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) = ax ^ 2 + BX + C
f(2)=-1,f(-1)=-1,
4a+2b+c=-1
a-b+c=-1
The maximum value of F (x) is 8
(4ac-b^2)/4a=8
Three simultaneous, find ABC
Let f (x) = a (X-B) ^ 2
Because the maximum value of F (x) is 8, so a
Limx tends to 0 (tanx-x) / (x ^ 2 * SiNx)
Can denominator x ^ 2 * SiNx be directly transformed into x ^ 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
F(2)=-1,F(-1)=-1
That is, f (2) = f (- 1), so the axis of symmetry is x = (2-1) / 2 = 1 / 2
Let f (x) = a (x-1 / 2) ^ 2 + 8
F(2)=a*(2-1/2)^2+8=-1
a*9/4=-9
a=-4
So, f (x) = - 4 (x-1 / 2) ^ 2 + 8 = - 4x ^ 2 + 4x + 7
limx→0 (tanx-sinx)/x
lim(x→0) (tanx-sinx)/x
=lim(x→0) tanx(1-cosx)/x
=lim(x→0) (1-cosx)
=0
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
Because it is a quadratic function and has a maximum value of 8, that is, the value of the downward vertex of F (x) is 8
Let f (x) = a (x + b) ^ 2 + 8
Because f (x) satisfies f (2) = - 1, f (- 1) = - 1
Substituting the formula of assumption, we get - 1 = a (2 + b) ^ 2 = 8 and - 1 = a (- 1 + b) + 8
The solution is a = - 4, B = - 1 / 2
So f (x) = - 4x ^ 2 + 4x + 7
What is the value of X when SiNx = TaNx
What is x equal to when cosx = TaNx
sinx=tanx=sinx/cosx
--sinx*cosx=sinx
-sin0*cos0=sin0=0
If the quadratic function y = x ^ 2 + BX + C passes through (1,0) and the image is symmetric with respect to the line x = 1 / 2, the analytic expression of the quadratic function is obtained
The result of my solution is y = x ^ 2-x
The first reason is over point (1,0)
So generation can have B + C + 1 = 0
Second, because the image is symmetrical about the line x = 1 / 2, it has the maximum value
The derivation can be y '= 2x + B. if x = 1 / 2 Generation Y' = 2x + B = 0, B = - 1 can be obtained
Then B = - 1 is replaced by B + C + 1 = 0, and C = 0 is solvable
So the result is y = x ^ 2-x
Over (1,0), then 0 = 1 + B + C; and the axis of symmetry is x = 1 / 2, so
-If B / (2a) = 1 / 2 and a = 1, then B and C can be obtained. It's your business to calculate.
The limit value of SiNx than TaNx. X tends to 0
sin x / tan x =sin x /(sinx/cos x)=cos x
X tends to 0, cos x tends to 1
=lim(cosx)
=cos0
=1
＝lim（cosx）＝cos0＝1
If the quadratic function y = x ^ + BX + C passes through (1,0) and the image is symmetric with respect to the straight line x = 1 / 2, the coordinates of the quadratic function are obtained
Substituting point (1,0) into the equation, B + C + 1 = 0
Because the function image is symmetric about the line x = 1 / 2, so (- b) / (2a) = 1 / 2
b=-1
Then substitute B + C + 1 = 0
C=0
Function expression y = x ^ 2-x
The axis of symmetry is x = - B / 2 = 1 / 2, B = - 1
Substituting point (1, 0) into
1-1+C=0，C=0
The function expression is y = x & sup2; - X
Using vertex formula: y = (x-1 / 2) ^ 2 + C
Substituting the point (1,0), we get: 0 = 1 / 4 + C, C = - 1 / 4
So, y = (x-1 / 2) ^ 2-1 / 4
y=x^2-x
TaNx = cos (π / 2 + x), then the value of SiNx is?
It's two PI + X
tanx=cos(π/2+x)=-sinx
sinx/cosx=-sinx
cosx=-1
sinx=0