Given the quadratic function f (x), the vertex of the image is (- 1,2), and through the origin, find f (x)

Given the quadratic function f (x), the vertex of the image is (- 1,2), and through the origin, find f (x)

The vertex is (- 1,2)
f(x)=a(x+1)²+2
If it passes through the origin, then f (0) = 0
So a (0 + 1) ² + 2 = 0
a=-2
f(x)=-2x²-4x

If a quadratic function f (x) is known and its image vertex is (1,2) and passes through the coordinate origin, then f (x)=

The image vertex of quadratic function f (x) is (1,2), so f (x) = a (x-1) ^ 2 + 2
And through the coordinate origin, that is, f (0) = a (0-1) ^ 2 + 2 = 0
A = - 2
That is f (x) = - 2 (x-1) ^ 2 + 2 = - 2x ^ 2 + 4x

sina:sin Let {A / 2} = 8:5 find Tan {2A} =?

5*sina=8*sin(a/2); 5*2*sin(a/2)*cos(a/2)=8*sina(a/2) 5*cos(a/2)=4 cos(a/2)=4/5 cos(a/2)=√((1+cos(a))/2) cosa=2*(cos(a/2)^2)-1 cosa=(4/5)^2*2-1 cosa=7/25cosa=√((1+cos2a)/2) cos2a=2*(cos(a2)^2)-1 cos2...

The maximum value problem of conic curve (solved by polar coordinates)
It is known that the center of the ellipse is O, the major axis and minor axis are 2a and 2b respectively (a > b > 0), a and B are two points on the ellipse, and OA ⊥ ob
Find the maximum and minimum of △ AOB area

Let the length of OA be R1, the length of OB be R2, and the angle of OA be?, then the coordinates of a and B are (r1cos?, r1sin?), (- r2sin?, r2cos?). Substituting them into the elliptic equation, the two formulas are added together to obtain the fixed value: 1 / (R1) ^ 2 + 1 / (R2) ^ 2 = 1 / A ^ 2 + 1 / b ^ 2