there is no separate question in the general exam, it will have parameters and intervals, and the inequality will be examined on the basis of these conditions.

However, you need to remember one thing about these problems, they have something in common, and the idea of solving the problem is basically the same, that is, they can be easily solved by combining images and classifying them.

In the problem of quadratic functions with parameters in the first term, it is generally divided into two categories, the first term is 0 and the first term is not 0, and the solution sets solved by these two categories are combined, and in the second category, it is necessary to subdivide three categories, greater than 0, less than 0, and equal to 0, and the conditions of these three subclasses and the second major class intersect.

In the problem of quadratic functions with other terms containing parameters, it is also necessary to classify and discuss, and only use three categories: greater than 0, less than 0, and sum equal to 0.

In the problem of moving axis intervaling, it is necessary to divide the symmetry axis to the left, right and middle of the interval, and each of the three categories is divided into three subcategories: greater than 0, less than 0, and sum equal to 0.

In the problem of fixed axis moving interval, the classification method is the same as that of the moving axis interval problem, which is divided into three categories: the interval is on the left, right and middle of the axis of symmetry, and the other classifications are the same as the moving axis interval problem.

When a>0, when the axis of symmetry is on the left side of the interval, the function is a subtraction function, when the axis of symmetry is on the right side of the interval, the function is an increasing function, and when the axis of symmetry is in the middle of the interval, the function goes to the maximum value at the vertex.

The problem of constant formation, generally at any time, the problem of the constant establishment of a function should be turned into a maximum value problem, to explain, the minimum value of a function is greater than 0, then the function is greater than 0, on the contrary, if the maximum value of a function is less than 0, then the function is constant less than 0, by extension, if the minimum value of a function is greater than the maximum value of another function, then this function is greater than the maximum value of another function, and conversely, if the maximum value of a function is less than the minimum value of another function, Then this function is invariably smaller than the other function.

The ideas for solving these problems, you can remember them, may be useful to you, remember that the function is analytic geometry, and you need to combine the image to understand it better.

The above is a summary of the solution ideas of quadratic functions in high school, and remembering what they have in common is the breakthrough to solve the problem of quadratic functions. As an extension of the content of junior high school, quadratic functions will not appear alone in large-scale examinations such as joint entrance examinations, and often combine sequences, inequalities, geometric problems, moving point problems, etc., so the principle of solving difficult problems is to gradually simplify, simplify the difficult, and decompose the problem.

What kind of problem, please list the topic, maybe I can help you.

1.The function is even, so the four monotonic intervals must be two on the left and right, so only the right side function needs to be studied.

2.On the right side of the function is x>0, then the function can be written as f(x)=ax 2+bx+c(x>0), and it can be seen that the monotonicity turning point is x=-b 2a, that is, there are two different monotonic intervals on both sides of the point.

3.So to satisfy the question, then the point x=-b 2a must be on the right side of the y-axis, then there is, -b 2a>0

4.In the same way, you can study the left-hand side function and come to the same conclusion.

To sum up, the correct answer is B

When the axis of symmetry is to the right of x=0, the image of the function can be divided into 4 parts

I made a mistake just now when I chose b, it should be x=0 on the right side of the image symmetrical to the left, so the image on the right side should be more complicated, that is, the axis of symmetry is on the right.

Choose b first, the function f(x)=ax 2+b|x |+c(a≠0) is an even function, so there should be two monotonic intervals from 0 to positive infinity, regardless of whether the x-axis has an intersection or not, so only -b 2a 0 is needed to choose b

The downstairs is very detailed, I won't write it.

This is an even function, so as long as the right part is studied, where f(x)=ax bx c, then two monotonic intervals are to be separated in this section, then the axis of symmetry must be on the right side of the y-axis, i.e., b (2a)>0

f(x) is an even function, and there are two of the four monotonic intervals on each side of the y-axis 2, so as long as the symmetry axis is not 0, and x > 0, the ax 2 + bx + c symmetry axis is on the right side of the y-axis, and b can be obtained

This is an even function, so choose b

Theorem: f(x) is continuous on the interval [a,b], if f(a)f(b) < 0, then f(x)=0 has a solution in the interval (a,b).

Let f(x)=ax 2 2+bx+c

f(x1)f(x2)=(ax1^2/2+bx1+c)(ax2^2/2+bx2+c)

Because of the condition, ax1 2+bx1+c=0, that is, bx1+c=-ax1 2-ax2 2+bx2+c=0, that is, bx2+c=ax2 2 are substituted into the above equation, and f(x1)f(x2)=(ax1 2 2-ax1 2)(ax2 2+ax2 2)=-3a 2x1 2x2 2 4<0

So the equation has a root between x1 and x2.

High school students have better grades to understand.

It can only be 0 or 1 because the opening is up and the definition range is 0-1; Then when b=0, only a=0 can guarantee that the value range is 0-1, and when b=1, f(0)=1 and f(1)=0 must also be satisfied, so a=-2 at this time

2.In three cases, when the abscissa of the vertex is on the right side of the range of x (i.e., -k 4>=1), x=1 is the minimum value; When the abscissa of the vertex is on the left side of the range of x (i.e., -k 4<=-1), the minimum value is obtained when x=-1; When the vertex is -1,1, the ordinate of the vertex is the minimum value. Then find the extreme value of k < in K4 (don't write too much detail, you can think about it).

Why don't you have a topic、、I don't know if you don't understand、、、、

Quadratic functions can be understood as quadratic equations, trim, equations of roots, monotonicity, derivatives, ,

If the function passes through the point (0,1), it can be seen that if the function has two roots, it must be the same positive and negative, and if there is a solution in the defined domain (0,+), there must be one positive root or two positive roots. So:

Axis of symmetry" 0 gives 1+1 m<0, gets -1=0, has (1+1 m) -4>=0, gets 1+1 m>=2 or <=-2, and gets 0 as -1 3<=m<0

First, if the equation has a solution, then b 2 4ac 0, and secondly, b ( 2a) 0, then (1 1 m) 2 4 0, (1 1 m) 2> 0;Because m is the denominator, then m is not equal to 0, and m 1 is solved in summary

The discriminant formula is greater than 0, and the larger root is greater than 0, which is the most direct method.

A: There is no problem with your answer.

But it can also be like this:

f(x)=ax²+bx+c

f(x+1)=a(x+1)²+b(x+1)+cf(x+1)-f(x)=a(x+1)²-ax²+b(x+1-x)=2x

So: 2ax+a+b=2x

So: 2a = 2, a + b = 0

Solution: a=1, b=-1

So: f(x)=x -x+c

Because: f(0)=1

So: f(0)=c=1

So: f(x)=x -x+1

This way, you don't have to calculate too many values.

Yes, as long as you can solve the correct answer without flawlessness.

Yes, there is no problem with this idea.

y=4(x²-2x+1)+1

y=4(x-1)²+1

Vertex coordinates (1,1) axis of symmetry equation x=1 Monotonic interval y increases at ( 1) and decreases at (1).

2.Let the quadratic function be y=a(x-b) +k

The vertex (-2,4) y=a(x+2) +4 and (-1,5) is brought in to give y=a(-1+2) +4=5 a=1 and the analytic formula is y=x +4x+8

3.The quadratic function passes (-1,2),(1,3),(2,7)a-b+c=2

a+b+c=3 ②

4a+2b+c=7 ③

Solution.

a=7/6 b=1/2 c=4/3

Vertex (1,1) Axis of symmetry equation x 1 Monotonic increase interval 1] Monotonic decrease interval [1,

Let the quadratic function y ax bx c axis of symmetry equation x 2 i.e. b 4a

Substituting the points (-2,4) and points (-1,5) into the equation yields 4a 2b c 4 a b c 5

A 1 b 4 c 8 substitut the points (-1,2),(1,3),(2,7 into the function y=ax +bx+c.

1) -2a/b=1 4a-b 2=1 vertex coordinates ( monotonic interval x<1 monotonically decreasing x>1 monotonically increasing 2)-2a-b=2 b=-4a Let y=ax 2+bx+c bring in two point coordinates a=-1 7 b=4 7 c=40 7 y=-1 7x 2+4 7x+40 7 3) Same as 2) bring in three coordinates a-b+c=2 a+b+c=3 4a+2b+c=7 a=7 6 b=1 2 c=4 3

1.Finding y= 4x -8x + 5 vertex coordinates, axis of symmetry equations, and monotonic intervals?

y = 4x -8x + 5 = 4(x - 1) 1 vertex coordinates: (1, 1).

Axis of symmetry: x = 1

Monotonic interval: x >1 increment function, x 1 subtract function 2The image of a known quadratic function takes the vertex of the point (-2,4) and passes the point (-1,5) to find the analytic expression of this quadratic function?

y = ax² +bx + c

b/(2a) = -2

4ac - b²) / (4a) = 4a - b + c = 5

a = 1，b = 4，c = 8

y = x² +4x + 8

3.Knowing that the quadratic function y=ax +bx+c crosses the point of the image (-1,2),(1,3),(2,7), find the value of a,b,c?

a - b + c = 2

a + b + c = 3

4a + 2b + c = 7

a = 7/6，b = 1/2，c = 4/3