It is known that the quadratic coefficient of the quadratic function f x is a, and the solution set

The idea is for you, knowing that the coefficient of the quadratic term is a, then let f(x)=ax +bx+c

By f(x) -4x, substituting the set f(x) into the equation ax +(b+4)x+c 0, we know that the solution set of the inequality is (1,3), we can see that the image opening is downward, then a 0, at the same time, the symmetry axis of the quadratic function is x=2, and 1 and 3 are the two roots of the equation ax + (b+4)x+c=0, substituting 1,3 into it, we can get b=-4a-4, c=3a

At this time, substituting b and c into f(x), since his maximum value is greater than 3, that is, the vertex ordinate of its quadratic function image is greater than 3, it can be solved by a representation, and the result of my solution is a 0 and not equal to 1, which may not be right, but the idea is like this, you can calculate it yourself!

The solution set of f(x) -4x is (1,3), i.e., the solution set of f(x)+4x>0 is (1,3).

a<0, g(x)=f(x)+4x, then the solution of g(x)=0 is x=1, x=3∴g(x)=a(x-1)(x-3)=ax²-4ax+3a

f(x)=ax -4(a+1)x+3a, its maximum value is 3a-16(a+1) 4a>3, 3a-4a-8-4 a=-a-8-4 a>3, i.e., a+11+4 a<0 and a>0, a +11a+4>0

A ranges from A<-(11+ 105)2 or A>(-11+ 105)2

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

Since the set of solutions of the inequality f(x) -2x is (1,3), we know a<0 and for the equation ax 2+(b+2)x+c =0 is given by the relation of the root to the coefficient.

x1+x2 = -(b+2)/a = 4

x1x2 = c/a=3

Substituting b=-(4a+2), c=3a into the equation f(x)+6a=0 with two equal real roots, = b 2-4ac = b 2-4a(6a+c) =0, we get (2a+1) 2 -9a 2 = 0

i.e. (5a+1)(1-a)=0

The solution gives a=1 (rounded), a=-1 5

So a=-1 5 , b= -6 5, c=-3 5 then the analytical formula for f(x) is f(x) = -1 5x 2 -6 5x -3 5

Solution: By the question condition:

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

Then: a+b+2+c=0

9a+3b+6+c=0, and a<0

Solution: a=c3, b=-2-4c3

Question 1: ax 2+bx+c+6a=0 has two equal roots, discriminant = 0b 2-4a*(c+6a)=0

Substitution simplification: 5c 2-12c-9=0

So c1 = -3 and c2 = -3 5

From a=c 3<0, so c=-3 5,a=-1 5,b=-6 5, i.e., f(x) is -1 5*x 2-6 5*x-3 5 Question 2:

Maximum (4ac-b2) 4ac>0

Substituting is simplified: (-4c 2-48c-36) 12c 2>0, that is, c 2+12c+9=(c+6) 2-27<0, so the value range of -3 root number 3-6a is -root number 3-2 to root number 3-2

Answer: The solution set of f(x)<0 is 10 with the parabolic opening pointing upwards.

And x1=1 and x2=3 are the abscissa of the intersection of the parabola and the x-axis.

Let the quadratic function be f(x)=a(x-1)(x-3).

1) f(x)+1=a(x-1)(x-3)+1=0 have two equal real roots, which means that the vertices of f(x) (2,-a) and the parabola f(x) fall exactly on the x-axis after translating upwards by 1 unit.

So: -a=-1

So: -a=-1

So: f(x) is parsed as f(x) = (x-1)(x-3) = x 2-4x+3

2) g(x)=f(x)+5=a(x-1)(x-3)+5 has a zero point.

So: g(2)=-a+5<=0

So: a>=5

3) When a=1, f(x)=a(x-1)(x-3)=x 2-4x+3, y=mx

When x>=3, f(x)>y=mx

A simple graph shows that when x>=3, f(x)>=0 is always above y=mx.

So: m<0

1. f(x) can be factored into f(x)=a*(x-x1)*(x-x2)=a*(x-1)*(x-3), and f(x)<0 solution set is interval (1,3), then a>0

2、g(x)=a*x^2-4*a*x+3*a+5

b -4ac=16*a2-4*a*(3*a+5)=4*a(a-5)>=0 and a>0 can be obtained from 1

then a>=5

3、f(x)=(x-1)*(x-3)=x^2-4x+3 x∈[3，＋∞

f(x) above y=mx, i.e. f(x)-mx>0 when x [3,

Set to h(x)= f(x)-mx=x 2-(4+m)x+3

i.e. h(x) has two roots to the left of 3 or no roots.

i.e. (4+m+(m 2+8*m+4) >=3 or =m 2+8*m+4<0

Solution. 1. There is no solution.

or two, -4-2*3 i.e. -4-2*3

Solution: 1) Let f(x)=ax 2+bx+c, since the solution set of f(x)>-2x is (1,3).

ax 2+(b+2)x+c>0, x=1,3 are ax 2+(b+2)x+c=0, and a<0

So, 1+3=-(b+2) 2a, 1*3=c sprig destroys a

And since f(x)+6a=0 has two equal roots, i.e., ax 2+bx+c+6a=0 has two equal roots, therefore, b 2-4a(c+6a)=0

Solution: a1=-1 7, a2=-1

So, b1 = -6 7, c1 = -3 7; b2=6,c2=-3

f(x)=-1 talk about sedan 7x 2-6 7x-3 7 or f(x)=-x 2+6x-3

2) The maximum value of f(x) is (4ac-b 2) 4a>0

Because b = -8a-2, c = 3a

Therefore, Dai Meng prepared to enter the solution: a no solution.

Let f(x)=ax +bx+c, and the inequality f(x)>-2x, i.e., ax +bx+c>-2x, i.e., ax +(b+2)x+c>0, and its solution set is (1,3), which means a<0, and 1+3=-(b+2) a, 1 3=c a, so b=-4a-2, c=3a, then f(x)=ax -(4a+2)x+3a=0

The equation f(x)+6a=0 is: ax -(4a+2)x+9a=0, there are two equal real roots, then δ=(4a+2) -4 9a =0, i.e., 5a -4a-1=0, i.e. (5a+1)(a-1)=0, and a<0, so a=-1 5, then f(x)=-1 5*x -6 5*x-3 5=-1 5*(x+3) +6 5

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

f(x)>-2x,ax 2+(b+2)x+c>0 solution set is (1,3) to judge a<0, and x=1 and x=3 are the two roots of the equation ax 2+(b+2)x+c=0.

b+2)/a=4

c a=31)f(x)+6a=0 has two equal roots.

ax^2+bx+c+6a=0

b^2-4a(c+6a)=0

Solution (1), (2), (3).

a=-1 5, or a=1 (rounded).

b=-6/5,c=-3/5

f(x)=-(1/5)x^2-(6/5)x-3/5

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

The solution set of f(x)+2x>0 is (1,3), so we can see that a<0 uses Veda's theorem 1+3=-(b+2) a, 1*3=c a, so that the expression of f(x) is composed of a and x.

In the same way, reuse δ=0 to solve a, note that a should be less than 0 and may be discarded a...

The solution set is (1,3), i.e., 10 sides divided by a and then the unequal sign is reversed.

So a<0

Because the solution set is (1,3) has two ends, and is greater than 0, the opening direction will naturally go downward, and downward of course a 0

Obviously, the opening is downward, otherwise it can't be a value between (1,3), and if the opening is up, then the range of f(x)+2x>0 will be a solution of the type of x<1 or x>3.

1.Let y=ax 2+bx+c(a<0).

So, y+2x=a(x-1)(x-3).

So, y=ax 2-(4a+2)x+3a

Then, f(x)+6a=ax 2-(4a+2)x+9a, discriminant = (4a+2) 2-4a*9a=0, so, a=-1 5 (rounded a=1).

y=-1/5x^2+6/5 *x-3/5

2.The original recipe gets:

y=a[x-(2a+1) 2] 2-(2a+1) 2 a +3a Since a<0, so:

2a+1) 2 a +3a is the maximum.

So the value is greater than 0, and the solution is:

0>x>-2 + root number 3, or x<-2 - root number 3

1) Let f(x)=a(x-1)(x-3)-2x [from the inequality f(x) -2x to (1,3), know a<0;

If a>0, then the set of inequalities f(x)—2x is (- 1) (3,+).

The equation f(x)+6a=0 has two equal real roots.

then =(4a+2) -36a =0

Solution: a=-1 5

So f(x)=-(1 5)x -(6 5)x-3 52) If the maximum value of f(x) is positive, the condition starts with **?