High School Mathematics Quadratic Functions, High School Mathematics Quadratic Functions

The answer should be d

f（x）=ax²+bx+c

Its axis of symmetry is the straight line x=-2a b

For the solution of the equation m[f(x)] nf(x)+p=0, let's take them as y1, y2

Then there must be y1=ax +bx+c, y2=ax +bx+c

Then from the image, y=y1, y=y2 is a straight line parallel to the x-axis.

They have an intersection point with f(x).

Due to the symmetry, the two solutions of the equation y1=ax +bx+c x1,x2 are symmetrical with respect to the straight line x=-2a b.

That is, 2(x1+x2)=-2a b

In the same way, the two solutions of the equation y2=ax +bx+c x3 and x4 should also be symmetrical with respect to the straight line x=-2a b.

Then we get 2(x3+x4)=-2a b

In answer c, we can find the axis of symmetry line x=, i.e. 1,4 is the solution of an equation and 2,3 is the solution of an equation.

So the set of solutions obtained can be.

In answer d, we can't find the axis of symmetry, which means that no matter how we group them, we can't make the sum of two equal to the sum of the other two.

Therefore the answer d is no.

What kind of question did you write, I can't see clearly, how can I help you do it?

1 All y=f(x) is a quadratic function, and f(0)=1 is set f(x)=ax 2+bx+1

f(x-1)=ax 2+(b-2a)x+(a-b+1)f(x-1)-f(x)=-2ax+(a-b)=2x, so -2a=2

a-b=0a=-1

The explanatory formula of b=-1y=f(x) f(x)=-x 2-x+1 is in the interval [-1,1], and the image of y=f(x) is always above the image of y=2x+m, which is -x 2-x+1-(2x+m)=-x 2-3x+1-m, and in the interval [-1,1], Evergrande is at zero.

Axis of symmetry x = -3 2

So -x 2-3x+1-m decreases monotonically in the interval [-1,1] only f(1)>0

Solution m<-3

Upstairs ground cow! I'm going to be a general solution, and I want to be praised for the above.

The axis of symmetry = -a (c=-2a-2 from the problem) 1 -a is less than or equal to a and a is greater than or equal to 0 and x=a is substituted into 3a 2-2a-2>-1 and in summary, a>1 is obtained

2.-a is greater than or equal to a+2 to get a less than or equal to -1 substituting x=a+2 into 3a 2+6a+2>-1 In summary, a<-1 is obtained

3 a<-a1

f(x)=x^2+2ax+c

axis of symmetry x=-a

a,a+2<1<-a

A<-1

a+2=-a, a=-a (rounded).

So it's a<-1 or a>1

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.

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

y=(m-1)x^2+(m-2)x-1=[(m-1)x-1](x+1)

Therefore, when x=-1, no matter what the value of M is, the quadratic function of hand hunger is 0, so no matter what the value of m, there is a zero point in the quadratic function.

From the above, we already know that x=-1, using the distance formula, we can know |x1-x2|=2, so x2 = 1 or -3

When x2=1, m=1 (does not match the title, so it is housed).

When 2=-3, m=2 3

One way to teach you is to draw and write constraints:

Let f(x)=x 2+(2k-1)x+k 2 then f(1)=k 2+2k>0

k<-2 or k>0

If the two roots of the equation are greater than 1, then x1+x2=1-2k>2 k<-1 2x1*x2=k 2>1 k>1 or k<-1δ=(2k-1) 2-4k 2 0 k 1 4 or more to obtain k<-2

Let f(x)=x2+(2k-1)x+k2

The axis of symmetry is large by 1 and f(1) > 0 and >0

It can be judged by the axis of symmetry and the opening of the image.

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

1.Let x1 be the root of the equation f(x)=0, then there is f(x1)=0 because f(3+x)=f(3-x).

So f(x1)=f[3+(x1-3)]=f[3-(x1-3)]=f(6-x1)=0

So x=6-x1 is also the root of the equation f(x)=0, i.e. x2=6-x1 so x1+x2=6

x^2-2kx+1-k^2=0

x1+x2=2k

x1x2=1-k^2

x1^2+x2^2

x1+x2)^2-2x1x2

4k^2-2(1-k^2)

4k^2-2+2k^2

6k^2-2

Discriminant = 4k2-4(1-k2).

4k^2-4+4k^2

8k^2-4>=0

k^2>=1/2

k<=- root number 2 2 k> = root number 2 2

x1 2 + x2 2 2 minimum 1

1f(3+x)=f(3-x)

f(x+3)=f(3-x)

x+3)+(3-x)]/2=3

f(x) is symmetrical with respect to x=3.

The axis of symmetry of the 2nd function f(x) x=-b 2a

x1+x2=-b/a

x1+x2)/2=3

x1+x2=6

2(x-k)^2=2k^2-1 2k^2-1>=0 k^2>=1/2

x1+x2=2k

x1x2=1-k^2

x1^2+x2^2=(x1+x2)^2-2x1x2=4k^2-2(1-k^2)=6k^2-2

x1 2 + x2 2 min = 6 * (1 2) - 2 = 1