Given the quadratic function f (x) = ax square + BX + C, if f (- 1) = 0, f (x) = 0, find the zero point of function f (x)

Given the quadratic function f (x) = ax square + BX + C, if f (- 1) = 0, f (x) = 0, find the zero point of function f (x)

When a = 0, it is a straight line, f (x) = BX + C, f (- 1) = 0, so there is only one 0 point!
When a is not zero, it is a parabola, and there are at most two intersections with the X axis!
Now we know that f (- 1) = 0, if there is only one intersection,
Yes: - B / 2A = - 1
b^2-4ac=0
a-b+c=0
The solution is a = C = B / 2
If there are two intersections!
b^2-4ac>0
a-b+c=0
(a+c)^2-4ac=(a-c)^2>0
A is not equal to c!
So: when a = 0, there is 1 intersection!
A is not 0 and a = C has 1 intersection!
A is not 0 and a is not equal to C, there are two intersections!
Because f (- 1) = o
So - A-B + C = o
From the meaning of the title
-a-b+c=0
X-1 = - A / b (this is a system of equations)
-x=a/c
The solution is a = - C
So x = 1
So the zeros of F (x) are - 1, 1
Given that the solution of the system of equations 2x + y = a 2 (1) x + y = 2A (2) is x = 3, y = m, find the value of a, M
(1) - (2) we get x = a2-2a, a2-2a = 3, and the solution is A1 = - 1, A2 = 3. Substituting A1 = - 1, x = 3 into (2) we get 3 + y = - 2, y = - 5, namely M1 = - 5; substituting A2 = 3, x = 3 into (2) we get 3 + y = 2 × 3, y = 0, namely M2 = 3.. A1 = - 1, M1 = - 5; A2 = 3, M2 = 3
The zeros of quadratic function f (x) = ax ^ 2 + BX + C are - 2 and 3. When x ∈ (- 2,3), f (x)
The faster way to do it is to,
Because the zero point of this function is - 2.3, f (x) = a (x + 2) (x-3) is expanded to get f (x) = a (x ^ 2-x-6), and then f (- 6) = 36 is brought in to get 36a = 36, a = 1 of the solution
That is, f (x) = x ^ 2-x-6. The advantage of this method is that it doesn't need to solve B and C at all, because the problem has been said to be a quadratic function
According to Vader's theorem, - B / a = 1, C / a = - 6 and f (- 6) = 36, the conventional method obtains the solution
When x ∈ (- 2,3), f (x)
The zeros are - 2 and 3, and f (- 6) = 36. Simply, substitute (- 2, 0), (3, 0) and (- 6, 36) into the function:
4a+（-2）b+c=0
9a+ 3b + c =0
36a+（-6）b+c=36
By solving the linear equation of three variables, a, B and C are obtained.
When x ∈ (- 2,3), f (x)
If x + y + Z is not equal to 0 and 2Y + Z / x = 2x + Y / z = 2Z + X / y = k
2y+z/x=2x+y/z=2z+x/y=k
∴2y+z=kx
2x+y=kz
2z+x=ky
Add the three formulas
3（x+y+z)=k(x+y+z)
∵x+y+z≠0
∴k=3
The quadratic function f (x) = ax ^ 2 + BX + C (a > 0) f (1) = - A / 2 has two zeros
f(1)=a+b+c=-a/2
b=-3a/2-c
△=b²-4ac=(-3a/2-c)²-4ac=9a²/4+3ac+c²-4ac=9a²/4-ac+c²=a²/4-ac+c²+2a²
=2a²+(a/2-c)²>0
So a quadratic function has two zeros
Given that the solutions of the equations MX + 2Y = n, 4x-ny = 2m-1 are x = 1, y = 1, then the values of M and N are
m=1 n=3
A new system of equations is obtained by taking x = 1, y = 1 generation MX + 2Y = n, 4x NY = 2m-1
m+2=n
4-n=2m-1
The results are as follows
M=1
N=3
So m = 1, n = 3
m=1,n=3
m=1,n=3
On the mathematical problems of the first Olympiad of higher education with known quadratic function f (x)
It is known that the quadratic function f (x) = x · x (the square of x) + m · x + 1 (M is an integer) and the equation f (x) = 2 has two different real roots in the interval (- 3,1 / 2)
(1) Find the analytic formula of F (x) (maybe more than one)
(2) If x belongs to interval [1, t], there is always f (x-4)
f(x)=2
x^2+mx+1=2
x^2+mx-1=0
Opening up
Both roots are in (- 3,1 / 2)
So there are four conditions
(1) The discriminant is greater than 0
M ^ 2 + 4 > 0
(2)
The axis of symmetry is within (- 3,1 / 2)
-33/2
To sum up
3/2
This is also an Olympian mathematical problem
period
It is known that the equations {MX + 3NY = 1,25x NY = n + 2} and {3x-y = 1,4x + 2Y = 8} about X and y have the same solution, so we can find the value of M and n
It is known that the equations {MX + 3NY = 1,25x NY = n + 2 and {3x-y = 1,4x + 2Y = 8 have phase relations
The same solution, find the value of M, n
3x-y = 1,4x + 2Y = 8 to get x = 1, y = 2
Substitute MX + 3NY = 1,25x NY = n + 2 to get
m+6n=1
25-2n=n+2
M = - 45, n = 23 / 3
Hope to adopt!
From 3x-y = 1, 4x + 2Y = 8, we can get x = 1, y = 2. Take X and Y into {MX + 3NY = 1,25x NY = n + 2 as M + 6y = 1,25-2n = n + 2, so we can get m = 45, n = 23 / 3!!!
It is known that the quadratic function f (x) satisfies f (2) = - 1, f (- 1) = - 1, and the maximum value of F (x) is 8
It is known that the quadratic function f (x) satisfies f (2) = - 1, f (- 1) = - 1, and the maximum value of F (x) is 8?
Let f (x) = AX2 + BX + C, the symmetry axis is x = 0.5, so - 2A / b = 0.5,
The formula is 4A + 2B + C = - 1, A-B + C = - 1, 0.25A + 0.5B + C = 8, a = - 1 / 4B,
Solve the equation, don't help you
f(x)=-4x^2+4x+7
Let f (x) = ax & # 178; + BX + C. f (x) have the maximum value, so the opening of the image is downward, f (x) = 4ac-b ^ 2 / 4A = 8
f(2)=4a+2b+c=8 f(-1)=a-b+c=8
It can be obtained from three formulas
a=-5,b=5,c=9
So f (x) = - 5x ^ 2 + 5x + 9
A system of bivariate quadratic equations x (x + y) = 12 and Y (x-2y) = 1
x(x+y)=12 （1）
y(x-2y)=1 （2）
From (2) we get x = 1 / y + 2Y
Substitute (1) to get 6y ^ 2 + 1 / y ^ 2-7 = 0
Then (2Y + 1 / y) (3y-1 / y) = 0
therefore
Y = positive and negative root sign three