If a quadratic inequality ax ^ 2 + BX + C with respect to X The last point is the solution set of ax ^ 2-bx + C > 0

If a quadratic inequality ax ^ 2 + BX + C with respect to X The last point is the solution set of ax ^ 2-bx + C > 0

Is the final inequality greater than or less than
ax^2+bx+c
(1) If x ∈ R, the inequality ax ^ 2 + ax + 1 > 0 is constant
(1) The solution set of one variable quadratic inequality ax ^ 2 + BX + 2 > 0 is (- 1 / 2, - 1 / 3), find the value of a + B? (2) if x ∈ R, the inequality ax ^ 2 + ax + 1 > 0 holds, find the value range of a
The solution set of ax ^ 2 + BX + 2 > 0 is - 1 / 2
1, the two roots of ax square + BX + 2 = 0 are X1 = - 1 / 2, X2 = 1 / 3,
Weida theorem: X1 + x2 = - B / A, X1 * x2 = 2 / A,
-1 / 2 + 1 / 3 = - B / A, - 1 / 2 * 1 / 3 = 2 / A, a = - 12, B = - 2, a + B = - 14;
2,
Ax ^ 2 + ax + 1 = 0, no real number solution, △ = a square - 4 * a * 1
If the solution set of the quadratic inequality ax + BX + 2 > 0 is (-,,,,), then A-B is equal to?
That is to say, the solution of ax ^ 2 + BX + 2 = 0 is - ½, ⅓
So the original formula is equivalent to (x-1 / 2) (x + 1 / 3) = 0
That is x ^ 2 - 1 / 6x - 1 / 6 = 0
So B / a = - 1 / 6
2/a=-1/6
Because a is negative
So a = - 12, B = 2
The solution set of A0 is {x | - 1 / 2 > > b / a = 5 / 6
x1x2=2/a =========>>>>> 2/a=1/6
The solution is: a = 1 / 12, B = 5 / 72
Then: A-B = 1 / 12-5 / 72 = 1 / 72
Take the equal sign, - 0.5 and 1 / 3 are the two roots of the equation. X1 + x2 = - B / A, X1 * x2 = C / A, the solution is a = 12, B = - 2, the answer is 14
Given that f (x) is an odd function on R, when x > 0, f (x) = - 2x ^ 2 + 3x + 1, find the analytic expression of F (x)
F (x) is an odd function defined on R
f(-x)=-f(x)
When x > 0, f (x) = - 2x ^ 2 + 3x + 1
X0
f(-x)=-2x^2-3x+1=-f(x)
=>f(x)=2x^2+3x-1
So f (x) = {- 2x ^ 2 + 3x + 1 x > 0
{0 x=0
{2x^2+3x-1 x
It is known that the quadratic function f (x) = AX2 + BX + C (a ≠ 0) has two zeros of 1 and 2, and f (0) = 2
It is known that the quadratic function f (x) = AX2 + BX + C (a ≠ 0) has two zeros 1 and 2, and f (0) = 2
The expression of finding f (x)
Substituting three points (1,0) (2,0) (0,2) into the function f (x) = AX2 + BX + C
F (1) = a + B + C = 0, f (2) = 4A + 2B + C = 0, f (0) = C = 2
So a = 1; b = - 3; C = 2
The expression of F (x) is f (x) = x2-3x + 2
Given f (x + 2) = 2x-3x + 1, find the analytic expression of function f (x)
Let t = x + 2
∴x=t-2
F (x + 2) = 2x-3x + 1
∴f(t)=2(t-2)²-3(t-2)+1
=2t²-8t+8-3t+6+1
=2t²-11t+15
∴f(x)=2x²-11x+15
Change element method to solve! Let x + 2 = t, then x = T-2, then f (T) = 2t-11t + 15, so f (x) = 2t-11x + 15
If the zero point of function f (x) = ax + B is 2, then the zero point of function g (x) = bx2 ax is ()
A. 0，2B. 0，12C. 0，-12D. 2，12
∵ function f (x) = ax + B has a zero point of 2, ∵ 2A + B = 0, {B = - 2A, ∵ g (x) = bx2 AX = - 2ax2 AX = - ax (2x + 1), ∵ ax (2x + 1) = 0 {x = 0, x = - 12 ∵ function g (x) = bx2 ax has zero points of 0, - 12
(1) If f (2x + 1) = x square + 2x + 3, find the analytic formula of F (x). (2) if f (3x + 5) = 9x square + 6, find the analytic formula of F (x)
(1) If f (2x + 1) = x ^ 2 + 2x + 3, let 2x + 1 = t, then x = (t-1) / 2, then f (T) = (t-1) ^ 2 / 4 + 2 (t-1) + 3 = T ^ 2 / 4-T / 2 + 1 / 4 + 2t-2 + 3 = T ^ 2 / 4 + 3t / 4 + 13 / 4
(1)f(2x+1)=x^2+2x+3
Let 2x + 1 = t
f(t)=(t-1)/2f(t)=(t-1)^2/4+2(t-1)+3=t^2/4-t/2+1/4+2t-2+3=t^2/4+3t/4+13/4.
f(x)=x^2/4+3x/4+13/4.
(2)f(3x+5)=9x^2+6
Let 3x + 5 = T. then x = (T-5) / 3
f(t)=9(t-5)^2/9+6=(t-5)^2+6=t^2-10t+31.
f(x)=x^2-10x+31
If the two zeros of the function f (x) = ax & # 178; + BX + 10 are 5 and 1 respectively, then a =? B =?
Which is right? I wonder
When x = 50 = 25A + 5B = 101
When x = 1 0 = a = B + 10 2
B = - a-1o 3 from 2
20a-40 = 0 by substituting 3 into 1
A=2
∴b=-12
The solution of F (x) = ax & # 178; + BX + 10 has two zeros of 5 and 1 respectively
We know that the two of ax & # 178; + BX + 10 = 0 are 5 and 1
From the relationship between root and coefficient
5+1=-b/a
5×1=10/a
The solution is a = 2, B = - 12
25a+5b+10=0
a+b+10=0
Solve the equations, a = 2, B = - 12
f（5）=25a+5b+10=0
f（1）=a+b+10=0
The solution is: a = 2, B = - 12
Sorry, the first time a was wrong
If f (2x-1) = 3x + 2, then f (x)=______ ,f(2)=_______ .
f(2x-1)=3x+2
=3/2*(2x-1)+2+3/2
=3/2*(2x-1)+7/2
f(x)=3x/2+7/2
f(2)=13/2